VCE mATHEmATiCS UNiTS 3 & 4
mATHS QUEST 12 mathematical methods CAS
TiNSPiRE 2.0 EDiTiON
VCE mATHEmATiCS UNiTS 3 & 4
mATHS QUEST 12 mathematical methods CAS TiNSPiRE 2.0 EDiTiON
BRiAN HODGSON NiCOLAOS KARANiKOLAS BEVERLY LANGSFORDWiLLiNG mARK DUNCAN TRACY HERFT LiBBY KEmPTON JENNiFER NOLAN GEOFF PHiLLiPS
CONTRiBUTiNG AUTHORS RUTH BAKOGiANiS  ANDREW mENTLiKOWSKi  mARK BARNES  KYLiE BOUCHER JENNY WATSON  CAROLiNE DENNEY  SONJA STAmBULiC  ELENA iAmPOLSKY ROSS ALLEN  ROBERT CAHN  RODNEY EBBAGE
First published and revised 2010 by John Wiley & Sons Australia, Ltd 42 McDougall Street, Milton, Qld 4064 Typeset in 10/12pt Times LT © John Wiley & Sons Australia, Ltd 2010 The moral rights of the authors have been asserted. National Library of Australia CataloguinginPublication data Title:
Maths quest 12 mathematical methods CAS: TINspire 2.0 edition / Brian Hodgson . . . [et al.]. ISBN: 978 1 7424 6465 7 (student pbk.) 978 1 7424 6466 4 (student ebook) 978 1 7424 6468 8 (teacher pbk.) 978 1 7424 6469 5 (teacher ebook) Notes: Includes index. Target Audience: For secondary school age. Subjects: Mathematics — Textbooks. Other Authors/ Contributors: Hodgson, Brian. Dewey Number: 510
Reproduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this work, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that the educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL). Reproduction and communication for other purposes Except as permitted under the Act (for example, a fair dealing for the purposes of study, research, criticism or review), no part of this book may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All inquiries should be made to the publisher. Screenshots from TINspire CAS calculator reproduced with permission of Texas Instruments Illustrated by Aptara and the Wiley Art Studio Typeset in India by Aptara Printed in Singapore by Craft Print International Ltd 10 9 8 7 6 5 4 3 2 1
Contents Introduction viii About eBookPLUS x Acknowledgements xi CHAPTER 1
Graphs and polynomials 1 1a The binomial theorem 1 1B 1C 1D 1E 1F 1G
Exercise 1A 6 Polynomials 6 Exercise 1B 10 Division of polynomials 11 Exercise 1C 15 Linear graphs 15 Exercise 1D 20 Quadratic graphs 22 Exercise 1E 28 Cubic graphs 29 Exercise 1F 34 Quartic graphs 38 Exercise 1G 45
Summary 47 Chapter review 50 eBookPLUS activities 56
CHAPTER 2
Functions and transformations 57 2a Transformations and the parabola 57
Exercise 2A 63 2B The cubic function in power form 65
Exercise 2B 69 2c The power function (the hyperbola) 71 2D 2E 2F 2G 2h 2I
Exercise 2C 76 The power function (the truncus) 78 Exercise 2D 84 The square root function in power form 86 Exercise 2E 91 The absolute value function 92 Exercise 2F 95 Transformations with matrices 96 Exercise 2G 101 Sum, difference and product functions 103 Exercise 2H 106 Composite functions and functional equations 108 Exercise 2I 110
2J Modelling 111
Exercise 2J 117 Summary 120 Chapter review 123 eBookPLUS activities 129
Exam Practice 1 Based on Chapters 1–2 130
CHAPTER 3
Exponential and logarithmic equations 132 3a The index laws 132
Exercise 3A 136 3B Logarithm laws 138
Exercise 3B 141 3C Exponential equations 142
Exercise 3C 146 3D Logarithmic equations using any base 147
Exercise 3D 149 3E Exponential equations (base e) 151
Exercise 3E 153 3F Equations with natural (base e)
logarithms 154 Exercise 3F 155 3G Inverses 156 Exercise 3G 158 3H Literal equations 159 Exercise 3H 162 3I Exponential and logarithmic modelling 162 Exercise 3I 164 Summary 167 Chapter review 168 eBookPLUS activities 171
CHAPTER 4
Exponential and logarithmic graphs 172 4a Graphs of exponential functions
with any base 172 Exercise 4A 181 4B Logarithmic graphs to any base 182 Exercise 4B 190 4c Graphs of exponential functions with base e 191 Exercise 4C 196
4d Logarithmic graphs to base e 197 4E
4F 4G
4H
Exercise 4D 201 Finding equations for graphs of exponential and logarithmic functions 202 Exercise 4E 205 Addition of ordinates 206 Exercise 4F 211 Exponential and logarithmic functions with absolute values 212 Exercise 4G 215 Exponential and logarithmic modelling using graphs 216 Exercise 4H 217
Summary 220 Chapter review 222 eBookPLUS activities 228
6i Trigonometric functions with
an increasing trend 311 Exercise 6I 313
Summary 314 Chapter review 317 eBookPLUS activities 321
Exam Practice 2 Based on Chapters 1–6 322
CHAPTER 7
Differentiation 324 7a Review — gradient and rates of change 324
Exercise 7A 328 7B Limits and differentiation from
CHAPTER 5
Inverse functions 229 5a Relations and their inverses 229
Exercise 5A 233 5b Functions and their inverses 234 Exercise 5B 239 5c Inverse functions 240 Exercise 5C 243 5d Restricting functions 245 Exercise 5D 249 Summary 252 Chapter review 253 eBookPLUS activities 257
CHAPTER 6
Circular (trigonometric) functions 258 6A Revision of radians and the unit circle 258 6B 6c 6D 6e 6f
6g 6h
vi
Exercise 6A 263 Symmetry and exact values 264 Exercise 6B 271 Trigonometric equations 273 Exercise 6C 281 Trigonometric graphs 282 Exercise 6D 288 Graphs of the tangent function 291 Exercise 6E 295 Finding equations of trigonometric graphs 295 Exercise 6F 297 Trigonometric modelling 299 Exercise 6G 301 Further graphs 303 Exercise 6H 310
Contents
7C 7d 7E 7F 7g
7h 7i 7J
first principles 332 Exercise 7B 336 The derivative of xn 338 Exercise 7C 340 The chain rule 341 Exercise 7D 343 The derivative of ex 344 Exercise 7E 347 The derivative of loge (x) 347 Exercise 7F 349 The derivatives of sin (x), cos (x) and tan (x) 351 Exercise 7G 354 The product rule 355 Exercise 7H 356 The quotient rule 357 Exercise 7I 359 Mixed problems on differentiation 360 Exercise 7J 364
Summary 366 Chapter review 368 eBookPLUS activities 372
CHAPTER 8
Applications of differentiation 373 8a Equations of tangents and normals 373
Exercise 8A 375 8b Sketching curves 376 Exercise 8B 384 8c Maximum and minimum problems when the function is known 386 Exercise 8C 389 8d Maximum and minimum problems when the function is unknown 390 Exercise 8D 394
8e Rates of change 396
10D Measures of variability of discrete random
Exercise 8E 398 8f Related rates 401 Exercise 8F 404 8G Linear approximation 404 Exercise 8G 406
Summary 529 Chapter review 531 eBookPLUS activities 536
Summary 408 Chapter review 410 eBookPLUS activities 415
CHAPTER 9
Integration 416 9a Antidifferentiation 416
Exercise 9A 423 9b Integration of e x, sin (x) and cos (x) 426
Exercise 9B 428 9c Integration by recognition 430
Exercise 9C 433 9d Approximating areas enclosed by
9e
9f 9g 9H 9I 9J
functions 435 Exercise 9D 438 The fundamental theorem of integral calculus 441 Exercise 9E 446 Signed areas 448 Exercise 9F 452 Further areas 455 Exercise 9G 458 Areas between two curves 460 Exercise 9H 464 Average value of a function 466 Exercise 9I 468 Further applications of integration 468 Exercise 9J 471
Summary 474 Chapter review 476 eBookPLUS activities 481
Exam Practice 3 Based on Chapters 1–9 482
CHAPTER 10
Discrete random variables 484 10a Probability revision 484
Exercise 10A 498 10b Discrete random variables 501 Exercise 10B 507 10c Measures of centre of discrete random distributions 510 Exercise 10C 516
distributions 518 Exercise 10D 526
CHAPTER 11
The binomial distribution 537 11a The binomial distribution 537
Exercise 11A 548 11b Problems involving the binomial distribution
for multiple probabilities 552 Exercise 11B 557 11c Markov chains and transition matrices 559 Exercise 11C 571 11D Expected value, variance and standard deviation of the binomial distribution 573 Exercise 11D 577 Summary 581 Chapter review 583 eBookPLUS activities 588
CHAPTER 12
Continuous distributions 589 12a Continuous random variables 589
Exercise 12A 594 12b Using a probability density function to
12C
12D 12e 12f 12G
find probabilities of continuous random variables 595 Exercise 12B 602 Measures of central tendency and spread 605 Exercise 12C 611 Applications to problem solving 613 Exercise 12D 616 The normal distribution 619 Exercise 12E 623 The standard normal distribution 626 Exercise 12F 633 The inverse cumulative normal distribution 636 Exercise 12G 640
Summary 643 Chapter review 646 eBookPLUS activities 651
Exam Practice 4 Based on Chapters 1–12 652
Answers 654 Index 730
Contents
vii
Introduction Maths Quest 12 Mathematical Methods CAS is specifically designed for the VCE Mathematical Methods CAS course and based on the awardwinning Maths Quest series. This resource contains: • a student textbook with accompanying eBookPLUS • a teacher edition with accompanying eGuidePLUS • a solutions manual.
Student textbook Full colour is used throughout to produce clearer graphs and headings, to provide bright, stimulating photos and to make navigation through the text easier. Clear, concise theory sections contain worked examples and highlighted important text and remember boxes. Icons appear for the eBookPLUS to indicate that interactivities and eLessons are available online to help with the teaching and learning of particular concepts. Worked examples in a Think/Write format provide clear explanation of key steps and suggest presentation of solutions. Many worked examples have eBookPLUS icons to indicate that a Tutorial is available to elucidate the concepts being explained. Worked examples also contain CAS calculator instructions and screens to exemplify judicious use of the calculator. Exercises contain many carefully graded skills and application problems, including multiplechoice questions. Cross references to relevant worked examples appear with the first ‘matching’ question throughout the exercises. Exercises also contain questions from past VCE examination papers along with relevant exam tips. A selection of questions is tagged as technologyfree to indicate to students that they should avoid using their calculators or other technologies to assist them in finding a solution. Exam practice sections contain examinationstyle questions, including time and mark allocations for each question. Fully worked solutions are available on the eBookPLUS for students. Each chapter concludes with a summary and chapter review exercise containing examinationstyle questions (multiple choice, short answer and extended response), which help consolidate students’ learning of new concepts. Also included are questions from past VCE exams along with relevant exam tips. Technology is fully integrated (in line with VCE recommendations).
Student website — eBookPLUS The accompanying eBookPLUS contains the entire student textbook in HTML plus additional exercises. Students may use the eBookPLUS on laptops, school or home computers, and cut and paste material for revision or the creation of notes for exams. Career profiles and History of mathematics place mathematical concepts in context. Investigations, often suggesting the use of technology, provide further discovery learning opportunities.
viii
Introduction
WorkSHEET icons link to editable Word documents that may be completed on screen or printed and completed by hand. SkillSHEET icons link to printable pages designed to help students revise required concepts, and contain additional examples and problems. Interactivity icons link to dynamic animations which help students to understand difficult concepts. eLesson icons link to videos or animations designed to elucidate concepts in ways that are more than what the teacher can achieve in the classroom. Tutorial icons link to oneway engagement activities which explain the worked examples in detail for students to view at home or in the classroom. Test yourself tests are also available and answers are provided for students to receive instant feedback.
Teacher website — eGuidePLUS The accompanying eGuidePLUS contains everything in the eBookPLUS and more. Two tests per chapter, fully worked solutions to WorkSHEETs, the work program and other curriculum advice in editable Word format are provided. Maths Quest is a rich collection of teaching and learning resources within one package. Maths Quest 12 Mathematical Methods CAS provides ample material, such as exercises, analysis questions, investigations, worksheets and technology files, from which teachers may set school assessed coursework (SAC).
Introduction
ix
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About eBookPLUS
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Acknowledgements
xi
1
1A 1B 1C 1D 1E 1F 1G
The binomial theorem Polynomials Division of polynomials Linear graphs Quadratic graphs Cubic graphs Quartic graphs
Graphs and polynomials AreAS oF STudy
• Review of algebra of polynomials, solution of polynomial equations with real coefficients of degree n having up to and including n real solutions
• Graphs of polynomial functions including axis intercepts, stationary points and points of inflection, domain (including maximal domain), range and symmetry
eBook plus
1A
Digital doc
The binomial theorem
10 Quick Questions
In Maths Quest 11 Mathematical Methods CAS we learned the following binomial expansions: (x + a)2 = x2 + 2xa + a2 (x + a)3 = x3 + 3x2a + 3xa2 + a3 These are called binomial expansions because the expressions in the brackets contain two terms, ‘bi’ meaning 2. By continuing to multiply successively by a further (x + a), the following expansions would be obtained: (x + a)4 = (x3 + 3x2a + 3xa2 + a3)(x + a) = x4 + 4x3a + 6x2a2 + 4xa3 + a3 (x + a)5 = (x4 + 4x3a + 6x2a2 + 4xa3 + a3)(x + a) = x5 + 5x4a + 10x3a2 + 10x2a3 + 5xa4 + a5 The coefficients associated with each term can be arranged in a triangular shape as shown: (x + a)0 (x +
1
a)1
1
(x + a)2
1
(x + a)3
1
(x + a)4 (x + a)5
1 1
2 3
4 5
1 1 3 6
10
1 4
10
1 5
Chapter 1
1
Graphs and polynomials
1
Notes 1. The first and last numbers of each row are 1. 2. Each other number is the sum of the two numbers immediately above it. This triangle is known as Pascal’s triangle. Each number can also be obtained using combinations, as follows. Row 0 0
0 1 0
1 2 0
2 3 0
3 4 0
4
1 1 2 1
3 1 4 1
2 2 3 2
4 2
3 3 4 3
4 4
n n n! Note: r = Cr = n − r ! r ! ( ) n Remember that nCr is another way of writing . r For example, the expansion of (x + a)6 can be written using combinations and then evaluated: 6 6 6 6 6 6 6 ( x + a)6 = x 6 a 0 + x 5 a1 + x 4 a 2 + x 3 a3 + x 2 a 4 + x1a 5 + x 0 a6 5 2 4 1 3 0 6 = x 6 + 6 x 5 a + 15 x 4 a 2 + 20 x 3 a3 + 15 x 2 a 4 + 6 xa 5 + a6 Now the binomial theorem can be formally stated. n n n n ( ax + b) n = ( ax ) n b0 + ( ax ) n − 1 b1 + . . . + ( ax )1 bn − 1 + ( ax )0 bn n 1 n − 1 0 Notes 1. The indices always sum to n, that is, the powers of (ax) and b sum to n. 2. The power of ax decreases from left to right while the power of b increases. 3. The number of terms in the expansion is always n + 1. n (ax)n − r br. 4. The (r + 1)th term is r The binomial theorem can also be stated using summation notation: ( ax + b) n =
n
r=0
2
n
∑ r ( ax)n − r br
n
, where
∑ means the sum of terms from r = 0 to r = n.
r= 0
Maths Quest 12 Mathematical Methods CAS for the TINspire
Worked exAMple 1
Use the binomial theorem to expand ( 2 x − 3)4. ThINk
WrITe
Method 1: Technologyfree 1
Complete the binomial theorem expansion where ax is the 1st term, b is the 2nd term and n is the index, using the appropriate row of Pascal’s triangle to assist.
4 4 (2 x − 3)4 = (2 x )4 (− 3)0 + (2 x )3 (− 3)1 1 0 4 4 + (2 x )2 (− 3)2 + (2 x )1 (− 3)3 2 3 4 + (2 x )0 (− 3)4 4
2
Evaluate the combinations and the powers.
3
Simplify.
= 1(16x4) + 4(8x3)(−3) + 6(4x2)(9) + 4(2x)(−27) + 1(81) = 16x4 − 96x3 + 216x2 − 216x + 81
Method 2: Technologyenabled 1 On a Calculator page, press: • MENu b • 3:Algebra 3 • 2:Expand 2 Complete the entry line as: expand((2x − 3)4) Then press ENTER ·.
2
(2x − 3)4 = 16x4 − 96x3 + 216x2 − 216x + 81
Write the result.
Worked exAMple 2 5
2 Expand the binomial expression + x . 2 x ThINk 1
Complete the binomial expansion where 2 ax = 2 , b = x and n = 5, using row 5 of Pascal’s x triangle to assist.
2
Evaluate the powers.
3
Simplify.
WrITe 5
5
4
3
2 2 2 2 2 x 2 + x = x 2 + 5 x 2 x + 10 x 2 x 2 2 3 2 + 10 2 x + 5 2 x 4 + x 5 x x 32 16 8 + 5 8 x + 10 6 x 2 x x x10 4 2 + 10 4 x 3 + 5 2 x 4 + x 5 x x 32 80 80 40 = 10 + 7 + 4 + + 10 x 2 + x 5 x x x x =
Chapter 1
Graphs and polynomials
3
Worked exAMple 3
State the coefficient of i
eBook plus
x2
and
ii
x4
in
( 3 − 2 x )8
, without the use of technology.
Tutorial
int0516
ThINk i
ii
1
WrITe
Find the appropriate term by using the binomial theorem.
3
Evaluate the term.
4
State the coefficient.
1
Find which term gives a power of x4.
2
i x0, x1, x2
The powers of the 1st term decrease and the powers of the 2nd term increase 0, 1, 2, . . . use this to find which term gives a power of x2.
2
Worked example 3
The third term gives a power of x2.
Third term =
8 6 − 2 3 ( 2x) 2
= 28 × 729 × 4x2 = 81 648x2 The coefficient of x2 is 81 648. i i x0, x1, x2, x3, x4
The fifth term gives a power of x4. 8 4 − 4 3 ( 2x) Fifth term = 4
Evaluate the term.
= 70 × 81 × 16x4 = 90 720x4 3
State the coefficient.
The coefficient of the fifth term is 90 720.
Worked exAMple 4
Find the fourth term in the expansion of ( x − 2 y)5. ThINk 1
Find the fourth term by using the binomial theorem.
2
Evaluate the term.
WrITe
5 Fourth term = x2(2y)3 3 = 10 × x2 × 8y3 = 80x2y3
Worked exAMple 5 5
1 Find and evaluate the term that is independent of x in the expansion of x 3 + . x2 ThINk 1
Find how the powers of x are generated in the expansion from left to right.
WrITe
1 Powers of x are (x3)5 = x15, (x3)4 2 = x10, x 2 3 1 1 (x3)3 = x5, (x3)2 = x0 . . . x2 x2 that is, x15, x10, x5, x0.
4
Maths Quest 12 Mathematical Methods CAS for the TINspire
2
Find the required term.
The fourth term is independent of x.
3
Evaluate.
Fourth term =
5 3 2 1 3 ( x ) x 2
3
1 x 6
= 10 x 6 = 10 4
State the solution.
The term that is independent of x is the fourth term, 10.
Worked Example 6
Find the coefficient of y4 in the expansion of ( y + 3) 3 ( 2 − y)5. Think
Write
1
y4 terms will result when multiplying from the first and second brackets respectively: terms 1 and 2, terms 2 and 3, terms 3 and 4 and terms 4 and 5.
2
Write down the sum of these 4 products, using Pascal’s triangle to assist.
y4 terms = y3[5(2)4(−y)] + 3y2(3)[10(2)3(−y)2] + 3y(3)2[10(2)2(−y)3] + 33[5(2)(−y)4] = −80y4 + 720y4 − 1080y4 + 270y4 = −170y4
3
Evaluate.
4
State the solution.
The coefficient of y4 is −170.
REMEMBER
1. Pascal’s triangle: 1 1 1 1 1
1 2
3 4
1 5 2. Binomial theorem:
1 3
6 10
1 4
10
1 5
1
n n n n (ax + b) n = (ax ) n b 0 + (ax ) n − 1 b + .. . + (ax )b n − 1 + (ax )0 b n 1 n 0 n − 1 Notes 1. The powers of (ax) and b sum to n. 2. There are n + 1 terms in the expansion. n n − r br . 3. The (r + 1)th term is r (ax )
Chapter 1 Graphs and polynomials
5
exerCISe
1A eBook plus
The binomial theorem 1 We 1
a (x + 3)2 d (2x + 3)4
Digital doc
SkillSHEET 1.1 Binomial expansions
use the binomial theorem to expand each of the following. b (x + 4)5 e (7 − x)4
c (x − 1)8 f (2 − 3x)5
2 We2 Expand each of the following binomial expansions. 3
a x + 1 x
b 3x − 2 x
7
c x2 + 3 x
6
d 3 − 2 x x 2
5
State the coefficient of i x2 ii x3 and iii x4 in each of the following.
3 We3
c 2 + 3x x
b (2x + 1)5
a (x − 7)3
5
d x2 − 3 x
6
e 7x + 3 x 2
6
3
5 4 MC The coefficient of x3 in 3 x 2 − is: x −135 − B 45 C −75 A 5 MC Which of the following does not have an B 3x 2 − 1 x
A (x + 6)8
7
D 45 x5
E 135
term when expanded?
C 6x + 5 x
8
D (8 − 3x)5
E 2x − 1 x 2
8
5
e f 2 6 MC If x 3 + 2 = ax15 + bx10 + cx 5 + d + 5 + 10 , then a + b + c + d + e + f equals: x x x A 15 B 31 C 63 D 243 E 127 7 MC Which one of the following expressions is not equal to (2x − 3)4?
( 2 x − 3)6 ( 3 − 2 x )2
A (3 − 2x)
B (2x − 3)(2x −
D 16x4 − 24x3 + 36x2 − 54x + 81
E 16x4 − 96x3 + 216x2 − 216x + 81
4
3)3
C
8 We4 Find the fourth term in the expansion (x + 3y)6. 9 10 11 12 13
Find the third term in the expansion of 3 −
9
x , assuming ascending powers of x. 4 6 2 We5 Find and evaluate the term that is independent of x in the expansion of 3 x + 2 . x 5 2 4 Find and evaluate the term independent of x in the expansion of x − 3 . x 4 3 Find and evaluate the term that is independent of x in the expansion of x 2 + 2 . x We6 Find the coefficient of p4 in the expansion of (p + 3)5 (2p − 5).
14 In the expansion of (2a − 1)n, the coefficient of the second term is −192. Find the value of n.
1B
polynomials A polynomial in x is an expression that consists of terms which have nonnegative integer powers of x only. P(x) is a polynomial in x if: P(x) = an xn + an − 1 xn − 1 + . . . + a2 x2 + a1 x + a0
6
Maths Quest 12 Mathematical Methods CAS for the TINspire
where n is the degree (or highest power) of the polynomial and is a nonnegative integer. The values of an, an − 1, . . ., a2, a1 and a0 are called the coefficients of their respective power of x terms.
Worked Example 7
Which of the following expressions are not polynomials?
9
a x6 − 4 x4 + 2 x3 + 7 x c 7 − 3 xy + 4 x 2 − x 3 +
e 3 x2 −
b x 2 + x3 − x2 + 6 x − 5 d 8 + 2 x − 3 x2 + 9 x3 − x4
x
2 x 2
Think 1
Write
a and d are polynomials because they are
expressions with nonnegative integer powers of x only. 2
b is not a polynomial as it has a power of 92 , which
is not an integer. 3
c is not a polynomial as it has a power of 1 2
( ), which is not an integer, and it also
b, c and e are not polynomials.
has one term, −3xy, which is not a power of x only. 4
2 = 2 x 2 and so x2 has a power that is not a positive integer. e is not a polynomial because
−
Polynomials can be added and subtracted by collecting like terms.
Worked Example 8
Given that P( x ) = 6 − 2 x + 3 x 2 + x 4 , Q( x ) = x 5 − 2 x 4 + x 2 − 5 x − 2 and R( x ) = x 2 − 4, find: a P ( x ) + Q( x ) b P ( x ) − R( x ). Think
Write/display
Method 1: Technologyfree a
b
a P(x) + Q(x) = 6 − 2x + 3x2 + x4 + x5 − 2x4
1
Add the polynomials.
2
Collect like terms.
1
Subtract the polynomials.
2
Remove brackets.
= 6 − 2x + 3x2 + x4 − x2 + 4
3
Collect like terms.
= x4 + 2x2 − 2x + 10
+ x2 − 5x − 2
= x5 − x4 + 4x2 − 7x + 4 b P(x) − R(x) = 6 − 2x + 3x2 + x4 − (x2 − 4)
Chapter 1 Graphs and polynomials
7
Method 2: Technologyenabled a
b
1
On a Calculator page, define the polynomials P(x), Q(x) and R(x). To do this, press: • Menu b • 1:Actions 1 • 1:Define 1 Complete the entry lines as: Define p(x) = 6  2x + 3x2 + x4 Define q(x) = x5  2x4 + x2  5x  2 Define r(x) = x2  4 Press ENTER · after each entry.
2
To calculate P(x) + Q(x), complete the entry line as: p(x) + q(x) Press ENTER ··
To calculate P(x)  R(x), complete the entry line as: p(x)  r(x) Press ENTER ·.
Evaluating polynomials A value for a polynomial, P(x), can be found for a particular value of x by simply substituting the given value of x into the polynomial expression and evaluating. That is, polynomial functions are evaluated in the same way as any function.
8
Maths Quest 12 Mathematical Methods CAS for the TINspire
Worked Example 9
For the polynomial P( x ) = 2 x 4 − x 3 + 5 x 2 − 6 x + 4, find: a its degree b P(1) c P(−2). Think
Write
a
The degree of the polynomial is the highest power of x. b 1 Substitute the given value of x into the polynomial expression. 2
c
a The degree of P(x) is 4. b P(1) = 2(1)4 − (1)3 + 5(1)2 − 6(1) + 4
=2−1+5−6+4=4
Evaluate.
1
Substitute the given value of x into the polynomial expression.
2
Evaluate.
P(−2)
c
= 2(−2)4 − (−2)3 + 5(−2)2 − 6(−2) + 4 = 32 + 8 + 20 + 12 + 4 = 76
Worked Example 10
If p( x ) = ax 5 + x 4 − 3 x 3 + bx − 5, p( − 1) = −5 and p(2) = − 65 , find the values of a and b. Think
Write/display
Method 1: Technologyfree 1
Substitute a given value of x into the polynomial and equate it to the given answer.
2
Simplify the equation.
3
Make b the subject of the equation and call this equation [1].
4
Substitute a given value of x into the polynomial and equate it to the given answer.
5
Simplify the equation.
P(−1) = a(−1)5 + (−1)4 − 3(−1)3 + b(−1) − 5 = −5 −a
+ 1 + 3 − b − 5 = −5 −a + 4 − b = 0 b = 4 − a
P(2) = a(2)5 + (2)4 − 3(2)3 + b(2) − 5 = −65 32a + 16 − 24 + 2b − 5 = −65 32a + 2b − 13 = −65 32a + 2b = −52
6
Substitute [1] into [2].
Substituting b = 4 − a: 32a + 2(4 − a) = −52
7
Solve this equation for a.
8
Substitute the value of a into equation [1].
Substituting a = −2 into equation [1]: b = 4 − −2
9
Find the value of b.
10
State the solution.
[1]
[2]
32a + 8 − 2a = −52 30a = −60 a = −2
=6 Therefore, a =
−2
and b = 6.
Chapter 1 Graphs and polynomials
9
Method 2: Technologyenabled 1 On a Calculator page, define the polynomial P(x) by completing the entry line as: Defi ne p(x) = a × x5 + x4 − 3x3 + b × x − 5 Then press ENTER ·. To calculate the values of a and b, complete the entry line as: solve (p(−1) = −5 and p(2) = −65,a) Then press ENTER ·.
2
Given p(x) = ax5 + x4 − 3x3 + bx − 5 and solving p(−1) = −5 and p(2) = −65 gives a = −2 and b = 6.
Write the answer.
reMeMBer
1. If P(x) = a nx n + an − 1xn − 1 + . . . + a2x2 + a1x + a0 and n is a nonnegative integer then P(x) is a polynomial of degree n and an, an − 1, . . . a2, a1 are called coefficients and ∈ R. 2. A polynomial P(x) is evaluated in the same way as any function. exerCISe
1B
polynomials 1 We 7
Which of the following are not polynomial expressions?
i x3 − 2x
ii x4 + 3x2 − 2x + x
iii x7 + 3x6 − 2xy + 5x
iv 3x8 − 2x5 + x2 − 7
v 4x6 − x3 + 2x − 3
vi 2 x 5 + x 4 − x 3 + x 2 + 3 x −
2 x
2 We8 Given that P(x) = 8 − 3x + 2x2 + x4, Q(x) = x5 − 3x4 − 4x2 − 1 and R(x) = 8x3 + 7x2 − 4x then find: a P(x) + Q(x) b Q(x) − R(x) c 3P(x) − 2R(x) d 2P(x) − Q(x) + 3R(x). eBook plus
3 We9 For each of the following polynomials, find: i its degree ii P(0) iii P(2) and iv P(−1).
Digital doc
Spreadsheet 043 Evaluating polynomials
eBook plus Digital doc
SkillSHEET 1.2 Simultaneous equations
a P(x) = x6 + 2x5 − x3 + x2
b P(x) = 3x7 − 2x6 + x5 − 8
c P(x) = 5x6 + 3x4 − 2x3 − 6x2 + 3
d P(x) = −7 + 2x − 5x2 + 2x3 − 3x4
4 MC If P(x) = x8 − 3x6 + 2x4 − x2 + 3, then P(−2) is equal to: A 479 B 95 C 31 D 481 5 We 10 6 7 8 9
If P(x) =
2x7
+
ax5
+
3x3
+ bx − 5, P(1) = 4 and P(2) = 163, find a and b.
Find a and b, given that f (x) = + bx3 − 3x2 − 4x + 7, f (1) = −2 and f (2) = −5. For Q(x) = x5 + 2x4 + ax3 − 6x + b, Q(2) = 45 and Q(0) = −7. Find a and b. Find a and b if P(x) = ax6 + bx4 + x3 − 6, 3P(1) = −24 and 3P(−2) = 102. ax4
MC
If P(x) = ax4 − x3 + 3x2 − 5 and P(1) = −1, then a is equal to:
a A 1
B 0
C 2
D
−3
b If f (x) = xn − 2x3 + x2 − 5x and f (2) = 10, then n is equal to: A 4 B 6 C 7 D 5 10
E 103
Maths Quest 12 Mathematical Methods CAS for the TINspire
E
−2
E
−1
1C
division of polynomials
eBook plus
Interactivity When sketching cubic or higher order graphs, it is necessary to int0246 factorise the polynomials in order to find the xintercepts. As will be Division of polynomials shown later in this section, division of polynomials can be used to factorise an expression. When one polynomial, P(x), is divided by another, D(x), the result can be expressed as: P( x) R( x ) = Q( x ) + D( x ) D( x ) where Q(x) is called the quotient, R(x) is called the remainder, and D(x) is called the divisor.
Worked exAMple 11
eBook plus
Find the quotient, Q(x), and the remainder, R(x), when x 4 − 3 x 3 + 2 x 2 − 8 is divided by the linear expression x + 2.
Tutorial
int0517 Worked example 11
ThINk
WrITe
Method 1: Technologyfree 1
Set out the long division with each polynomial in descending powers of x. If one of the powers of x is missing, include it with 0 as the coefficient.
2
Divide x into x4 and write the result above.
3
4
5 6
7 8
Multiply the result underneath.
x3
by x + 2 and write the result
Subtract and bring down the remaining terms to complete the expression.
)
x + 2 x 4 − 3x3 + 2 x 2 + 0 x − 8 x3 x + 2 x − 3x 3 + 2 x 2 + 0 x − 8
)
x3 − 3x 3 + 2 x 2 + 0 x − 8 x+2 4 +2 3 x x
)
The polynomial x3 − 5x2 + 12x − 24, at the top, is the quotient. The result of the final subtraction, 40, is the remainder.
x4
x3 x+2 − 3x3 + 2 x 2 + 0 x − 8 − (x 4 + 2 x 3 ) − 5x3 + 2 x 2 + 0 x − 8
)
Divide x into −5x3 and write the result above. Continue this process to complete the long division.
4
x+2
)
x4
x 3 − 5 x 2 + 12 x − 24 x4
− 3x3 + 2 x 2 + 0 x − 8
− (x 4 + 2x3 ) − 5x3 + 2 x 2 + 0 x − 8 − − ( 5 x 3 − 10 x 2 ) 12 x 2 + 0 x − 8 − (12 x 2 + 24 x ) − 24 x − 8 − − ( 24 x − 48) 40 The quotient, Q(x), is x3 − 5x2 + 12x − 24. The remainder, R(x), is 40.
Chapter 1
Graphs and polynomials
11
Method 2: Technologyenabled 1
On a Calculator page, press: • Menu b • 2:Number 2 • 8:Fraction Tools 8 • 1:Proper Fraction 1 Complete the entry line as: x 4 − 3x3 + 2 x 2 − 8 propFrac ( ) x+2 Then press ENTER ·.
2
Write the answer.
Dividing x4  3x3 + 2x2  8 by x + 2 gives a quotient, Q(x), of x 3 − 5 x 2 + 12 x − 24 and a remainder, R(x), of 40.
Note: P(−2) = (−2)4 − 3(−2)3 + 2(−2)2 − 8 = 16 + 24 + 8 − 8 = 40 The remainder when P(x) is divided by (x + 2) is P(−2). This leads to the remainder theorem, which states: When P(x) is divided by ( x − a), the remainder is P(a) or − when P(x) is divided by ( ax + b), the remainder is P b . a Furthermore, if the remainder is zero, then (x − a) is a factor of P(x). This leads to the factor theorem, which states: If P( a) = 0, then ( x − a) is a factor of P(x) or − b if ( ax + b) is a factor of P(x), then P = 0. a Note: If (x − a) is a factor of P(x) and a is an integer, then a must be a factor of the term independent of x. For example, if (x − 2) is a factor of P(x), then the term independent of x must be divisible by 2. Therefore, (x − 2) could be a factor of x3 − 2x2 − x + 2, but (x + 3) could not be a factor.
Worked Example 12
Determine whether or not D( x ) = ( x − 3) is a factor of P ( x ) = 2 x 3 − 4 x 2 − 3 x − 8. Think
Write
Method 1: Technologyfree 1
12
Evaluate P(3).
P (3) = 2(3)3 − 4(3)2 − 3(3) − 8 = 54 − 36 − 9 − 8 =1
Maths Quest 12 Mathematical Methods CAS for the TINspire
2
If P(3) = 0 then (x − 3) is a factor of P(x), but if P(x) ≠ 0, (x − 3) is not a factor of P(x).
P (3) ≠ 0 so (x − 3) is not a factor of P(x).
Method 2: Technologyenabled 1
On a Calculator page, define P(x) by completing the entry lines as: Define p( x ) = 2 x 3 − 4 x 2 −3 x − 8 p(3) Press ENTER · after each entry.
2
Since p(3) = 1, (x  3) is not a factor of P(x).
p(3) ≠ 0 so ( x − 3) is not a factor of P(x).
Worked Example 13 a Factorise P ( x ) = 2 x 3 − x 2 − 13 x − 6. b Solve 2 x 3 − x 2 − 13 x − 6 = 0. Think
Write
Method 1: Technologyfree a
1
2
Use the factor theorem to find a value for a where P(a) = 0 and a is a factor of the numerical term. Try a = 1, −1, 2, −2, 3, −3, 6, −6 until a factor is found.
Divide P(x) by the divisor (x + 2) using long division.
3
Express P(x) as a product of linear and quadratic factors.
4
Factorise the quadratic, if possible.
a
P(1) = 2(1)3 − (1)2 − 13(1) − 6 = −18 ≠0 P(−1) = 2(−1)3 − (−1)2 − 13(−1) − 6 =4 ≠0 P(2) = 2(2)3 − (2)2 − 13(2) − 6 = −20 ≠0 P(−2) = 2(−2)3 − (−2)2 − 13(−2) − 6 =0 So (x + 2) is a factor.
)
2 x 2 − 5x − 3
x + 2 2 x 3 − x 2 − 13 x − 6 − (2 x 3 + 4 x 2 ) − 5 x 2 − 13 x − 6 − (− 5 x 2 − 10 x ) − 3x − 6 −(− 3 x − 6) 0 P(x) = (x + 2)(2x2 − 5x − 3) = (x + 2)(2x + 1)(x − 3)
Chapter 1 Graphs and polynomials
13
b
1
Rewrite the equation in factorised form, using the answer to part a .
b 2x3 − x2 − 13x − 6 = 0
(x + 2)(2x + 1)(x − 3) = 0 − x = −2, 1 or 3 2
2
Use the Null Factor Law to state the solutions. Method 2: Technologyenabled a
b
1
On a Calculator page, define the polynomial P(x) by completing the entry line as: Define p( x ) = 2 x 3 − x 2 − 13 x − 6 Then press ENTER ·. To factorise P(x), complete the entry line as: factor (p(x)) Then press ENTER ·.
2
Write the answer.
1
To solve P(x) = 0, complete the entry line as: solve ( p( x ) = 0, x ) Then press ENTER ·.
2
Write the answer.
a
Factorising p(x) = 2x3  x2  13x  6 gives P ( x ) = ( x − 3)( x + 2)(2 x + 1) b
Solving 2x3  x2  13x  6 = 0 gives − 1 x = − 2, x = , or x = 3 2
REMEMBER
P( x) R( x ) = Q( x ) + D( x ) D( x ) where Q(x) is called the quotient, R(x) is called the remainder, D(x) is called the divisor. 2. Remainder theorem: If P(x) is divided by (x − a), then the remainder is P(a). 3. Factor theorem: If P(a) = 0, then (x − a) is a factor of P(x). −b If (ax + b) is a factor of P(x), then P = 0. a 4. If (x − a) is a factor of P(x), then a must be a factor of the term independent of x. 1.
14
Maths Quest 12 Mathematical Methods CAS for the TINspire
exerCISe
1C
division of polynomials 1 We 11 Find the quotient, Q(x), and the remainder, R(x), when each of the following polynomials are divided by the given linear expression. a x3 − 2x2 + 5x − 2, x − 4 b x5 − 3x3 + 4x + 3, x + 3 c 6x4 − x3 + 2x2 − 4x, x − 3 d 3x4 − 6x3 + 12x, 3x + 1 2 a For each corresponding polynomial in question 1, evaluate: i P(4) ii P(−3) iv P(
iii P(3)
−1 ) 3
b Compare these values to R(x) in question 1 and comment on the result. 3 We 12 a b c d eBook plus Digital doc
Spreadsheet 096 Finding factors of polynomials
In each of the following determine whether or not D(x) is a factor of P(x).
P(x) = x3 + 9x2 + 26x − 30, D(x) = x − 3 P(x) = x4 − x3 − 5x2 − 2x − 8, D(x) = x + 2 P(x) = 4 − 9x + 6x2 − 13x3 − 12x4 + 3x5, D(x) = 4 − x P(x) = 4x6 + 2x5 − 8x4 − 4x3 + 6x2 − 9x − 6, D(x) = 2x + 1
4 MC Examine the equation f (x) = x4 − 4x3 − x2 + 16x − 12. a Which one of the following is a factor of f (x)? A x+1 B x C x+2 D x+3 b When factorised, f (x) is equal to: A (x + 1)(x − 3)(x + 4) C (x + 2)(x − 4)(x + 3)(x + 1) E x(x − 1)(x + 2)(x + 3)
E x−4
B (x + 2)(x − 2)(x − 3)(x − 1) D (x − 1)(x + 1)(x − 3)(x − 4)
5 We 13a Factorise the following polynomials. b P(x) = 3x3 − 13x2 − 32x + 12 a P(x) = x3 + 4x2 − 3x − 18 c P(x) = x4 + 2x3 − 7x2 − 8x + 12
d P(x) = 4x4 + 12x3 − 24x2 − 32x
6 We13b Solve each of the following equations. b 2x4 + 10x3 − 4x2 − 48x = 0 a 3x3 + 3x2 − 18x = 0 c 2x4 + x3 − 14x2 − 4x + 24 = 0 7 If (x − 2) is a factor of
x3
+
ax2
8 If (x − 1) is a factor of
x3
+
x2
eBook plus
d x4 − 2x2 + 1 = 0
Digital doc
WorkSHEET 1.1
− 6x − 4, then find a.
− ax + 3, then find a.
9 Find the value of a if (x + 3) is a factor of 2x4 + ax3 − 3x + 18. 10 Find the value of a and b if (x + 1) and (x − 2) are factors of ax3 − 4x2 + bx − 12. 11 If (2x − 3) and (x + 2) are factors of 2x3 + ax2 + bx + 30, find the values of a and b.
1d
linear graphs Linear graphs are polynomials of degree 1. Graphs of linear functions are straight lines and may be sketched by finding the intercepts.
revision of properties of straight line graphs
y
1. The gradient of a straight line joining two points is: y − y1 m= 2 x2 − x1
B (x2, y2) A (x1, y1) 0
Chapter 1
x
Graphs and polynomials
15
2. The general equation of a straight line is: y = mx + c where m is the gradient and c is the value of the yintercept. 3. The equation of a straight line passing through the point (x1, y1) and having a gradient of m is: y − y1 = m(x − x1)
y
Gradient = m
A (x1, y1) x
0
4. The intercept form of the equation of a straight line is:
y (0, b)
x y + = 1 or bx + ay = ab a b
(a, 0) 0
5. Parallel lines have the same gradient. 6. The product of the gradients of two lines that are perpendicular equals −1. −1 That is, m1 × m2 = −1 or m1 = m2 Worked Example 14
Sketch the graph of the linear function 3 x − 2 y = 6 by indicating the intercepts. Think
Write/Draw
1
Substitute y = 0 into the equation.
2
Solve the equation for x to find the xintercept.
3
Substitute x = 0 into the equation.
Therefore, the xintercept is 2.
4
Solve the equation for y to find the yintercept.
When x = 0, 3 × 0 − 2y = 6 y = −3
5
Draw a set of axes. Indicate the xintercept and yintercept and rule a line through these points.
Therefore, the yintercept is −3.
6
When y = 0, 3x − 2 × 0 = 6 x=2
y 0
3x − 2y = 6 x 2
−3
Worked Example 15
Find the equation, in the form ax + by + c = 0, of each straight line described below. a The line with a gradient of 2 and passing through ( 3, − 2 ) b The line passing through (0, 8) and ( − 2, 2 ) c The line that passes through (3, 4) and is parallel to the line with equation y − 2 x − 5 = 0 d The line that passes through (1, 3) and is perpendicular to the line with equation y + 2 x − 3 = 0 16
Maths Quest 12 Mathematical Methods CAS for the TINspire
x
ThINk a
b
c
d
WrITe
1
Write the rule for the point–gradient form of the equation of a straight line, y − y1 = m(x − x1).
2
Substitute the value of the gradient, m, and the coordinates of the point (x1, y1), into the equation.
3
Expand the brackets.
4
Express the equation in the form required.
1
Write the rule for the gradient, m, of a straight line, given 2 points.
2
Substitute the values of (x1, y1) and (x2, y2) into the rule and evaluate the gradient.
3
Substitute the values of m and (x1, y1) into the rule for the point–gradient form of the equation of a straight line. (Coordinates of either point given may be used.)
4
Expand the brackets.
5
Express the equation in the form required.
1
State the gradient of the given line, which is the same as the gradient of the parallel line.
y − y1 = m(x − x1)
a
y − (−2) = 2(x − 3)
y + 2 = 2x − 6 y − 2x + 8 = 0 or 2x − y − 8 = 0 y − y1 b m= 2 x2 − x1 2−8 −0 −6 = − 2 =3 =
⇒
−2
y − y1 = m(x − x1) y − 8 = 3(x − 0)
y − 8 = 3x 3x − y + 8 = 0 c y − 2x − 5 = 0 becomes y = 2x + 5.
The gradient of the parallel lines is 2.
2
Write the rule for the point–gradient form of the equation of a straight line.
y − y1 = m(x − x1)
3
Substitute the values of m and the coordinates (x1, y1) = (3, 4).
y − 4 = 2(x − 3)
4
Simplify and write in the required form.
1
Find the gradient of the given line.
y − 4 = 2x − 6 2x − y − 2 = 0 d y = −2x + 3
The gradient of the line is −2.
2
Find the gradient of the perpendicular line.
3
Write the rule for the point–gradient form of the equation of a straight line.
y − y1 = m(x − x1)
4
Substitute the values of m and the coordinates (x1, y1) = (1, 3).
y − 3 = 12 (x − 1)
5
Simplify and write in the required form.
The gradient of the perpendicular line is 12 .
2y − 6 = (x − 1) x − 2y + 5 = 0
Chapter 1
Graphs and polynomials
17
The domain and range of functions The domain of a function, y = f (x), is the set of values of x for which the function is defined (that is, all xvalues that can be substituted into f (x) and an answer found). The range of f (x) is the set of values of y for which the function is defined. If the rule and the domain of a function are given, then the function is completely defined. For example, y = −4x, x ≤ 0 f (x) = −4x, x ≤ 0 or f : (−∞, 0] → R, f (x) = −4x
Interval notation Restricted domains or ranges can be represented by interval notation in three forms. 1. The closed interval. 2. The open interval. 3. The halfopen interval. a
b
[a, b] = {x : a ≤ x ≤ b}
a
b
a
(a, b) = {x : a < x < b}
b
[a, b) = {x : a ≤ x < b}
If the domain or range is unrestricted, it can be denoted as R or (−∞, ∞). R+ = (0, ∞) R+ ∪ {0} = [0, ∞) − − R = ( ∞, 0) R− ∪ {0} = (∞, 0] Worked exAMple 16
Sketch the graph of each of the following functions, stating the domain and range of each. a 4 x − 2 y = 8, x ∈[ − 3, 3] b f ( x ) = 1 − 2 x, x ∈ ( − ∞ , − 1) ThINk a
1
WrITe/drAW/dISplAy
Substitute the smallest value of x into the equation.
a When x = −3, −12
− 2y = 8 −2y
= 20 y = −10
2
Solve the equation for y, to find an end point of the straight line.
3
State the coordinates of the end point.
(−3, −10) is a closed end of the line.
4
Substitute the largest value of x into the equation.
When x = 3, 12 − 2y = 8
5
Solve the equation for y, to find the other end point of the line.
−2y
6
State the coordinates of the 2nd end point.
7
Plot the two points on a set of axes with closed circles (since both points are included).
= −4 y=2
(3, 2) is the other closed end of the line. y 2 –3
0 –4
(–3, –10)
18
Maths Quest 12 Mathematical Methods CAS for the TINspire
–10
(3, 2) 2 3
x
4x – 2y = 8, x ∈ [–3, 3]
8
Draw a straight line between the two points.
9
Find the intercepts and mark them on the graph.
When x = 0, y = −4 When y = 0, x = 2 The xintercept is 2 and the yintercept is −4.
10
State the domain, which is given with the rule.
The domain is [−3, 3].
11
State the range from the graph.
The range is [−10, 2].
Method 1: Technologyfree b
1
There is no smallest value of x, so substitute the largest value of x into the equation and find y.
b When x = −1,
y = f (−1) =3
2
State the coordinates of the upper end point.
(−1, 3) is an open end of the line.
3
Substitute another value of x within the domain into the equation (that is, a value of x < −1, since x ∈ (−∞, −1)) and find y.
When x = −2, y = f (−2) =5
4
State the coordinates of the point.
(−2, 5) is another point on the line.
5
Plot the 2 points on a set of axes and mark the point (−1, 3) with an open circle.
6
Rule a straight line from (−1, 3) to (−2, 5) and beyond. An arrow should be placed on the other end to indicate that the line continues.
f(x) = 1 – 2x, x ∈ (–∞, –1) (–2, 5) (–1, 3)
y 5 3
–2 –1 0
x
7
Note that there are no intercepts.
8
State the domain, which is given with the rule.
The domain is (−∞, −1).
9
State the range by examining the graph.
The range is (3, ∞).
Method 2: Technologyenabled n a Graphs page, complete the function entry O line as: f 1( x ) = 1 − 2 x  − ∞ < x < − 1 Then press ENTER ·.
Chapter 1 Graphs and polynomials
19
reMeMBer
Linear graphs 1. Linear equations are polynomials of degree 1. y − y1 2. Gradient, m = 2 x2 − x1 3. General equation is ax + by + c = 0 or y = mx + c where m = gradient and c = yintercept. 4. Equation if a point and the gradient are known: y − y1 = m(x − x1) 5. Equation if the intercepts are known: x y + =1 a b 6. Parallel lines have the same gradient. 7. If m1 and m2 are the gradients of perpendicular lines, then: m1 × m2 = −1 1 m1 = − or m2
exerCISe
1d
linear graphs 1 We 14 Sketch the graph of each of the following linear functions by indicating the intercepts. a 2x + 3y = 12 b 2y − 5x − 10 = 0 c 2x − y = 1 2 We 15a Find the equation, in the form ax + by + c = 0, of each straight line described below. a The line with a gradient of 3 and passing through (2, 1). b The line with a gradient of −2 and passing through (−4, 3).
eBook plus Digital doc
SkillSHEET 1.3 Gradient
eBook plus Digital doc
SkillSHEET 1.4 Using gradient to find the value of a parameter
20
3 We 15b Find the equation, in the form ax + by + c = 0, of each straight line described below. a The line passing through (−3, −4) and (−1, −10). b The line passing through (7, 5) and (2, 0). 4 MC Which one of the following points does not lie on the straight line with equation 2y − 3x − 6 = 0? A (2, 6) B (−2, 0) C (0, 3) D (1, 2) E (4, 9) 5 We 15c Consider the points A(−2, 5) and B(1, b). a Find b if: i the gradient of the straight line AB is −2 ii the equation of the straight line AB is y − x = 7. b Find the general equation of the straight line which passes through (4, 5) and is parallel to the line with equation y − 3x + 4 = 0. c We 15d Find the equation in the form ax + by + c = 0 that passes through (−2, 4) and is perpendicular to the line with equation 2y − x + 1 = 0.
Maths Quest 12 Mathematical Methods CAS for the TINspire
6 Match each of the following graphs with the appropriate rule below. b c a y y y (2, 4)
2
x
0 −1
e
y
y
x
0
x
0
3
−2
i x + 2y + 4 = 0 iv 3y + 2x = 6 eBook plus
x
0
ii y = 3
iii y − 2x − 2 = 0
v y − 2x = 0
vi x = −2
7 State the range for each function graphed below. a
b
y
c
y
y
(4, 3)
Digital doc
SkillSHEET 1.5
x
0
Interval notation
x
0
Digital doc
d
(−3, 3)
x
0
(−5, −2) eBook plus
x
3
f
y −4
−2
0
x
0
d
2
(−4, −2)
(6, −5)
e
y
f
y
y
(5, 6)
SkillSHEET 1.6 Domain and range for linear graphs
x
0
0
(5, −2)
0
x
4
x
8 We 16 Sketch the graph of each of the following functions, stating i the domain and i i the range of each. a 4y + 3x = 24, x ∈ [−12, 12] b 2x − 5y = 10, x < 5 c 4x − 3y − 6 = 0, x ∈ [2, 5) 9 Find the equation of the straight line which passes through the point (2, 5) and is: a parallel to the line with equation y = 3 − 2x b perpendicular to the line with equation y = 3x − 7. Write equations in the form ax + by + c = 0. 10 Find the equation of the straight line which passes through the point (−3, 1) and is: a parallel to the line with equation 4x − 2y = 13 b perpendicular to the line with equation 4x − 2y = 13. 11 MC If the straight lines 3x − y = −2 and ax + 2y = 3 are parallel then a = : A 6 B 2 C −2 D −3 E −6 12 MC If the straight lines 5x + y − 3 = 0 and bx − y − 2 = 0 are perpendicular, then b is equal to: A 5
B
1 5
C −5
D
−1 5
E 3
Chapter 1
Graphs and polynomials
21
1e
Quadratic graphs Quadratic functions are polynomials of degree 2. Graphs of quadratic functions are parabolas and may be sketched by finding the turning point and intercepts.
revision of quadratic functions 1. The general form of the quadratic function is y = ax2 + bx + c, x ∈ R. 2. The graph of a quadratic function is called a parabola and: (a) for a > 0, the graph has a minimum value (b) for a < 0, the graph has a maximum value (c) the yintercept is c − b (d) the equation of the axis of symmetry and the xvalue of the turning point is x = 2a (e) the xintercepts are found by solving the equation ax2 + bx + c = 0. 2 3. The equation ax + bx + c = 0 can be solved by either: (a) factorising or − b ± b 2 − 4 ac (b) using the quadratic formula, x = . 2a 4. The turning point can be found by ‘completing the square’ (see page 23). The turning point is located on the axis of symmetry, which is halfway between the xintercepts.
The discriminant The value of (b2 − 4ac), which is the value inside the square root sign in the quadratic formula, determines the number of solutions to a quadratic equation or the number of xintercepts on a quadratic graph. This value is called the discriminant. 1. If b2 − 4ac > 0, there are two solutions to the equation and there are two xintercepts on the graph. 2. If b2 − 4ac > 0 and is a perfect square, the solutions are rational; otherwise they are irrational. 3. If b2 − 4ac = 0, the two solutions are equal and there is one xintercept on the graph; that is, the graph has a turning point on the xaxis. 4. If b2 − 4ac < 0, there are no real solutions and there are no xintercepts on the graph.
Worked exAMple 17
eBook plus
Use the discriminant to determine the number of xintercepts for the quadratic function f ( x ) = 2 x 2 + 3 x − 10. ThINk
22
WrITe
Tutorial
int0518 Worked example 17
b = 3, c = −10
1
Find the values of the quadratic coefficients a, b and c using the general quadratic function, y = ax2 + bx + c.
a = 2,
2
Evaluate the discriminant.
b2 − 4ac = 32 − 4(2)(−10) = 9 + 80 = 89
3
If the discriminant is greater than 0, there are two xintercepts. If it is not a perfect square, the solutions are irrational.
b2 − 4ac > 0 So there are two xintercepts, which are both irrational.
Maths Quest 12 Mathematical Methods CAS for the TINspire
Worked Example 18
Sketch the graph of the function f ( x ) = 12 − 5 x − 2 x 2, showing all intercepts. Give exact answers. Think
Write/draw
f(0) = 12 − 5(0) − 2(0)2 = 12 The yintercept is 12. f(x) = 12 − 5x − 2x2 = 0 (4 + x)(3 − 2x) = 0
5
Evaluate f (0) to find the yintercept (or state the value of c). State the yintercept. Set f(x) = 0 to find the xintercepts. Factorise the quadratic (or use the quadratic formula). Solve the equation using the Null Factor Law.
6
State the xintercepts.
The xintercepts are −4 and 2 .
7
Draw a set of axes and mark the intercepts or the coordinates of the points where the graph crosses the axes.
1 2 3 4
8
4 + x = 0 or 3 − 2x = 0 x = −4 or x = 32 3
y 12 (0, 12) f(x) = 12 − 5x − x2
Sketch a parabola through the intercepts. (−4, 0) –4
( 3–2 , 0) x 0 1 2
The xcoordinate of the turning point of a quadratic function is exactly halfway between −4 + 3 − 5 − 2 = the two xintercepts, so for worked example 18, x = (or −1 14 ). Substitute x = 5 4 2 4 into the original equation to find the ycoordinate of the turning point. − b The xcoordinate of the turning point can also be found by using the formula x = , where 2a ax2 + bx + c = 0.
Finding turning points by completing the square Consider the general quadratic equation: y = ax2 + bx + c By completing the square, this equation may be manipulated into the form y = a(x − b)2 + c where the turning point is (b, c). This way of writing the function is known as the power form or turning point form. The transformations associated with this form will be discussed more fully in chapter 2. Worked Example 19
For the function y = − 2( x + 3)2 − 4, find: a the coordinates of the turning point b the domain and range. Think
Write
Write the general formula.
y = a(x − b)2 + c
Write the function.
y = −2(x + 3)2 − 4
a 1
Identify the values of a, b and c.
a a = −2, b = −3, c = −4
Chapter 1 Graphs and polynomials
23
2
b 1 2
State the coordinates of the turning point (b, c). Write the domain of the parabola. Write the range y ≤ c (as a < 0).
The turning point is (−3, −4). b The domain is R.
The range is y ≤ −4.
Worked Example 20
The function graphed at right is of the form y = x 2 + bx + c. Find: a the rule b the domain c the range. Write the answers to b and c in interval notation.
y (−5, 10)
0
x
(−1, −6) Think a
b
c
Write a y = a(x − b)2 + c
1
Write the general rule for a quadratic in turning point form.
2
Find the values of b and c using the given turning point.
Since the turning point is (−1, −6): b = −1, c = −6
3
State the value of a (given).
a=1
4
Substitute these values in the rule.
So y = 1(x + 1)2 − 6
5
Expand the brackets.
6
Simplify.
1
Use the graph to find the domain. Look at all the values that x can take.
2
State the domain in interval notation.
1
Use the graph to find the range. Look at all the values that y can take.
2
State the range in interval notation.
= x2 + 2x + 1 − 6 = x2 + 2x − 5 The rule is y = x2 + 2x − 5. b x ≥ −5
Domain = [−5, ∞) c y ≥ −6
Range = [−6, ∞)
Worked Example 21 1
Sketch the graph of y = 2 ( x − 1)2 + 2, clearly showing the coordinates of the turning point and the intercepts with the axes. State its range. Think
24
Write/Draw
1
Write the general equation of the parabola.
y = a(x − b)2 + c
2
Identify the values of the variables.
a = 12 , b = 1, c = 2
Maths Quest 12 Mathematical Methods CAS for the TINspire
3
Write a brief statement on the transformation of the basic parabola.
The graph of y = x2 is dilated in the y direction by the factor of 12 (that is, it is wider than the basic curve); it is translated 1 unit to the right and 2 units up.
4
State the shape of the parabola (that is, positive or negative).
a > 0; the parabola is positive.
5
State the coordinates of the turning point (b, c).
The turning point is (1, 2).
6
As both a and c are positive, only the yintercept needs to be determined. Find the yintercept by making x = 0.
yintercept: x = 0 y = 12 (0 − 1)2 + 2 = 12 (−1)2 + 2 = 12 + 2 = 2 12
7
Sketch the graph: Draw a set of axes and label them. Plot the turning point and the yintercept. Sketch the graph of the positive parabola, so that it passes through the points previously marked.
y
1
2 –2 2 0
Since y ≥ 2, that is the range.
8
x 1 y = 1–2 (x − 1)2 + 2
The range is y ∈ [2, ∞).
Worked exAMple 22
Sketch the graph of y = 3 + 8 x − 2 x 2, showing the turning point and all intercepts, rounding answers to 2 decimal places where appropriate. ThINk
WrITe/drAW
Method 1: Technologyfree 1
Find y when x = 0.
When x = 0, y = 3
2
State the yintercept.
The yintercept is 3.
3
Let the quadratic equal zero.
When y = 0, 3 + 8x − 2x2 = 0
4
Solve for x using the quadratic formula. x= = = =
−8 ±
82 − 4(− 2)(3)
−8 ±
2(− 2) 88
−4 −8 ± 2
22
−4 −4 ±
=2−
22
−2
22 2
or
Chapter 1
2+
22 2
Graphs and polynomials
25
5
State the xintercepts, rounding to 2 decimal places.
6
Use the formula for the xvalue of the turning − b point, x = . 2a
The xintercepts are −0.35 and 4.35.
x=
−8
2(− 2) x=2
7
To calculate the ycoordinate of the turning point, substitute x = 2 into the function.
y = 2(2)2 + 8(2) + 3 y = 11
8
State the turning point.
The turning point is (2, 11).
9
Draw a set of axes and mark the coordinates of the turning point and the points where the graph crosses the axes. Sketch a parabola through these points.
10
y 12
(2, 11)
9 f(x) = 3 + 8x –
2x2
6
3 (0, 3) (–0.35, 0) (4.35, 0) x 0 –1 4 5
Method 2: Technologyenabled On a Graphs page, complete the function entry line as: f(x) = 3 + 8x  2x2 Then press ENTER ·. To label the coordinates of the intercepts and turning point, press: • Menu b • 6:Analyze Graph 6 Select the appropriate action.
Note: Function notation includes the rule, the domain and the codomain. For example, f (x): [−2, 1] → R, where f (x) = x2 − 3, is a parabola with rule f (x) = x2 − 3 and domain [−2, 1]. The range is a subset of the codomain, R.
Worked Example 23
The weight of a person t months after a gymnasium program is started is given by the 2 function: W ( t ) = t − 3 t + 80, where t ∈ [0, 8] 2 and W is in kilograms. Find: a the minimum weight of the person b the maximum weight of the person.
26
Maths Quest 12 Mathematical Methods CAS for the TINspire
Think 1
Write
Complete the square to find the turning point.
t2 − 3t + 80 2 1 = 2 [t 2 − 6t + 160]
W=
1
= 2 [t 2 − 6t + 9 + 160 − 9] 1
= 2 [(t − 3)2 + 151] 1
= 2 (t − 3)2 + 75.5 2 3 4 5 6 7 8 9
State the minimum turning point. Find the end point value for W when t = 0. State its coordinates. Find the end point value of W when t = 8. State its coordinates. On a set of axes, mark the end points and turning point. Sketch a parabola between the end points. Locate the maximum and minimum values of W on the graph.
The turning point is (3, 75.5). When t = 0, W = 80 One end point is (0, 80). When t = 8, W = 88 The other end point is (8, 88). W (kg) Maximum (8, 88)
90 80 (0, 80) 70
Minimum (3, 75.5) 0 1 2 3 4 5 6 7 8 t (months)
a b
State the minimum weight from the graph. State the maximum weight from the graph.
a b
The minimum weight is 75.5 kg. The maximum weight is 88 kg.
REMEMBER
Quadratic graphs 1. Quadratic equations are polynomials of degree 2. 2. The general equation is y = ax2 + bx + c. 3. The quadratic formula is given by the equation − b ± b 2 − 4 ac 2a 4. The discriminant is b2 − 4ac and if: (a) b2 − 4ac > 0, there are two xintercepts. If b2 − 4ac is a perfect square, the intercepts are rational. (b) b2 − 4ac = 0, there is one xintercept, which is a turning point. (c) b2 − 4ac < 0, there are no xintercepts. 5. The turning point form of the quadratic graph or parabola is: y = a(x − b)2 + c and the turning point is (b, c). 6. The equation of the axis of symmetry of a parabola and the xvalue of the turning −b point is given by the expression x = . 2a 7. The axis of symmetry is halfway between the xintercepts. x=
Chapter 1 Graphs and polynomials
27
exerCISe
1e eBook plus Digital doc
Spreadsheet 103 Discriminant
Quadratic graphs 1 We17 use the discriminant to determine the number of xintercepts for each of the following quadratic functions. a f (x) = x2 − 3x + 4 b f (x) = x2 + 5x − 8 c f (x) = 3x2 − 5x + 9 2 2 d f (x) = 2x + 7x − 11 e f (x) = 1 − 6x − x f f (x) = 3 + 6x + 3x2 2 We18 Sketch the graphs of each of the following functions, showing all intercepts. Give exact answers. a f (x) = x2 − 6x + 8 c f (x) = 10 + 3x − x2
eBook plus
b f (x) = x2 − 5x + 4 d f (x) = 6x2 − x − 12
Digital doc
Spreadsheet 107 Quadratic graphs
3 Find the turning point for each of the functions in question 2. Give exact answers.
eBook plus
4 We19 For each of the following functions find: i the coordinates of the turning point ii the domain iii the range. 2 b y = (x − 6)2 a y=2−x c y = −(x + 2)2 d y = 2(x + 3)2 − 6
Digital doc
SkillSHEET 1.7 Domain and range for quadratic graphs
5 We20 Each of the functions graphed below is of the form y = x2 + bx + c. For each function, give: i the rule ii the domain iii the range. Write the answers to b and c in interval notation. a
b
y
c
y
y
(−1, 6) x
0 (1, −2)
0
x
0
(1, 9)
(2, −3)
4
x
(−4, −16)
6 We21 Sketch the graphs of the following, clearly showing the coordinates of the turning point and the intercepts with the axes. a y = 2x2 + 3 b y = (2x − 5)(2x − 3) c y = (2x − 3)2 − 8 Consider the function with the rule y = x2 − 2x − 3. 7 MC a It has xintercepts: A (1, 0) and (3, 0) B (−1, 0) and (3, 0) C (1, 0) and (−3, 0) −1, 0) − E (0, 1) and (0, 3) D (2, 0) and ( b It has a turning point with coordinates: A (−1, 0)
B (2, −3) −(x
+ 8 MC The function f (x) = A (3, ∞) B (−∞, −3]
C (1, −4) 3)2
D (−1, −4)
+ 4 has a range given by: C [4, ∞) D (−∞, 4]
9 MC The range of the function y = (x − 4)2, x ∈ [0, 6] is: A [0, 16] B [4, 16] C [0, 4] D (4, 12]
E [0, 16)
10 We22 Sketch the graph of each of the following functions, showing the turning point and all intercepts. Round answers to 2 decimal places where appropriate. a f (x) = (x − 2)2 − 4 b f (x) = −(x + 4)2 + 9 2 c y = x + 4x + 3 d y = 2x2 − 4x − 6
28
Maths Quest 12 Mathematical Methods CAS for the TINspire
eBook plus Digital doc
Spreadsheet 041 Function grapher
E (1, 0) E R−
eBook plus Digital doc
Spreadsheet 108 Quadratic graphs — turning point form
11 Sketch the graph of each of the functions below and state i the domain and ii the range of each function. b y = −x2 + x − 1, x ∈ R+ a y = x2 − 2x + 2, x ∈ [−2, 2] c f (x) = x2 − 3x − 2, x ∈ [−10, 6] d f (x) = 5 + 6x − 3x2, x ∈ [−5, 3) 12 We23 The volume of water in a tank, V m3, over a 10 month period is given by the function V(t ) = 2t 2 − 16 t + 40, where t is in months and t ∈ [0, 10]. Find: Maximum height a the minimum volume of water in the tank b the maximum volume of water in the tank. Tower
13 A ball thrown upwards from a tower attains a height above the ground given by the function h(t) = 12t − 3t2 + 36, where t is the time in seconds and h is in metres. Find: a the maximum height above the ground that the ball reaches b the time taken for the ball to reach the ground c the domain and range of the function.
eBook plus Digital doc
WorkSHEET 1.2
1F
Ball
h(t) = 12t − 3t2 + 36
Ground
14 A section of a rollercoaster at an amusement park follows the path of a parabola. The function h(t) = t2 − 12t + 48, t ∈ [0, 11], models the height above the ground of the front of one of the carriages, where t is the time in seconds and h is the height in metres. a Find the lowest point of this section of the ride. b Find the time taken for the carriage to reach the lowest point. c Find the highest point above the ground. d Find the domain and the range of the function. e Sketch the function.
Cubic graphs Cubic functions are polynomials of degree 3. In this section, we will look at how graphs of cubic functions may be sketched by finding intercepts and recognising basic shapes.
Forms of cubic functions Cubic functions may take several forms. The three main forms are described below.
General form The general form of a cubic function is y = ax3 + bx2 + cx + d If a is positive (that is, a > 0), the function is called a ‘positive cubic’. Several positive cubics appear below. y
y
y
x x
x
Chapter 1
Graphs and polynomials
29
If a is negative (that is, a < 0), the function is called a ‘negative cubic’. Several negative cubics appear below. y
y
x
y
x x
Basic form Some (but certainly not all) cubic functions are transformations of the form y = x3, which has a point of inflection at the origin. These may be expressed in the power form y = a(x − b)3 + c where (b, c) is the point of inflection. For example, y = 2(x − 3)3 + 5 is the graph of y = x3 translated +3 from the yaxis, +5 in the y direction and dilated by a factor of 2 from the xaxis. This form, called basic form or power form, works in the same way as a quadratic equation expressed in turning point form or power form: y = a(x − b)2 + c where (b, c) is the turning point and a is the dilation factor. The power form and its transformations will be discussed in more detail in chapter 2. y
y
y = x3
x
y = a(x − b)3 + c
(b, c) x
Factor form Cubic functions of the type y = a(x − b)(x − c)(x − d) are said to be in factor form, where b, c and d are the xintercepts. Often a cubic function in general form may be factorised to express it in factor form. y
y = a(x − b)(x − c)(x − d) where a > 0
y = −(x + 2)(x − 1)(x − 3) y
–2 b
30
c d
x
Maths Quest 12 Mathematical Methods CAS for the TINspire
1
3 x
repeated factors y
A twice only repeated factor in a factorised cubic function indicates a turning point that just touches the xaxis.
a
b
x
y = (x − a)2 (x − b)
Worked exAMple 24
eBook plus
For each of the following graphs, find the rule and express it in factorised form. Assume that a = 1 or a = − 1 . a
y
−4
0
b
f(x)
3
Tutorial
int0519
y
f(x)
x −2 0
ThINk a
b
Worked example 24
3
x
WrITe a The graph is a positive cubic, so a = 1.
1
Find a by deciding whether the graph is a positive or negative cubic.
2
use the xintercepts −4, 0 and 3 to find the factors.
The factors are (x + 4), x and (x − 3).
3
Express f(x) as a product of a and its factors.
f(x) = 1(x + 4) x(x − 3)
4
Simplify.
f(x) = x(x + 4)(x − 3)
1
Find a by deciding whether the graph is a positive or negative cubic.
2
use the xintercept −2, which is also a turning point, to find the repeated factor.
(x + 2)2 is a factor.
3
use the other xintercept, 3, to find the other factor.
(x − 3) is also a factor.
4
Express f(x) as a product of a and its factors.
f(x) = −1(x + 2)2(x − 3)
5
Simplify.
f(x) = (3 − x)(x + 2)2
b The graph is a negative cubic, so a = −1.
Worked exAMple 25
Sketch the graph of y = x 3 − x 2 − 10 x − 8, showing all intercepts. ThINk 1
Find y when x = 0.
WrITe/drAW
When
x = 0, y = −8
Chapter 1
Graphs and polynomials
31
2
State the yintercept.
The yintercept is −8.
3
Let P(x) = y.
Let P(x) = x3 − x2 − 10x − 8
4
Use the factor theorem to find a factor of the cubic P(x) = x3 − x2 − 10x − 8.
P(1) = 13 − 12 − 10(1) − 8 = −18 ≠0 P(−1) = (−1)3 − (−1)2 − 10(−1) − 8 =0 so (x + 1) is a factor.
5
Use long division, or otherwise, to find the quadratic factor.
By long division: x2 − 2x − 8
)
x + 1 x 3 − x 2 − 10 x − 8 (x 3 + x 2 ) − 2 2 − 10 − 8 x x (− 2 x 2 − 2 x ) − 8x − 8 − −( 8 x − 8) 0 y = (x + 1)(x2 − 2x − 8) = (x + 1)(x − 4)(x + 2)
6
Factorise the quadratic, if possible.
7
Express the cubic in factorised form and let it equal 0 to find the xintercepts. Solve for x using the Null Factor Law. Alternatively, use a CAS calculator to solve for x.
If (x + 1)(x − 4)(x + 2) = 0
State the xintercepts. Sketch the graph of the cubic.
The xintercepts are −2, −1, and 4.
8
9 10
x = −1, 4 or −2
y −2 −1 0
y = x3 − x2 − 10x − 8 4
x
−8 Exam tip When sketching graphs, ensure that they are smooth, that relevant turning points and intercepts are labelled, and that they are drawn within the correct domain (end points should be shown using • or °) using [Authors’ advice] an appropriate scale.
32
Maths Quest 12 Mathematical Methods CAS for the TINspire
Restricting the domain of cubic functions 1. If the domain is R then the range is also R. 2. To find the range if the domain is restricted, it is necessary to look at the end points and turning points, then find the highest and lowest yvalues. y (6, 8) For example: The range can not be stated for the diagram at right because the ycoordinate of the local minimum is not known. Recall that cubic functions that do not have any turning points can have only one xintercept.
x
0 (–4, –3)
Coordinate of local minimum required
Worked Example 26
Sketch the graph of y = − x 3 − 5 x, where x ∈ ( − 2, − 1], using the unrestricted function as a guide. State the domain and range, without the use of technology. Think
WRITE/DRAW
1
Decide whether it is a positive or negative cubic by looking at the coefficient of x3.
Negative cubic
2
Find the xintercept/s.
When y = 0, −x3 − 5x = 0 −x(x2 + 5) = 0 x = 0 (x2 + 5 ≠ 0) The xintercept is 0.
3
Find the yintercept.
When x = 0, y = −(0)3 − 5(0) =0 The yintercept is 0.
4
Find y when x has the value of the lower end point of the domain.
When x = −2, y = −(−2)3 − 5(−2) = 18
5
State the coordinates of this end point and decide whether it is open or closed.
The open end point is (−2, 18).
6
Find y when x has the value of the upper end point.
When x = −1, y = −(−1)3 − 5(−1) =6
7
State the coordinates of this end point and decide whether it is open or closed.
The closed end point is (−1, 6).
8
Mark these points on a set of axes.
9
Sketch the part of the cubic between the end points.
10
Verify this graph using a graphics calculator.
y
(−2, 18)
(−1, 6) 0
x
11
State the domain, which is given with the rule.
The domain is (−2, −1].
12
From the graph, state the range. Note that the intercept is not included in the domain.
The range is [6, 18).
Chapter 1 Graphs and polynomials
33
reMeMBer
Cubic graphs 1. The general equation is y = ax3 + bx2 + cx + d. 2. Basic shapes of cubic graphs: (a) If a > 0: Positive cubic
Power form
y
y
y = a(x − b)3 + c
x
(b, c) x
Factor form
Repeated factor
y
y y = a(x − b)(x − c)(x − d) where a > 0
b
c d
a
b
x
y = (x − a)2 (x − b)
x
(b) If a < 0, the reflections through the xaxis of the types of graph in the above figures are obtained. exerCISe
1F eBook plus
Cubic graphs 1 We24 For each of the following graphs, find the rule and express it in factorised form. Assume that a = 1 or a = −1. b a y y
Digital doc
Spreadsheet 013 Cubic graphs — factor form
−6
0
5
0
−2
x
1
x
−4
2
Match each of the following graphs to the most appropriate rule below. a
−3
34
b
y
0 1
4
x
c
y
−2 0
Maths Quest 12 Mathematical Methods CAS for the TINspire
5
x
y
−3
0
1
4
x
d
e
y
0
g
−4
x
3
y 0
−2
1
x
0
1
x
ii iv vi viii
y = (x + 3)(1 − x)(x − 4) y = (x + 2)2(5 − x) y = (x + 4)(x + 2)(x − 1) y = (x + 2)2(x − 5)
0
−2
x
3
y = (x − 3)3 y = (x + 4)(x + 2)(1 − x) y = (x + 3)(x − 1)(x − 4) y = (3 − x)3
3 We25 Sketch the graph of each of the following, showing all intercepts. a y = x3 + x2 − 4x − 4 b y = 2x3 − 8x2 + 2x + 12 3 c y = 24 + 26x − 2x d y = 18 − 21x + 8x2 − x3 Verify your answers by using a calculator. 4 MC a Fully factorised, x3 + 6x2 + 12x + 8 is equal to: A (x + 3)3 B (x + 2)3 3 D (x − 3) E (x + 2)(x − 2)2 b The graph of y = x3 + 6x2 + 12x + 8 is: B A y y
−2
0
D
2
0
−2
x
E
y
−2
x
5
y
−4 0
i iii v vii
−2
h
y
f
y
2
eBook plus Digital doc
Spreadsheet 014 Cubic graphs
C (x − 2)3
C
x
0
y
−2
0
2
x
y
x
−2
x
0
5 MC The function graphed in the figure could have the following rule: A y = (x − 2)3 + 2 y B y = (x + 2)3 + 2 C y = (2 − x)3 + 2 10 D y = (x + 2)3 − 2 3 E y = (x − 2) (2, 2)
0
x
Chapter 1
Graphs and polynomials
35
eBook plus Digital doc
6 MC The graph of f (x) = 5(x + 1)3 − 3 is best represented by: A
B
y
C
y
y
Spreadsheet 015 Cubic graphs — y = a( x − b ) 3 + c form
x
0
(−1, −3)
0
x
0
(−1, −3)
D
E
y
x (1, −3)
y (1, 3)
(−1, 3) x
0
x
0
The graph of f (x) = 2(x − 1)2 (x + 3) is best represented by:
7 MC A
B
y
y
(0, 6)
(−1, 0) 0
C
(0, 6)
x
(3, 0)
(−3, 0)
D
y
0
x
(1, 0)
y
(1, 0) x
(−3, 0) 0
y (3, 0) (−1, 0)
0
x
(0, −6)
36
(1, 0)
(0, −6)
(0, −6)
E
(−3, 0) 0
Maths Quest 12 Mathematical Methods CAS for the TINspire
x
8 MC
The graph shown is best represented by the equation:
A y = (x − +b B y = −(x − a)3 + b C y = (a − x)3 + b D y = −(x + a)3 + b E y = (x + a)3 + b
y
a)3
(a, b) (0, c) x
0
9 MC A y = B y = C y = d y = E y =
If a < 0 and b, c > 0 then the graph shown is best represented by the equation: y b ( x + a ) 2 ( x − c) 2 a c −b ( x + a) 2 (c − x ) b a2c b ( x − a ) 2 ( x + c) a2c −b a c 0 ( x + a)2 ( x − b)(c − x ) a2c −b ( x − a ) 2 ( x − c) a2c
x
10 WE 26 Sketch the graph of each of the following restricted functions, using the unrestricted function as a guide. State i the domain and ii the range in each case. a f (x) = x3 + x2 − 10x + 8, x ∈ [2, ∞) b f (x) = 3x3 − 5x2 − 4x + 4, x ∈ [−2, −1] c f (x) = −3x3 + 4x2 + 27x − 36, x ∈ (0, 1] d f (x) = −3x − x3, x ∈ [−1, 2) e f (x) = x3 + 2x, x ∈ [−2, −1) ∪ (0, 3] f f (x) = −2x3 − x, x ∈ (−1, 1) ∪ [2, 3) 11 The function f (x) = x3 + ax2 + bx − 64 has xintercepts (−2, 0) and (4, 0). Find the values of a and b. 12 The functions y = x3 − 2x2 + ax + 10 and y = 6 + (a + b)x − 4x2 − x3 both have (−1, 0) as an xintercept. Find the values of a and b. 13 The crosssection of a glass vessel that is 6 cm high can be modelled by the cubic function f (x) and its reflection through the yaxis, g(x), as shown at right. y f(x) = a(x + b)3 + c a Find the values of a, b and c, and hence state the rule g(x) (4, 6) of f (x). b Find the rule for g(x) and state its domain and range. (3, 3) c What is the width of the vessel when the height is 3.375 cm? 0
(2, 0)
x
14 The distance of a group of hikers, d km, from their starting point t hours after setting off on a hike can be modelled by the function with the rule: d(t) = at2 (b − t) The hikers are 3 km from the start after 2 hours and return to the starting point after 5 hours. a Find the values of a and b. b Hence, give the rule for d(t ) stating its domain and range.
Chapter 1 Graphs and polynomials
37
c Sketch the graph of d(t ). d Find to the nearest 100 metres the maximum distance of the hikers from their starting point and the time, to the nearest minute, that it occurs.
1G
Quartic graphs Quartic functions are polynomials of degree 4. The general form of a quartic is: y = ax4 + bx3 + cx2 + dx + e When sketching the graphs of quartic functions, all axes’ intercepts can be found by factorisation and a sign diagram used to check the shape. If a sign diagram is not sufficient and the basic shape is not recognised, then a graphics calculator could be used to establish the shape of the graph.
Basic shapes of quartic graphs Positive quartics (a > 0) 1. y = ax4
2. y = ax4 + cx2, c ≥ 0 y
y
x
0
3. y = ax2(x − b)(x − c)
4. y = a(x − b)2(x − c)2 y
y
b
0
c x 0
b
The repeated factor x2 shows there is a turning point at the origin. The factors (x − b) (x − c) show xintercepts at x = b and x = c. 5. y = a(x − b)(x − c)3
0
6. y = a(x − b)(x − c)(x − d )(x − e) y
c
x b
The cubed factor (x − c)3 shows the graph as a point of inflection at x = c. 38
x
c
The repeated factors (x − b)2 and (x − c)2 show the graph touches the xaxis at x = b and x = c.
y
b
x
0
Maths Quest 12 Mathematical Methods CAS for the TINspire
c
0
d
ex
The factors show intercepts at x = b, c, d and e.
y
Negative quartics (a < 0) If a < 0, that is, each of the above rules is multiplied by −1, then the graphs are reflected through the xaxis. For example, the graph of y = −x4 (at right) is a reflection, through the xaxis, of the graph of y = x4. Similarly, the graph of y = −x4 + x2 = −(x4 − x2) is a reflection through the xaxis of the graph of y = x4 − x2. Note: The above graphs can be translated horizontally or vertically but this is considered in chapter 2. To find the xintercepts of a quartic function, let y = 0 and solve the equation for x. Repeated factors touch the xaxis as they do for cubic and quadratic functions.
x
0
y = −x 4 y x
0
y = −x 4 + x2 Worked exAMple 27
Sketch the graph of y =
eBook plus
x4
−
x3
− 7 x2
+ 5 x + 10, showing all intercepts.
Tutorial
int0520
ThINk
Method 1: Technologyfree 1 Find the yintercept.
3
Let y = P(x). Find two linear factors of the quartic expression, if possible, using the factor theorem.
4
Find the product of the two linear factors.
5
use long division to divide the quartic by the quadratic factor x2 − x − 2 (or use another method).
2
Worked example 27
WrITe/drAW
When x = 0, y = 10 The yintercept is 10. Let P(x) = x4 − x3 − 7x2 + 5x + 10 P(1) = (1)4 − (1)3 − 7(1)2 + 5(1) + 10 =8 ≠0 P(−1) = (−1)4 − (−1)3 − 7(−1)2 + 5(−1) + 10 =0 (x + 1) is a factor. P(2) = (2)4 − (2)3 − 7(2)2 + 5(2) + 10 =0 (x − 2) is a factor. (x + 1)(x − 2) = x2 − x − 2 x2
−x−2
)
x4
−
−7
x3
−(x − x 4
3
x2
− 5
x2
+ 5 x + 10
− 2x2 )
0 − 5 x 2 + 5 x + 10 − (− 5 x 2 + 5 x + 10) 0 6
Express the quartic in factorised form.
y = (x + 1)(x − 2)(x2 − 5)
7
Factorise the quadratic factor, x2 − 5, using difference of perfect squares.
y = (x + 1)(x − 2)(x + 5)(x − 5)
8
To find the xintercepts, set y equal to zero.
Let y = (x + 1)(x − 2)(x + 5)(x − 5) = 0
9
Solve for x using the Null Factor Law.
10
State the xintercepts.
x = −1, 2, ± 5 The xintercepts are −1, 2, − 5 and 5.
Chapter 1
Graphs and polynomials
39
11
Sketch the graph of the quartic.
y (0, 10) (−1, 0) (− 5, 0) −3 −2 −1 0
(2, 0) 1
( 5, 0) x 2 3
Method 2: Technologyenabled 1
On a Calculator page, define the polynomial P(x) by completing the entry line as: Define p( x ) = x 4 − x 3 − 7 x 2 + 5 x + 10 Then press ENTER ·. To find the yintercept, complete the entry line as: p(0) Then press ENTER ·. To find the xintercepts, complete the entry line as: solve ( p( x ) = 0, x ) Then press ENTER ·.
2
To sketch the graph of P(x), open a Graphs page. Complete the function entry line as: f 1( x ) = p( x ) Then press ENTER ·.
Worked exAMple 28
eBook plus
Sketch the graphs of each of the following equations, showing the coordinates of all intercepts. Use a CAS calculator to find the coordinates of the turning points, rounding to 2 decimal places as appropriate. a y = x 2 ( x − 1)( x + 2 ) b y = − ( x + 3)2 ( x − 1)2 ThINk a
40
WrITe/drAW
1
State the function.
2
Find the yintercept.
a y = x2(x − 1)(x + 2)
When x = 0, y = 0 The yintercept is 0.
Maths Quest 12 Mathematical Methods CAS for the TINspire
Tutorial
int0521 Worked example 28
3
Find the xintercepts.
When y = 0, 0 = x2(x − 1)(x + 2) x = −2, 0, 1
4
State the xintercepts, noting where the graph touches and where it cuts the xaxis.
The graph touches the xaxis at x = 0. The other xintercepts are −2 and 1.
5
State the coordinates of the turning points.
The minimum turning points are (−1.44, −2.83) and (0.69, −0.40). The maximum turning point is (0, 0).
6
Sketch the graph of the quartic, using a CAS calculator to assist.
y
(−2, 0)
(0, 0) 0
(1, 0) (0.69, −0.40)
x
(−1.44, −2.83) b
b y = −(x + 3)2(x − 1)2
1
State the function.
2
Find the yintercept.
When x = 0, y = −(3)2(−1)2 = −9 The yintercept is −9.
3
Find the xintercepts.
When y = 0, 0 = −(x + 3)2(x − 1)2 x = −3, 1
4
State the points where the graph touches the xaxis from the repeated factors.
The graph touches the xaxis at x = −3 and x = 1.
5
State the coordinates of the turning points.
The maximum turning points are (−3, 0) and (1, 0), and the minimum turning point is (−1, −16).
6
Sketch the graph of the quartic, using a graphics calculator to assist.
y (−3, 0)
(1, 0) 0
x
(0, −9)
(−1, −16)
Chapter 1 Graphs and polynomials
41
Worked Example 29
Determine the equation of the graph shown. y 3 −3
−1 0
1 2
Think
x
Write/display
Method 1: Technologyfree 1
State the xintercepts.
The xintercepts are −3, −1, 1, 2.
2
Write the equation using factor form with a dilation factor of a.
y = a(x + 3)(x + 1)(x − 1)(x − 2)
3
State the yintercept.
The yintercept is 3.
4
Substitute the coordinates of the point where the graph crosses the yaxis into the equation.
(0, 3) ⇒ 3 = a(0 + 3)(0 + 1)(0 − 1)(0 − 2)
5
Solve the equation to find a.
3=a×6 a=
6
Write the equation.
1 2
y = 12 ( x − 1)( x − 2)( x + 3)( x + 1)
Method 2: Technologyenabled
42
1
On a Calculator page, complete the entry line as: a × ( x + 3) × ( x + 1) × ( x − 1) × ( x − 2) → y Then press ENTER ·. To calculate the value of a, complete the entry line as: solve(y = 3, a)  x = 0 Then press ENTER ·. To find the equation of y, complete the entry line as: 1 y  a = 2 Then press ENTER ·.
2
Write the equation.
y=
( x − 2)( x − 1)( x + 1)( x + 3) 2
Maths Quest 12 Mathematical Methods CAS for the TINspire
Worked Example 30
Sketch the graph of y = − x 4 − 2 x 2 , ∈ x( − 1, 1] , using the unrestricted function as a guide. State the domain and the range in each case. Think
Write/Draw
1
State the function.
y = −x4 − 2x2, x ∈ (−1, 1]
2
Find the yintercept.
When x = 0, y = −(0)4 − 2(0)2 =0
3
State the yintercept.
The yintercept is 0.
4
Find the xintercepts.
When y = 0 −x4 − 2x4 = 0
5
Factorise the quartic expression.
6
Solve for x.
x = 0 is the only solution (as x2 + 2 ≠ 0).
7
State the xintercepts.
The only xintercept is 0.
8
Find y when x is one end point of the domain.
When x = −1, y = −(−1)4 − 2(−1)2 = −3
9
State the coordinates and whether it is an open or closed point.
(−1, −3) is an open end point.
10
Find y when x is the other end point of the domain.
When x = 1, y = −(1)4 − 2(1)2 = −3
11
State the coordinates and whether it is an open or closed point.
(1, −3) is a closed end point.
12
Sketch the graph of the quartic, using knowledge of basic shapes or a CAS calculator to assist, over the domain.
−x2(x2
+ 2) = 0
y (0, 0)
x
0 (−1, −3)
(1, −3) y = −x 4 − 2x2
13
State the domain, which is given with the rule.
The domain is (−1, 1].
14
From the graph, state the range.
The range is [−3, 0].
Chapter 1 Graphs and polynomials
43
REMEMBER
Quartic graphs 1. General equation is y = ax4 + bx3 + cx2 + dx + e. 2. Basic shapes of quartic graphs: (a) If a > 0: y = ax4
y
x
0
y = ax4 + cx2, c ≥ 0
y
x
0
y = ax2(x − b)(x − c)
y
b
c
0
x
y = a(x − b)2(x − c)2
y
0
b
x
c
y = a(x − b)(x − c)3
y
b
0
x
c
y = a(x − b)(x − c)(x − d)(x − e)
y
b
c 0
d
ex
(b) If a < 0, then the reflection through the xaxis of the types of graph in the figures above is obtained.
44
Maths Quest 12 Mathematical Methods CAS for the TINspire
exerCISe
1G eBook plus Digital doc
SkillSHEET 1.8 Solving quartic equations
Quartic graphs 1 We27 Sketch the graph of each of the following, showing all intercepts. a y = (x − 2)(x + 3)(x − 4)(x + 1) b y = 2x4 + 6x3 − 16x2 − 24x + 32 c y = x4 − 4x2 + 4 d y = 30x − 37x2 + 15x3 − 2x4 e y = 6x4 + 11x3 − 37x2 − 36x + 36
eBook plus Digital doc
Spreadsheet 105 Quartic graphs — factor form
2 We28 Sketch the graph of each of the following equations, showing the coordinates of all intercepts. use a calculator to find the coordinates of the turning points, rounding to 2 decimal places as appropriate. a y = x2 (x − 2)(x − 3) b y = −(x + 1)2 (x − 1)2 c y = (x − 1)2(x + 1)(x + 3) d y = (x + 2)3 (1 − x) 3 MC Consider the function f (x) = x4 − 8x2 + 16. a When factorised, f (x) is equal to: A (x + 2)(x − 2)(x − 1)(x + 4) C (x + 3)(x − 2)(x − 1)(x + 1) E (x − 2)2 (x + 2)2 b The graph of f (x) is best represented by: B A y −2
0
2
0
–2
E
y 16
C
y 16
x
−16
D
B (x − 1)(x − 4)(x + 4) D (x − 2)3 (x + 2)
−2
x
2
y 16
0
2
x
y 4
−2
0
2
x
−2
0
2
x
c If the domain of f (x) is restricted to [−2, 2], then the range is:
A [0, 16] D R+
B [0, 10] E [0, ∞)
C [−2, 12]
d If the range of f (x) is restricted to (0, 25) then the maximal domain is:
A [−2, 3)
B (−2, 3)
C (−3, 2)
D (−3, 3)
e If the domain of f (x) is restricted to (−1, 0), then the range is:
A (0, 16)
B (0, 4)
C (−1, 9)
D (9, 16)
f If the domain of f (x) is restricted to [0, ∞), then the range is:
A R
B R+
C [0, ∞)
D [0, 16)
E (−3, 4)
E [9, ∞) E [2, ∞)
Chapter 1
Graphs and polynomials
45
4 We29 Determine the equation of each of the following graphs. b a y y 8
6
−2 −1 0
1
3
x
−1 0
2
4
x
5 We30 Sketch the graph of each of the following restricted functions, using the unrestricted function as a guide. State i the domain and ii the range in each case. a y = (2 − x)(x2 − 4)(x + 3), x ∈ [2, 3] b y = 9x4 − 30x3 + 13x2 + 20x + 4, x ∈ (−2,−1] c y = −(x − 2)2(x + 1)2, x ∈ (−∞, −2]
d y = 4x2 − x4, x ∈ [−3, −2]
6 The function f (x) = x4 + ax3 − 4x2 + bx + 6 has xintercepts (2, 0) and (−3, 0). Find the values of a and b. 7 The function f (x) = x4 + ax3 + bx2 − x + 6 has xintercepts (1, 0) and (−3, 0). Find the values of a and b. 8 The functions y = (a − 2b)x4 − 3x − 2 and y = x4 − x3 + (a + 5b)x2 − 5x + 7 both have an xintercept of 1. Find the value of a and b. eBook plus Digital doc
Investigation Quartics and beyond
46
Maths Quest 12 Mathematical Methods CAS for the TINspire
Summary Pascal’s triangle
1 1 1 1 1 1
2 3
4 5
1 1 3 6
10
1 4
10
1 5
1
Binomial theorem
n n n n (ax + b) n = (ax ) n b 0 + (ax ) n − 1 b + .. . + (ax )b n − 1 + (ax )0 b n 1 n 0 n − 1 Notes 1. Indices add to n. 2. There are n + 1 terms in the expansion. n 3. The (r + 1)th term is (ax ) n − r br . r Polynomials
• If P(x) = an x n + an − 1 x n − 1 + . . . + a2 x2 + a1 x + a0 and n is a nonnegative integer then P(x) is a polynomial of degree n and an, an − 1, . . . , a2, a1 are called coefficients and ∈ R. • Remainder theorem: If P(x) is divided by (x − a), then the remainder is P(a). If P(x) is divided by (ax + b) then the remainder is −b P . a • Factor theorem: 1. If P(a) = 0, then (x − a) is a factor of P(x) or if (ax + b) is a factor of P(x), then −b P = 0 a 2. If (x − a) is a factor of P(x) then a must be a factor of the term independent of x. Linear graphs
• Linear equations are polynomials of degree 1. • General equation is ax + by + c = 0 or y = mx + c where m = gradient c = yintercept y − y1 m= 2 • The gradient x2 − x1 • Equation if a point and the gradient is known: y − y1 = m(x − x1)
Chapter 1 Graphs and polynomials
47
• Parallel lines have the same gradient. • If m1 and m2 are the gradients of perpendicular lines, then: m1 × m2 = −1 −1 or m1 = m2 Quadratic graphs
• Quadratic equations are polynomials of degree 2. • General equation is y = ax2 + bx + c x=
• Quadratic formula is
−b ±
b 2 − 4 ac 2a
• Discriminant = b2 − 4ac and 1. if b2 − 4ac > 0, there are 2 xintercepts (and if b2 − 4ac is a perfect square, the intercepts are rational) 2. if b2 − 4ac = 0, there is 1 xintercept 3. if b2 − 4ac < 0, there are no xintercepts. • The power form or turning point form of the quadratic is: y = a(x − b)2 + c and the turning point is (b, c). −b • The equation of the axis of symmetry and the xvalue of the turning point of a parabola is . 2a • The axis of symmetry is halfway between the xintercepts. Cubic graphs
• Cubic equations are polynomials of degree 3. • General equation is y = ax3 + bx2 + cx + d • Basic shapes of cubic graphs: 1. If a > 0: Basic form Positive cubic y
y
y = a(x − b)3 + c
(b, c)
x
x
Factor form y
Repeated factor y
y = a(x − b)(x − c)(x − d) where a > 0
b
c d
x
a
b
x
y = (x − a)2 (x − b)
2. If a < 0, then the reflections through the xaxis of the types of graph in the above figures are obtained. Quartic graphs
• Quartic equations are polynomials of degree 4. • General equation is y = ax4 + bx3 + cx2 + dx + e 48
Maths Quest 12 Mathematical Methods CAS for the TINspire
• Basic shapes of quartic graphs: 1. If a > 0: y
y
0
b x
0
y=
y = a(x  b)2(x  c)2
ax4 y
y
b x
0
y=
x
c
ax4
+
cx2,
0
y = a(x  b)(x  c)3
c≥0
y
b
0
x
c
y
c
y = ax2(x  b)(x  c)
x
b
c
0
d
e x
y = a(x  b)(x  c)(x  d)(x  e)
2. If a < 0, then reflection through the xaxis of the types of graph above is obtained. Note: It is possible to translate the cubic and quartic graphs shown in the cubic graphs and quartic graphs sections above. Functions
• • • •
A function is fully defined if the rule and domain are given. The domain of a function is the set of values of x for which the function is defined. The range of a function is the set of values of y for which the function is fully defined. Restricted domains can be represented by interval notation: [a, b] = {x: a ≤ x ≤ b} (a, b) = {x: a < x < b} [a, b) = {x: a ≤ x < b}
Chapter 1 Graphs and polynomials
49
chapter review Short answer
1 Expand each of the following: x 2 b − 2 x
a (2y − 3x)5
8
2 If a factor of P(x) = −7 + ax + 5x2 + 15x3 + bx4 is (x2 − 1), find the values of a and b. 3 Factorise each of the following expressions: a x3 − 12x2 + 17x + 90 b 2x4 + 7x3 − 31x2 + 36 4 Find the equation of each of the straight lines described below. a The line which passes through the points (−5, 6) and (1, −1). b The line which is perpendicular to the line with equation 2x − y + 10 = 0 and passes through the point (3, 3). 5 Sketch the graph of y = 8 − 2x − x2, by labelling the turning point and all intercepts. State its domain and range. 6 Sketch the graph of y = 3x2 + 8x − 3, x ∈ [−3, 0). State the range of this function. 7 a If (x + 3) is a factor of f (x) = −x3 + bx2 + ax − 18 and g(x) = ax2 + bx − 75, then find the values of a and b. b Sketch the graph of f (x) by labelling all intercepts. 8 Sketch the graph of f (x) = x4 − 7x3 + 12x2 + 4x − 16. Multiple choice
1 When expanded, (1 − 2x)5 is equal to: A 1 + 2x − 4x2 − 8x3 + 16x4 + 32x5 B 1 − 2x + 4x2 − 8x3 + 16x4 − 32x5 C 5 − 10x + 20x2 − 40x3 + 80x4 − 160x5 D −1 + 2x − 4x2 + 8x3 − 16x4 + 32x5 E 1 − 10x + 40x2 − 80x3 + 80x4 − 32x5 2 The coefficient of x5 in the expansion of 8 1 is: 4 x − x 2 A 4096 D −16 384
50
B −131 072 E 16 384
C −4096
3 Assuming descending powers of x, the fifth term of 10 1 the expansion of 3 x + is: x 81 2 A 153 090x B 243x4 C x2 81 E 5 D 729x2 x 4 Which of the following expressions is not a polynomial? A x3 + 3x − 1 B x4 − 5x3 + 3x2 − 6x C x 21 − x11 + x − 3 3
D x 4 + 5 x 3 − 2 x 2 + 5 x − 3 E x6 − x5 + 2x4 − x3 + 4x − 2 5 The value of P(−3) in the polynomial, P(x) = x5 − 4x3 − 3x2 + 10x + 1, is: B −139 C −191 A −31 D 6 E 1 6 The degree of the polynomial (5 − 6x + x3 + 7x6) (x2 − 3x4 + 2) when expanded is: A 24 B 8 C 10 D 16 E 21 7 The remainder when x5 + 2x4 + 4x3 − 5x + 3 is divided by (x + 3) is: B 51 C −171 A −271 D 3 E 108 8 For which one of the following polynomial expressions is (x − 2) not a factor? A x3 + 3x2 − 4x − 12 B x4 − 2x3 − 6x2 − 8x + 2 C x4 + 2x3 − 7x2 − 8x + 12 D x3 + x2 − 10x + 8 E 2x3 + 3x2 − 9x − 10 9 Which one of the following is a factor of 2x4 − 4x3 − 10x2 + 12x? A (x − 2) B (x + 3) C (x + 1) D (x − 4) E (x − 3) 10 The rule for the graph shown is: A 2x + y + 4 = 0 y B x − 2y − 4 = 0 2 C 2y − x − 4 = 0 D x + 2y − 4 = 0 0 E 4x + 2y = 0
Maths Quest 12 Mathematical Methods CAS for the TINspire
4
x
Questions 11 and 12 refer to the graph below, which has a gradient of 2. 11 The value of b must be: A 5 B 3 C 1 D −1 E 4
D
y (2, b) 0
2
x
−1 0 2 −2
−6
(−3, −5)
E
12 The yintercept is: A (0, 3) B (0, 2) 1 D (0, 1) C (0, 2 ) −1 E ( 2 , 0)
y
D − 2 3 E 76 Questions 14 and 15 refer to the function with the rule: y = 2x2 + 8x  10 where x ∈(−6, 2).
y
0
B
15 The range of this function is: A (−18, 14) B (−10, 14) − D [ 18, 14] E (−14, 10)
y
C
(−2, 14)
0
−1
2
x
0
x
D
y
y
x x
0
0
1
x
(2, 14)
0 12 −10
y
1
(2, 14)
E −6−5
C [−18, 14)
16 The graph of y = −3x3 could be: B A y y
−10
(−6, 14)
x
−10
C −6−5
(2, 10)
0 12
−6−5
14 Which one of the following graphs could represent this function? (−6, 14)
x (2, −3.6)
(−6, 6)
13 If 3x2 + 4x − 5 = 0, then the value of the discriminant is: A 76 B −44 C − 44
A
y
(−6, 22)
y
x
−1
0
1
x
(6, 14)
17 Which of the following intercepts does the graph of −2 −1 0 −10
56
x
f (x) = −6 + 11x + 3x2 − 2x3 have? A ( 12 , 0), (−2, 0), (3, 0) and (0, −6) B (−2, 0), (2, 0), (3, 0) and (0, −6) C (− 12 , 0), (−2, 0), (3, 0) and (0, 6)
D (−2, 0), (−1, 0), (3, 0) and (0, −6) E ( 12 , 0), (−3, 0), (2, 0) and (0, −6)
Chapter 1 Graphs and polynomials
51
18 The rule for the graph shown below could be: A f (x) = (x − 1)2(x + 3) y f(x) B f (x) = (x + 1)(x − 3)2 C f (x) = (x + 1)2(3 − x) x D f (x) = (x2 − 1)(x + 3) 3 −1 0 E f (x) = (x − 3)(x + 1)2
C
y
–3
−3
0
D 19 The rule for the graph shown below could be: A f (x) = x(x + 2)3 y f(x) B f (x) = −x(x − 2)2 C f (x) = x2(x − 2)2 D f (x) = x(x − 2)3 E f (x) = x(2 − x)2 0
2
−3
B
0
3
1
3
x
y
−3
0
x
E
0 1
3
x
y
x
21 The graph with equation y = x2 is translated 3 units down and 2 units to the right. The resulting graph has the equation: A y = (x  3)2 + 2 B y = (x  2)2 + 3 C y = (x  2)2  3 D y = (x + 2)2  3 E y = (x + 2)2 + 3 [© VCAA 2006]
Extended response
1 An empty parfait glass has been left on a table with the rim just touching a wall. Ants are marching in a line down into the parfait glass and then up the other side, following the path of a parabola. They begin their journey where the glass touches the wall, 18 cm above the table. a The stem of the glass is 4 cm long and the diameter of the top of the glass is 5 cm. Find the rule for the quadratic function that describes the shape of the glass. b State the domain and range of the function. c If there is fruit juice in the bottom of the glass to a depth of 1 cm, find the coordinates of the point where the ants first touch the juice. Round answers to the nearest whole number. d Using function notation, write the rule for the surface of the crosssection of the juice in the glass.
52
3
0
−3
1
1
x
x
y
−3
3
y
20 The graph of y = (x + 3)2(x − 1)(x − 3) is best represented by: A
1
Maths Quest 12 Mathematical Methods CAS for the TINspire
2 A ‘rogue satellite’ has its distance from Earth, d thousand kilometres, modelled by a cubic function of time, t days after launch. After 1 day it reaches a maximum distance from Earth of 4000 kilometres, then after 2 days it is 2000 kilometres away. It effectively returns to Earth after 3 days, then moves further and further away. a What is the satellite’s initial distance from Earth? b Sketch the graph of d versus t for the first 6 days of travel. c Express d as a function of t. d The moon is approximately 240 000 kilometres from Earth. Which is closer to Earth after 8 days, the satellite or the moon? By how far? e The satellite is programmed to selfdestruct. This happens when it is 490 000 kilometres from Earth. What is the ‘life span’ of the satellite? f State the domain and range of d(t). 3 A bridge spans a narrow canal as shown in the diagram at right. a Find the equation of a parabola that models the shape of the archway. b Show that a barge 3 m wide and carrying cargo with a total height of 1.7 m (with a rectangular crosssection) cannot fit under the arch. c How much cargo (in height, correct to 1 decimal place) must be removed for the barge to fit under the bridge?
y 3 2 1 Barge
x 4 In the town of Newtonia there is an annual 10 m race (for the Polynomial Cup) −2 −1 0 1 2 for mini robots that have been programmed with mathematical formulas by Surface of canal Professor Liebnitz. There is a lot of betting on the race as the professor keeps the formulas secret and is known to favour surprise winners. The three contestants were programmed as follows, where x is the distance from the start line in metres and t is the time in minutes: Liney x = 2.4 + 0.75t Quadder x = 0.2t(t  5.1) Cubric x = 0.2t(t  5.1)(t  9.1) Using a CAS calculator, describe the motions of the three contestants, specifically: a the direction they travelled in and how fast they were moving b where and when they changed direction c where and when they passed or met each other d who won the race and by how much. Sketch the graphs of their movements on the same set of axes, labelling all relevant points. You will need an extra graph to get a closeup of the finish.
5 The diagram at right shows a main road passing through O, A, C and E. The road crosses a river at point O and 3 kilometres further along the road at point C. Between O and C, the furthest the river is from the road is 8.54 kilometres, at a point D, 2.25 kilometres east of a north–south line through O. Point A is 1 kilometre east of point O. If point O is taken as the N origin and the road as the xaxis, then the path of the river can be modelled River W E by a quartic function, as shown in blue. S a Give the coordinates of C and D. O A C Main road b Find the rule for the quartic function, f (x). E c How far is the river from the main road along the track AB? d A canoeing race, of at least 17 kilometres in length, along the river is B being organised. It is suggested that the race could start at O and finish D at C. Is this course satisfactory? Why?
Chapter 1 Graphs and polynomials
53
6 Willie Wonkie, of Willie Wonkie’s Construction Company, makes a sketch of the symmetrical W for a large neon sign as shown below. The x and yaxes represent the supporting crosspieces. The width of the W along the xaxis is 6 metres and the point on the vertical support is 2 1 metres above the horizontal support. The W 4 can be modelled by a quartic function, with all xintercepts exactly evenly spaced. y
x
a b c d
Find the rule for the letter W. If the top of the W is 8 metres wide, find the coordinates of the highest points of the letter. State the domain of the function. Use a graphics calculator to find the coordinates of the lowest points of the W, giving values correct to 3 decimal places. Hence find the range of the function. e In order to test the strength of his design, Willie Wonkie moves the horizontal crosspiece so that it just touches the lowest points of the W. Find the new rule that describes the W now. f State the domain and range of the new function. Note: The following questions use differentiation of polynomials. 7 A plane cruising at 10 000 m is coming in to land at an airport at sea level, as can be seen in the diagram below. y Plane’s
10 km
flight path 50 km
10 000 m x
Airport
If the plane descends smoothly and makes no changes in direction, show that a possible model would be y = ax2(x  b). a Find the equation if the plane begins its descent when 50 km horizontally from the airport. b What is the altitude of the plane when it is 2 km horizontally from the airport? c How accurate do you think this model is?
54
Maths Quest 12 Mathematical Methods CAS for the TINspire
8 The diagram below shows a symmetrical skateboard ramp with horizontal platforms at A and B, and vertical supporting struts at C and D. y A
B E
F
C
G 2m
4m
x 3m
D 2m 4m
a Write an equation for a quartic function that models the ramp, assuming a smooth connection at A and B. b Show that the right half of the ramp can be modelled by a cubic equation y = a(x − b) (x − 4)2 and find its equation by evaluating a and b. c The righthand side can also be modelled by two smoothly connected parabolas. i If the strut DF is 1 m long, fi nd the equation of the lower parabola passing through F. ii Find the equation of the upper parabola if it meets the lower one at F, and show that the connection is not smooth (that is, their gradients are not equal at the point where they meet). iii Show that the two parabolas meet smoothly at (3, −0.75) provided the lower parabola passes through F. eBook plus d Which model is the closest to the actual ramp if the strut is really 1.6 m long? Digital doc
Test Yourself Chapter 1
Chapter 1
Graphs and polynomials
55
eBook plus
ACTIvITIeS
Chapter opener Digital doc
• 10 Quick Questions: Warm up with ten quick questions on graphs and polynomials. (page 1) 1A
The binomial theorem
Tutorial
• We 3 int0516: Watch a worked example on binomial expansion. (page 4) Digital doc
• SkillSHEET 1.1: Practise binomial expansions using Pascal’s triangle. (page 6) 1B
Polynomials
Digital docs
• Spreadsheet 043: Investigate evaluating polynomials. (page 10) • SkillSHEET 1.2: Practise solving simultaneous equations. (page 10) 1C
Division of polynomials
Interactivity
• Division of polynomials int0246: Consolidate your understanding of the division of polynomials and rational functions. (page 11) Tutorial
• We 11 int0517: Watch a worked example on the division of polynomials. (page 11) Digital docs
• Spreadsheet 096: Investigate finding factors of polynomials. (page 15) • WorkSHEET 1.1: Binomial expansion, division of polynomials and solving and factorising polynomial equations (page 15) 1D
Linear graphs
Digital docs
• SkillSHEET 1.3: Practise calculating the gradient of parallel and perpendicular lines. (page 20) • SkillSHEET 1.4: Practise using the gradient to find the value of a parameter. (page 20) • SkillSHEET 1.5: Practise using interval notation. (page 21) • SkillSHEET 1.6: Practise finding the domain and range for linear graphs. (page 21) 1E
Quadratic graphs
Tutorial
• We 17 int0518: Watch a worked example on using the discriminant. (page 22) Digital docs
• Spreadsheet 103: Investigate the value of the disciminant. (page 28) • Spreadsheet 107: Investigate quadratic graphs. (page 28) 56
• SkillSHEET 1.7: Practise recognising domain and range for quadratic graphs. (page 28) • Spreadsheet 041: Investigate graphs of functions. (page 28) • Spreadsheet 108: Investigate quadratic graphs in turning point form. (page 28) • WorkSHEET 1.2: Calculate gradients, axial intercepts and values of the discriminant, sketch graphs of polynomials, and determine equations for graphs. (page 29) 1F
Cubic graphs
Tutorial
• We 24 int0519: Watch a worked example on determining the rule of a cubic. (page 31) Digital docs
• Spreadsheet 013: Investigate cubic graphs in factor form. (page 34) • Spreadsheet 014: Investigate cubic graphs. (page 35) • Spreadsheet 015: Investigate cubic graphs of the form y = a(x − b)3 + c. (page 36) 1G
Quartic graphs
Tutorials
• We 27 int0520: Watch a worked example on sketching the graph of a quartic. (page 39) • We 28 int0521: Watch a worked example on finding the turning points of a quartic using a CAS calculator. (page 40) Digital docs
• SkillSHEET 1.8: Practise solving quartic equations. (page 45) • Spreadsheet 105: Investigate quartic graphs in factor form. (page 45) • Investigation: Quartics and beyond (page 46) Chapter review Digital doc
• Test Yourself Chapter 1: Take the endofchapter test to test your progress. (page 55) To access eBookPLUS activities, log on to www.jacplus.com.au
Maths Quest 12 Mathematical Methods CAS for the TINspire
2 Functions and transformations AREAS OF STUDY
• The behaviour of functions of a single real variable, including key features of their graphs such as axis intercepts, stationary points and points of inflection, domain (including maximal domain) and range, asymptotic behaviour, symmetry and the algebra of functions, including composition of functions. The behaviour of these functions is linked to applications in practical situations. • Identification of key features of graphs of: – power functions y = xn for n ∈ N and n = −1, −2, 1 and transformations of these to the 2 form y = a(x + b)n + c where a, b and c ∈ R – the modulus function y =  x  • Transformations from y = f (x) to y = Af (n(x + b)) + c (where A, n, b and c ∈ R and f is one of the functions specified above) and the relation between the original function and the graph of the transformed
• •
•
• •
2A 2B 2C 2D 2E 2F 2G 2H 2I
Transformations and the parabola The cubic function in power form The power function (the hyperbola) The power function (the truncus) The square root function in power form The absolute value function Transformations with matrices Sum, difference and product functions Composite functions and functional equations 2J Modelling
function (including families of transformed functions for a single transformation parameter) Graphs of sum, difference, product and composite functions of f and g, where f and g are functions of the types above Recognition of the general form of possible models for data presented in graphical or tabular form, using polynomial and power functions Applications of simple combinations of the above functions, and interpretation of features of the graphs of these functions in modelling practical situations The relationship of f (x ± y), f (xy) and f to values of f (x) and f (y) for different functions f Composition of functions, where f composition g is defined by f (g(x)), given rang ⊆ domf ; for 4 example x 2 + 5 , x 3 + 2 x −1 eBook plus
2A
Transformations and the parabola
Digital doc
10 Quick Questions
Transformations In this chapter we consider the basic graphs of the quadratic and cubic functions, the hyperbola and truncus, square root and absolute value functions. The following transformations of the above graphs are discussed: dilation, reflection and translation.
Dilation A dilation is the stretching or compressing of a graph. Let the basic graph be y = f (x).
Chapter 2
Functions and transformations
57
Dilation away from the xaxis: y = af (x (x) 1. Stretches or compresses the graph f (x) by a factor of a from the xaxis. 2. Each yvalue of the basic graph is multiplied by a factor of a, that is (x, y) → (x, ay). 3. When  a  > 1, the graph of f (x) is stretched and becomes narrower. 4. When 0 <  a  < 1, the graph of f (x) is compressed and becomes wider. Dilation away from the yaxis: y = f (nx) 1. Stretches or compresses the graph f (x) by a factor of 1 from the yaxis. n x 2. Each xvalue of the basic graph is multiplied by a factor of 1 , that is (x, y) → ( , y). n n 3. When n > 1, the graph of f (x) is compressed from the yaxis and becomes narrower. 4. When 0 < n < 1, the graph of f (x) is stretched from the yaxis and becomes wider. Note: For the graphs we will be looking at in this chapter a horizontal dilation can be expressed as a vertical dilation. For example, (2 (2x + 1)3 can be written as 23 ( x + 12 )3 = 8( 8( x + 12 )3. So in this 1 case a horizontal dilation from the yaxis by a factor of 2 is the same as a vertical dilation from the xaxis by a factor of 8. This can simplify the process of describing transformations for these particular graphs. The concept of dilation is illustrated in the following diagram: y
y
x
x
Original graph dilated from the xaxis
Original graph dilated from the yaxis
x
Original graph
y
Ref lection Reflection provides a ‘mirror image’ of a graph. Reflection can take place in one or both axes. Let the basic graph again be y = f (x). Ref lection in the xaxis: y = −f (x (x) 1. The mirror image of the original graph appears across the xaxis (the mirror line). 2. Each yvalue is the negative of the original, the xvalue is unchanged, that is (x, y) → (x, −y). Reflection in the yaxis: y = f (−x) 1. The mirror image of the original graph appears across the yaxis (the mirror line). 2. Each xvalue is the negative of the original, the yvalue is unchanged, that is (x, y) → (−x, y). Reflection in both axes: y = −f (−x) 1. The basic graph is reflected in the xaxis and then the yaxis (or vice versa). 2. Both the x and yvalues are the negatives of the original, that is (x, y) → (−x, −y). The concept of reflection is shown in the diagram below. The red star is the original graph. y
y
x
y
x
x
Reflection in the yaxis
Reflection in the xaxis
58
Maths Quest 12 Mathematical Methods CAS for the TINspire
Reflection in both axes
Translation: y = f (x − b) + c A translation slides the graph. Translation can be horizontal (to the right or left along the xaxis), or vertical (up or down along the yaxis). Consider our basic graph y = f (x). 1. If y = f (x – b) the basic graph is translated b units parallel to the xaxis: (a) in the positive direction (i.e. to the right) when b > 0 (b) in the negative direction (i.e. to the left) when b < 0. Each xvalue has b added to it, that is (x, y) → (x + b, y). 2. If y = f (x) + c, the basic graph is translated c units parallel to the yaxis: (a) in the positive direction (i.e. up) when c > 0, (b) in the negative direction (i.e. down) when c < 0. Each yvalue has c added to it, that is (x, y) → (x, y + c). 3. If y = f (x – b) + c the basic graph is translated both horizontally and vertically. y
y
y
x x Original graph
x Vertical translation up y
Vertical translation down
y
x
x
Horizontal translation to the left
Horizontal translation to the right
Naturally, the graph can be subject to a combination of two or more transformations. eBoo k plus eBook Digital doc
Spreadsheet 132 Transformations
Combination of transformations When describing transformations that have been applied to a basic graph f (x), it is best to put the graph into the format y = af (x − b) + c. The order of transformations is important as dilations and reflections are applied before translations, so ensure that you describe the transformations in this order (remember DRT). In this chapter we shall consider graphs, derived from basic curves, using single transformations — dilations, reflections or translations as well as combinations of those. Modelling of data will also be considered.
The quadratic function in power form The graph of y = x2 is a parabola with the turning point at the origin. The domain of the function is R and the range is R+ ∪ {0}. Throughout this section we refer to the graph of y = x2 as the basic parabola. Let us now consider the effect of various transformations on the graph of this basic parabola.
y
0
Chapter 2
x
Functions and transformations
59
Quadratic functions are also power functions. Power functions are functions of the form f (x) = xn, n ∈ R. The value of the power, n, determines the type of function. When n = 1, f (x) = x, and the function is linear. When n = 2, f (x) = x2 and the a=2 function is quadratic. Other power functions will be discussed later. y a=1 1 a = –2 Under a sequence of transformations of f (x) = xn, n ∈ R, the general form of a power function, is f (x) = a(x − b)n + c (where a, b, c and n ∈ R). All linear and quadratic polynomials are also linear and x quadratic power functions, because all linear and quadratic functions 0 are transformations of f (x) = x and f (x) = x2, respectively. When a quadratic function is written in turning point form it is written in power form. For example, the quadratic function y = ax2 y = x2 + 4x + 6 can also be represented as the power function y = (x + 2)2 + 2.
Dilation In power form, a is the dilation factor. It dilates the graph in the y direction. The larger  a  is, the thinner the graph of the parabola. If  a  is a proper fraction, that is, 0 <  a  < 1, the graph is wider than the basic parabola.
Reflection If a is negative, the graph of the basic parabola is reflected in the xaxis, that is, the graph is ‘flipped’ upside down. If x is replaced with −x, the graph of the basic parabola is reflected in the yaxis, that is, the graph is ‘flipped’ sideways. Due to its symmetry, this effect cannot be seen on the basic parabola, but it is more obvious with a parabola that has already been translated. For example, the graphs of y = (x − 3)2 and y = (−x − 3)2 are reflections of each other across the yaxis.
y = (−x − 3)2 y
y = (x − 3)2 (0, 9)
x
(−3, 0)0 (3, 0)
Translation Horizontal translation If b > 0, the graph of the basic parabola is translated horizontally to the right, and if b < 0, the graph of the basic parabola is translated horizontally to the left. For example, a graph with the equation y = (x − 2)2 is a basic parabola that has been translated 2 units to the right, and a graph with the equation y = (x + 3)2 is a basic parabola that has been translated 3 units to the left. If the coefficient of x is not 1, the equation must be rewritten in the form y = a(x − b)2 + c in order to be able to work out the value of b. For example, y = (4x + 3)2 is translated 43 of a unit to the left, since
y b=
−3
0 −3 2 y = (x − b)
b=2
2
x
y = (4 x + 3)2 3 = [4( x + 4 )]2 3
= 16( x + 4 )2 Vertical translation If c > 0, the graph is translated vertically upward, and if c < 0, the graph is translated vertically downward. For example, the graph with equation y = x2 + 2 is a basic parabola that has been translated 2 units up, and the graph with equation y = x2 − 1 is a basic parabola that has been translated 1 unit down.
60
Maths Quest 12 Mathematical Methods CAS for the TINspire
y
c=2 c = −1
2 x 0 −1 y = x2 + c
eBoo k plus eBook Digital doc
Spreadsheet 108 The quadratic function in power form
Combination of transformations
y y = a(x − b)2 + c
The graph of y = a(x − + c shows the combination of the transformations shown above. The turning point of the graph is (b, c). The domain of the parabola is R and the range is [c, ∞) if a > 0 or [−∞, c) if a < 0. b)2
(b, c) 0
x
WORKED EXAMPLE 1
State the changes required to transform the graph of y = x2 into the graph of y = 2(x 2( + 3)2 − 4. THINK
WRITE
1
Write the general formula for the parabola.
y = a(x − b)2 + c
2
Identify the value of a.
a=2
3
State the effect of a on the graph.
The graph of y = x2 is dilated by the factor of 2 from the xaxis.
4
Identify the value of b.
b = −3
5
State the effect of b on the graph.
The graph is translated 3 units to the left.
6
Identify the value of c.
c = −4
7
State the effect of c on the graph.
The graph is translated 4 units down.
We can use transformations to find the equation of the function from its graph by first examining the new position of the turning point. WORKED EXAMPLE 2
Use transformations to find the equation of this function.
y
(4, 2)
x
0 THINK
WRITE
1
Write the general formula of the parabola.
y = a(x − b)2 + c
2
From the graph state the horizontal translation and hence the value of b.
Translated 4 units to the right, so b = 4.
3
From the graph, state the vertical translation and hence the value of c.
Translated 2 units up, so c = 2.
4
Substitute the values of b and c into the general formula. y = a(x − 4)2 + 2
5
The graph of the parabola passes through the origin. Substitute x = 0 and y = 0 into the formula.
Using (0, 0): 0 = a(0 − 4)2 + 2
6
Solve for a, which is the dilation factor.
0 = 16a + 2 16a = −2 −2 a = 16 =
−1 8
Chapter 2
Functions and transformations
61
7
Substitute the value of a into y = a(x − 4)2 + 2 and write your answer.
The equation of the parabola shown is: y = − 81 ( x − 4)2 + 2
WORKED EXAMPLE 3
Given the equation y = kx2, determine the effect on the graph y = x2, when k = {2, 3, 4}. Sketch the graphs. THINK
WRITE/DISPLAY
1
On a Graphs page, complete the function entry line as: f 1(x) = x2 Then press ENTER · .
2
Complete the function entry lines as: f 2(x) = 2x2 f 3(x) = 3x2 f 4(x) = 4x2 Press ENTER · after each entry.
3
Answer the question by describing the changes in words.
As the value of k increases the graph becomes thinner and stretches away from the xaxis.
REMEMBER
1. The graph of y = x2 is called a basic parabola. 2. The graph of y = a(x − b)2 + c is the basic parabola, dilated by the factor of a from the xaxis, translated b units horizontally (to the right if b > 0 or to the left if b < 0) and c units vertically (up if c > 0 or down if c < 0). 3. If a is negative, the graph is reflected in the xaxis. 4. If x is replaced with −x, the graph is reflected in the yaxis. 5. The turning point of the parabola is (b, c). 6. The domain of the parabola is R. 7. The range is y ≥ c if a > 0 or y ≤ c if a < 0.
62
Maths Quest 12 Mathematical Methods CAS for the TINspire
y y = a(x − b)2 + c
(b, c) 0
x
EXERCISE
2A
Transformations and the parabola 1 WE 1 State the changes required to transform the graph of y = x2 into the graph of each of the following. a y = 2x b y = 13 x 2 2 c e g i k
y = −3x2 1 y = 1 − 2 x2 y = −(x + 3)2 y = (x + 2)2 − 1 y = 1 − 2(3 + x)2
y = x2 − 6
d f h j l
y = (x − 2)2 y = 2(3 − x)2 y = (x − 0.5)2 + 2 1
y = 3(2 x − 3)2 − 4
−1 2 MC The equation of a parabola is given by (2 − x )2 + 3 . Increasing m will result in the m graph being: A translated further to the left B translated further up C thinner D wider E reflected in the yaxis
3 Match the graphs of the parabolas with the following equations. a y = x2 + 2 b y = −2(xx − 2)2 c
y = 2 − (x + 2)2
d
e y = (2 + x)2 + 2
iii
ii y = x2
y
1
y = 2 (2 − x ) 2
iv
2
i
x
0 2 −2 −2 v
4 WE2 Use transformations to find the equation of each function. a
y
b
(2, 2)
y 0
x
0
x
(−1, −2) 2)
c
y
(1, 3)
0
2 x
d
y −4
0
x
(−2, −4)
5 MC The equation of the graph shown opposite is best given by: A y = (x − c)2 + d B y = c − (x − b)2 − 2 C y = (x + c) + b D y = (c − x)2 + d 2 E y = d − (x − c)
y d b a0
c
e
x
6 Find the equation of the image of y = x2 under each of the following transformations: a dilation by the factor of 12 from the xaxis b a reflection in the xaxis c a translation by 2 units to the right and 1 unit down
Chapter 2
Functions and transformations
63
d a dilation by the factor of 3 from the xaxis, followed by translation of 2 units down e a reflection in the xaxis, followed by translation of 3 units to the left. 7 Find the equations of these graphs. a
b
y
y 1
0
3
5
x
x
0 −1
−1
−4
c
d
y −3
y 2
x −1 0
x
0 2
−4
e
f
y
y 8
9 6 0
1
x
− 2 −2
−2
0 x
8 WE3 Find the equation of y = x2 under the following sequential transformations (in order): a dilation by a factor of 2 from the xaxis b reflection in the xaxis c translation of −1 parallel to the xaxis d translation of 3 parallel to the yaxis. 9 Find the image of the point ((x, y) under each of the following transformations: a reflection in the yaxis b reflection in the xaxis c dilation by a factor of 3 from the xaxis d dilation by a factor of 2 from the yaxis e dilation by a factor of 13 from the yaxis f translation of 2 units horizontally in the positive direction g translation of −1 unit parallel to the yaxis. 10
The parabola has a turning point at (z, ( −8); it intersects the yaxis at y = 10 and one of the xintercepts is x = 5. Find: a the value of z b the equation of the parabola.
11 For the parabola whose range is y ≤ 3, whose xcoordinate of the turning point is −4 and whose 1 yintercept is y = − 2 3 , find: a the ycoordinate of the turning point b the equation of the parabola c the coordinates of the xintercepts. 64
Maths Quest 12 Mathematical Methods CAS for the TINspire
12 The design shown in the diagram at right can be obtained by taking the red portion of the parabola and transforming it to form each of the other 9 fragments. (One or more transformations may be used to form each fragment.) If the highlighted fragment is given by f (x (x), −2 ≤ x ≤ 2, define the other 9 fragments in terms of f (x (x) and specify their domains.
2B
y
3 −2 0
2
x
The cubic function in power form The graph of the function y = x3 is shown at right: Both the domain and y range of the function are R. The function is constantly increasing and has a stationary point of inflection (where the gradient is 0) at the origin (0, 0). x 0 Throughout this section we shall refer to the shape of the graph of 3 y = x as a positive cubic, or a basic cubic curve. Cubic functions are also power functions. Power functions are functions of the form f (x) = xn, n ∈ R. The value of the power, n, determines the type of function. When n = 1, f (x) = x and the function is linear. When n = 2, f (x) = x2 and the function is quadratic. When n = 3, f (x) = x3 and the function is cubic. When n = 4, f (x) = x4 and the function is quartic. Other power functions will be discussed later. Under a sequence of transformations of f (x) = xn, n ∈ R, the general form of a power function is f (x) = a(x − b)n + c (where a, b, c and n ∈ R). All linear and quadratic polynomials are also linear and quadratic power functions, but this is not the case for cubic functions (or quartic functions). For example, a cubic power function in the form f (x) = a(x − b)3 + c has exactly one xintercept and one stationary point of inflection. A cubic polynomial in the form f (x) = ax3 + bx2 + cx + d can have one, two or three xintercepts and is therefore not a power function. All cubic power functions are also cubic polynomials, but not all cubic polynomials are cubic power functions. For example, the cubic function y = 2(x − 3)3 + 1 is a polynomial and a power function. It is the graph of y = x3 under a sequence of transformations.
Dilation
a=2 a=1 a = 1–2
y
The value a is the dilation factor; it dilates the graph from the xaxis. The larger a is, the thinner the graph. 0
x
y = ax3 eBoo k plus eBook Digital doc
Spreadsheet 015 Cubic function — y = a( x − b ) 3 + c
y = (−x − 1)3 y
Reflection If a is negative, the graph of the basic cubic is reflected in the xaxis, that is, the graph is ‘flipped’ upside down. If x is replaced with −x, the graph of the basic cubic is reflected in the yaxis, that is, the graph is ‘flipped’ sideways. For example, the graphs y = (x − 1)3 and y = (−x − 1)3 are reflections of each other across the yaxis.
Chapter 2
(−1, 0) 0
y = (x − 1)3
(1, 0)
x
(0, −1)
Functions and transformations
65
Translation Horizontal translation If b > 0, the graph of the basic cubic is translated horizontally to the right, and if b < 0, the graph of the basic cubic is translated horizontally to the left. For example, the graph with equation y = (x − 2)3 is a basic cubic translated 2 units to the right, and the graph of y = (x + 3)3 is a basic cubic, translated 3 units to the left, that is, parallel to the xaxis in the negative direction. If the coefficient of x is not 1, the equation must be rewritten in the form y = a(x − b)3 + c in order to be able to work out the value of b. For example, the graph of y = (2x (2 − 5)3 is translated 25 units to the 3 right, since y = (2x (2 − 5)
y
b = −3
b=2
0
−3
2
x
y = (x − b)3
= [2( x − 25 )]3 = 8( x − 25 )3 Vertical translation The value of c translates the graph vertically or along the yaxis. If c > 0, the graph is translated vertically up, and if c < 0, the graph is translated vertically down. The coordinates of the stationary point of inflection are (b, c). For example, if y = x3 is translated 1 unit up, the equation of the resulting graph is y = x3 + 1 and the point of inflection is (0, 1); if it is translated 2 units down, the equation of the resulting graph is y = x3 − 2 and the point of inflection is (0, −2).
Combination of transformations The graph of y = a(x − b)3 + c shows the combination of the transformations described above. Finally, the domain and range of y = a(x − b)3 + c are R (all real numbers).
y 1
c=1 c = −2 x
0 −22
y = x3 + c
y
y = a(x − b)3 + c
(b, c) 0
WORKED EXAMPLE 4
State the changes necessary to transform the graph of y = x3 into the graph of y = 2(x 2( + 1)3 − 4. THINK
66
WRITE
1
Write the general equation of the cubic function.
y = a(x − b)3 + c
2
Identify the value of a.
a=2
3
State the effect of a on the graph.
The graph is dilated by the factor of 2 in the y direction.
4
Identify the value of b.
b = −1
5
State the effect of b on the graph.
The graph is translated 1 unit to the left.
6
Identify the value of c.
c = −4
7
State the effect of c on the graph.
The graph is translated 4 units down.
Maths Quest 12 Mathematical Methods CAS for the TINspire
x
WORKED EXAMPLE 5
For each of the following graphs: i state the coordinates of the stationary point of inflection ii find the x and yintercepts iii sketch the graph iv state the transformations that the graph of y = x3 has undergone to form each new equation. a y = −(x ( + 3)3 − 1 b y = (4 − x)3 + 6 THINK a Write the equation.
WRITE/DRAW a y = −(x + 3)3 − 1 i b = −3, c = −1
i Since the rule is of the form y = a(x − b)3 + c,
Stationary point of inflection: (−3, −1)
identify the values of b and c and hence write the coordinates of the stationary point of inflection (b, c).
ii Find the yintercept by letting x = 0.
i i yintercept: x = 0,
y = −(0 + 3)3 − 1 = −27 − 1 = −28 xintercept: y = 0 (x + 3)3 − 1 = 0 (x + 3)3 = −1 x + 3 = −1 x = −4
Find the xintercept by letting y = 0.
iii To sketch the graph on a set of labelled axes,
iii
mark the stationary point of inflection and the x and yintercepts, then sketch the positive cubic passing through the points marked.
(−4, 0) 0 (−3, −1)
iv State the kind of reflection and the vertical and
Write the equation. i Since the rule is of the form y = a(x − b)3 + c,
identify the values of b and c and hence write the coordinates of the stationary point of inflection (b, c).
ii Find the yintercept by letting x = 0.
Find the xintercept by letting y = 0. Note: Do not round off until the very last step; for graphing purposes, round off your final answer to 1 decimal place.
x
i v The graph is reflected in the xaxis.
horizontal translations.
b
y
There is a horizontal translation of 3 units to the left and a vertical translation of 1 unit down. b
y = (4 − x)3 + 6 i b = 4, c = 6
Stationary point of inflection: (4, 6)
i i yintercept: x = 0,
y = (4 − 0)3 + 6 = 64 + 6 = 70 xintercept: y = 0 (4 − x)3 + 6 = 0 (4 − x)3 = −6 4−x=
−
3
6 3
x=4+ 6 ≈ 5.8
Chapter 2
Functions and transformations
67
iii To sketch the graph on a set of labelled
iii
y
axes, mark the stationary point of inflection and the x and yintercepts, then sketch the positive cubic passing through the points marked.
(4, 6)
0 iv State the kind of reflection and the vertical
(5.8, 0)
x
i v The graph is reflected in the yaxis.
and horizontal translations.
There is a horizontal translation of 4 units to the right and a vertical translation of 6 units up.
To find the equation of the curve from a given graph, we need to establish exactly what transformations were applied to the basic cubic curve. This is best done by observing the shape of the graph and the position of the stationary point of inflection. WORKED EXAMPLE 6
Find the equation of the curve, if it is of the form y = a(x ( − b)3 + c. (x
y 5 3
THINK
0
1
1
Write the general equation of the cubic function.
y = a(x − b)3 + c
2
Write the coordinates of the stationary point of inflection (b, c) and hence state the values of b and c. Substitute the values of b and c into the general formula.
The stationary point of inflection is (1, 3). So b = 1, c = 3.
4
The graph passes through the point (0, 5) (yintercept). Substitute the coordinates of this point into the equation.
Using (0, 5): 5 = a(0 – 1)3 + 3
5
On a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve(5 = a × (0 – 1)3 + 3,a) Then press ENTER · .
6
Write the solution for the equation.
a = −2
7
Substitute the value of a into y = a(x − 1)3 + 3.
y = −2(x − 1)3 + 3
3
68
WRITE/DISPLAY
y = a(x − 1)3 + 3
Maths Quest 12 Mathematical Methods CAS for the TINspire
x
REMEMBER
1. The graph of y = x3 is a basic cubic curve. 2. The graph of y = a(x − b)3 + c where a > 0 is the basic cubic, dilated by the factor of a from the xaxis, translated b units along the xaxis (to the right if b > 0, or to the left if b < 0) and c units along the yaxis (up if c > 0, or down if c < 0). y = a(x − b)3 + c
y
(b, c)
x
0
3. If a < 0, the graph is reflected in the xaxis. 4. The stationary point of inflection is at (b, c). 5. Both the domain and range are R. EXERCISE
2B
The cubic function in power form 1 WE4 State the changes necessary to transform the graph of y = x3 into the graph of each of the following. b y=
a y = 7x3 c e g i
y = x3 + 4 y = (x − 1)3 y = 4(2 − x)3 y = 3(x + 3)3 − 2
k y = 1 (2 x + 5)3 4 2
−2 3
x3
d f h j
y = 6 − x3 y = −(x + 3)3 y = −6(7 − x)3 1 y = 6 − 2 ( x − 1)3
l
y = 3 − 2(4 + 2 x )3
1
Which of these transformations were applied to the graph of y = x3 to obtain each of the graphs below? i reflection in the xaxis i i translation to the left i i i translation to the right i v translation up v translation down a
b
y
c
y x
0 x
0
d
y
0
e
y
0
x
f
y
0
y
0
x
Chapter 2
x
x
Functions and transformations
69
3 WE5 For each of the following graphs: i find the stationary point of inflection ii find the x and yintercepts iii state the transformation(s) that the graph y = x3 has undergone to produce the given graph iv sketch the graph.
EXAM TIP Be careful when sketching graphs — use appropriate scales on the axes, clearly draw the graph and use the correct domain. Show at least one scale point on each axis and two coordinate points on each curve (including intercepts). [Authors’ advice]
eBoo k plus eBook
a y = 3 x3 4
b y = 1 − 2x 2 3
c y = 2 x3 − 6 3
d y = 2(x − 4)3
−1
f y = 4(1 − x)3
e y=
2
( x − 2)3
g y = (x − 1)3 + 2 i y = 2(x + 1)3 − 6
Digital doc
Spreadsheet 236 Function grapher
h y = 3 − (x + 2)3
Questions 4 to 6 refer to the function y = 2(mx − 4)3 − 3. 4 MC The coordinates of the stationary point of inflection are: − A ( 4 , 3) m
4 −3 C ( , ) m m
B (4m, −3)
4 E ( , − 3) m
4 D ( , − 3m3 ) m
5 MC The graph of y = x3 is dilated in the y direction by the factor of: A 2
C 2m3
B 2m
2 m
D
2 m3
E
6 MC If m > 1, increasing m will cause the graph to become: A wider B thinner and translated not as far to the right C shifted further to the left D shifted further to the right E shifted further down 7 Find the equation of the graph resulting from each of the following transformations of the graph of y = x3: a a dilation by the factor of 12 from the xaxis b a reflection in the xaxis and a translation by 5 units to the left c a translation by 3 units to the right and 1 unit down d a dilation in the y direction by the factor of 2, followed by the vertical translation of 3 units e a reflection in the xaxis, then a translation of 1 unit to the left and 1 unit down. 8 Find the equation of the graph resulting from the following sequential transformations of the graph of y = x3: a dilation by a factor of 2 from the xaxis b reflection in the yaxis c translation of 2 in the positive direction parallel to the xaxis d translation of 1 in the negative direction parallel to the yaxis. 9 WE6 Find the equations of these curves, if they are of the form y = a(x ( − b)3 + c. (x a
y
b
y
4
y
c
1
(1, 2) −1
0
70
2
x
0
d
x
Maths Quest 12 Mathematical Methods CAS for the TINspire
−2
0
y
e
9
x 0
3
x
y − 1–2 −1 1 0 − –2
x
10 MC The graph of y = 2(x 2( + 3)3 + 1 has been reflected in the xaxis, shifted 3 units to the right and 1 unit up. The equation of the resulting graph is: A y = 2(3 − x)3 + 2 B y = −2(x + 2)3 + 2 C y = 2(2 − x)3 D y = −2(x − 3)3 + 1 E y = −2x 2 3 11 The graph of a cubic function of the form y = a(x ( − b)3 + c has a stationary point of inflection (x at (−1, −4) and cuts the yaxis at y = −2. Find the equation of the function. 12
The graph of y = a(b − x)3 + c has a stationary point of inflection at (2, 1) and passes through the point (1, 12 ). a Find the equation of the curve. b State the shape of the curve (that is, whether it is positive or negative cubic).
13 The graph of y = a(x ( − b)3 + c cuts the xaxis at x = −4 and the yaxis at y = 28. If it is known (x that the dilation factor is equal to 1: a find the position of the stationary point of inflection b sketch the graph.
2C
The power function (the hyperbola) The graph shown at right is called a hyperbola and is given by the y equation y = 1 . x Power functions are functions of the form f (x) = xn, n ∈ R. The y=0 value of the power, n, determines the type of function. We saw earlier x 0 that when n = 1, f (x) = x and the function is linear. When n = 2, f (x) = x2 and the function is quadratic. When n = 3, f (x) = x3 and the function is cubic. When n = 4, f (x) = x4 and the function is quartic. x=0 The power function that produces the graph of a hyperbola has a 1 value of n = −1. Thus, the function f ( x ) = can also be expressed as x the power function f (x) = x−1. The graph exhibits asymptotic behaviour. That is, as x becomes very large, the graph approaches the xaxis, but never touches it, and as x becomes very small (approaches 0), the graph approaches the yaxis, but never touches it. So the line x = 0 (the yaxis) is a vertical asymptote and the line y = 0 (the xaxis) is the horizontal asymptote. Both the domain and the range of the function are all real numbers, except 0; that is, R\{0}. The graph of y = 1 can be subject to a number of transformations. x a − Consider y = + c or y = a(x − b) 1 + c x−b
Dilation
y
The value a is a dilation factor. It dilates the graph from the xaxis.
y=0 0
Reflection If a is negative, the graph of the basic hyperbola is reflected in the xaxis. If x is replaced with −x, the graph of the basic hyperbola is reflected in the yaxis.
Chapter 2
x=0
a=2 a=1 a = 1–2 x y = a–x
Functions and transformations
71
y
1 1 and y = − x−3 x−3 are reflections of each other across the yaxis. For example, the graphs of y =
x = −3 y=
Translation
x=3 y = x −1 3
1
−x − 3
0
Horizontal translation x y=0 The value b translates the graph b units horizontally, that is, parallel to the xaxis. If b > 0, the graph is translated to the right, and if b < 0, the graph is (0, − 13 ) translated to the left. For example, the graph with 1 equation y = is a basic hyperbola translated x−3 3 units to the right. This graph has a vertical asymptote of x = 3 and domain R\{3} (and a horizontal asymptote y = 0). If a basic hyperbola is translated 3 units to the left, it becomes 1 y= , with a vertical asymptote of x = −3 and domain R\{−3}. Hence, the equation of the x+3 vertical asymptote is x = b and the domain is R\{b}. The horizontal asymptote and the range remain the same, x = 0 and R \ {0}, respectively. Vertical translation The value c translates the graph c units vertically, that is, parallel to the yaxis. If c > 0, the graph is translated upward, and if c < 0, the graph is translated c units downward. The graph 1 with equation y = + 3 is a basic hyperbola translated 3 units up. This graph has a horizontal x asymptote of y = 3 and a range of R\{3} (and a vertical asymptote x = 0). If a basic hyperbola is 1 translated 3 units down, it becomes y = − 3, with a horizontal asymptote of y = −3 and a range x of R\{−3} (and a vertical asymptote x = 0). Hence the equation of the horizontal asymptote is y = c and the range is R\{c}. Always draw the asymptote as a dotted line and label it with its equation (for example, y = 3) at the end of the asymptote. Ensure that the graph continues to approach the asymptote getting closer but not touching or crossing the asymptote or bouncing away from the asymptote.
Combination of transformations The graph of y =
a + c shows the combination of these transformations. x−b y
eBoo k plus eBook Digital doc
Spreadsheet 051 The hyperbola
y=c
a
y = x— c −b+
c 0
b
x
x=b
Finally, if the coefficient of x is a number other than 1, to obtain the value of h the equation should be rearranged first. For example, 4 4 y= = 3 x + 6 3( x + 2) Therefore, b = −2 (not −6 as it may seem at first); that is, the graph is translated 2 units to the left.
72
Maths Quest 12 Mathematical Methods CAS for the TINspire
WORKED EXAMPLE 7
1 State the changes that should be made to the graph of y = in order to obtain the graph of x − y = 4 − 1. x+2 THINK
WRITE
a +c x−b
1
Write the general equation of the hyperbola.
y=
2
Identify the value of a.
a = −4
3
1 State the changes to y = , caused by a. x
1 The graph of y = x is dilated by the factor of 4 from the xaxis and reflected in the xaxis.
4
Identify the value of b.
b = −2
5
State the effect of b on the graph.
The graph is translated 2 units to the left.
6
Identify the value of c.
c = −1
7
State the changes to the graph, caused by c.
The graph is translated 1 unit down.
WORKED EXAMPLE 8
For the graph of y =
2 + 2, state: x−3
a the equations of the asymptotes c the range.
b the domain
THINK a
WRITE
1
Write the general equation of the hyperbola.
2
Identify the values of b and c and hence write the equations of the asymptotes: Horizontal asymptote: y = c Vertical asymptote: x = b
a y=
a +c x−b
b = 3, c = 2 Horizontal asymptote: y = 2 Vertical asymptote: x = 3
b
State the domain of the hyperbola: R\{b}.
b Domain: R\{3}
c
State the range of the hyperbola: R\{c}.
c Range: R\{2}
Sketching the graph of the hyperbola by hand can be easily done by following these steps: 1. Find the position of the asymptotes. 2. Find the values of the intercepts with the axes. 3. Decide whether the hyperbola is positive or negative. 4. On the set of axes draw the asymptotes (using dotted lines) and mark the intercepts with the axes. 5. Treating the asymptotes as the new set of axes, sketch either the positive or negative hyperbola, making sure it passes through the intercepts that have been previously marked.
Chapter 2
Functions and transformations
73
WORKED EXAMPLE 9
eBoo k plus eBook
2 −4 , clearly showing the intercepts x+2 with the axes and the position of the asymptotes. Sketch the graph of y = THINK 1
a +c Compare the given equation with y = x − b and state the values of a, b and c. Write a short statement about the effects 1 of a, b and c on the graph of y = . x
3
Write the equations of the asymptotes. The horizontal asymptote is at y = c. The vertical asymptote is at x = b. Find the value of the yintercept by letting x = 0.
5
int0522 Worked example 9
WRITE/DRAW
2
4
Tutorial
Find the value of the xintercept by making y = 0.
b = −2,
a = 2,
c = −4
1 is dilated by the factor of x 2 from the xaxis, translated 2 units to the left and 4 units down. Asymptotes: x = −2; y = −4 The graph of y =
yintercept: x = 0 2 y= −4 0+2 = 1− 4 = −3 Point (0, −3) xintercept: y = 0 2 0= −4 x+2 2 =4 x+2 2 = 4( x + 2) = 4x + 8 4x = 2 − 8 = −6 x= = Point
6
(
−6 4 −3 2
−3 2
,0
)
To sketch the graph:
y
a Draw the set of axes and label them. b Use dotted lines to draw the asymptotes. The
(− 3–2 , 0) 0
asymptotes are x = −2 and y = −4.
(0, −3)
c Mark the intercepts with the axes. The
intercepts are y = −3 and x =
−3 . 2
d Treating the asymptotes as your new set of
x y = −4
x = −2
axes, sketch the graph of the hyperbola (as a is positive, the graph is not reflected); make sure the upper branch passes through the x and yintercepts previously marked. The next example shows how to find the equation of the hyperbola from its graph.
74
Maths Quest 12 Mathematical Methods CAS for the TINspire
WORKED EXAMPLE 10
Find the equation of the graph shown.
y 6 3 0 2
THINK
4
x
WRITE
a +c x−b
1
Write the general equation of the hyperbola.
y=
2
From the graph, identify the values of b and c Remember that the equation of the horizontal asymptote is y = c and of the vertical asymptote is x = b.
b = 2,
3
Substitute the values of b and c into the formula.
y=
4
Substitute the coordinates of any of the 2 known points of intersection with the axes into the formula (say, xintercept).
5
Solve for a.
Substitute (4, 0): a 0= +3 4−2 a 0= +3 2 a − = 3 2 a = −6
6
Substitute the value of a into y =
7
Transpose (optional).
a + 3. x−2
y=
c=3
a +3 x−2
−
6 +3 x−2
y = 3−
6 x−2
REMEMBER
1 is called a hyperbola. x a 2. The graph of y = + c is the graph of the basic x−b hyperbola, dilated by the factor of a from the xaxis, translated b units horizontally (to the right if b > 0, or to the left if b < 0) and c units vertically (up if c > 0, or down if c < 0). If a < 0, the graph is reflected in the xaxis. The equations of the asymptotes are: x = b and y = c. The domain of the function is R\{b} and its range is R\{c}. 1. The graph of y =
Chapter 2
y
y= a
x−b
+ c y=c x
0 x=b
Functions and transformations
75
EXERCISE
2C
The power function (the hyperbola) 1
2
1 State the changes that should be made to the graph of y = in order to obtain the x graph of each of the following. −3 1 2 a y= b y= c y= x x−6 x 2 1 2 d y= e y= +7 f y= −5 x+4 x x −4 1 2 g y= −3 +6 h y= i y= −4 4+x x−3 x −1 WE7
1 Which of the following transformations were applied to the graph of y = to obtain each x of the graphs shown below? i translation to the right i i translation to the left iii translation up i v translation down v reflection in the xaxis a
0
d
g
0
0
h
x
0
x
f
y
x
y
y
x
0
y
x
0
x
WE8 For each of the following, state:
i the equations of the asymptotes
a y=
76
0
e
y
c
y
x
y
0
3
b
y
−2
x
i i the domain
i i i the range.
b y=
1 x+6
c y=
d y=
2 3− x
e y=
3 +4 x
f
g y=
4 −2 x+6
h y=
5 +1 2− x
i
Maths Quest 12 Mathematical Methods CAS for the TINspire
−
3 x−2 −
1 −5 x 1 y= −m n+ x y=
x
4 For each of the following graphs, state: i the equations of the asymptotes i i the domain i i i the range. a
b
y
c
y
y 2
0
d
4
e
y −1 0
2 0
x
−1
x
f
y
y
n 0
a m
x
0
b
eBoo k plus eBook
−4 1 2 3 5 On the same set of axes sketch the graphs of y = , y = , y = and y = . x x 3x 3x
Digital doc
6
Spreadsheet 236 Function grapher
7
x
WE9 Sketch each of the following, clearly showing the position of the asymptotes and the intercepts with the axes. Check your answers, using a CAS calculator. a y= 1 b y = 1 −1 c y= 3 −3 x+3 x+2 x −1 4 −2 − 6 d y= e y= f y = 3 +6 −3 1− x x+5 x−2 1 2 4 g y = 1− h y= + i y= 1 +4 2− x 5 1+ x 2x + 3 2 x + 3 j y= k y= l y = 4x + 3 −1 x−2 x −1 3 − 4x MC The equation of the graph shown is likely to be:
1 x−4 1 C y = 3− 4−x A y = 3+
E 8
x
3
0
x
y = 3−
1 +4 x−3 1 D y= −3 4−x
y
B y=
3 0
1 x−4
MC Which of the following is a true statement for the graph of y =
A B C D E
The domain is R\{1}. The range is R\{3}. The equation of the horizontal asymptote is y = −3. The equation of the vertical asymptote is x = 2. None of the above.
Chapter 2
4
x
2 − 3? x +1
Functions and transformations
77
9 WE10 Find the equation for each of the following hyperbolas, if they are of the form a y= + c. x−b a
b
y
−1
d
0
1
eBoo k plus eBook Digital docs
WorkSHEET 2.1 History of mathematics The history of some major curves
2D
1
x
0
e
y
−4
10
2
c
y
0
−4
x
f
y 2 11–2 0
x
3
y
x
− 3–4
y 5
3 4
x
−11 −1
5
x
1 If a function is given by f ( x ) = , sketch each of the following, labelling the asymptotes x and the intercepts with the axes. a f (x + 2) b f (x) − 1 c −f (x) − 2 − d f (1 − x) + 2 e f (x − 1) − 1 f 1 − f (x − 2)
11 Sketch the graph of yx − 3x + 1 = 0, and state its domain and range. (Hint: First transpose the equation to make y the subject.)
EXAM TIP When you are asked to give the domain and range in a question, be sure to demonstrate clearly which answer is which. For example, for y = x2, dom f : R, ran f : [0, ∞]. [Authors’ advice]
The power function (the truncus) The graph shown at right is known as a truncus. The equation of the graph is given by:
y
y = 12 x Power functions are functions of the form f (x) = xn, n ∈ R. y=0 x 0 The value of the power, n, determines the type of function. We saw earlier that when n = 1, f (x) = x and the function is linear. x=0 When n = 2, f (x) = x2 and the function is quadratic. When n = 3, f (x) = x3 and the function is cubic. When n = 4, f (x) = x4 and the function is quartic. When n = −1, f (x) = x−1 and the power function produces the graph of a hyperbola. The power function that produces the graph of a truncus has a value of n = −2. Thus, the function f ( x ) = 12 can also be expressed as the power function f (x) = x−2. x The function is undefined for x = 0. Hence, the equation of the vertical asymptote is x = 0 and the domain of the function is R\{0}.
78
Maths Quest 12 Mathematical Methods CAS for the TINspire
We can also observe that the graph approaches the xaxis very closely, but never touches it. So y = 0 is the horizontal asymptote. Since the whole graph of the truncus is above the xaxis, its range is R+ (that is, all positive real numbers). 1 Similar to the graphs of the functions, discussed in the previous sections, the graph of y = 2 x can undergo various transformations. a + c , or y = a (x − b)−2 + c. Consider the general formula y = ( x − b) 2
Dilation
y
The value a is the dilation factor. It dilates the graph from the xaxis. The dilation factor does not affect the domain, range or asymptotes.
y=0
a=3 a=2 x
0
x=0 y
Reflection If a is negative, the graph of a basic truncus is reflected in the xaxis. The range becomes R− (that is, all negative real numbers).
a y = –2 x
y = 12 x
y=0
x
0
− y = 12
If x is replaced with −x, the graph of the basic truncus is reflected in the yaxis. The effect of this reflection cannot be seen in the basic graph, but it becomes more obvious if the graph has been translated horizontally first. For example, 1 1 the graphs of y = and y = (− x − 3)2 ( x − 3)2 are reflections across the yaxis. The vertical asymptote changes from x = 3 to x = −3 and the domain changes from R \{3} to R \{−3}.
x
x=0 y 1 ( x − 3)2
1 y= − ( x − 3)2
y=
y=0
x
0
x = −3
x=3
Translation Horizontal translation The value b translates the graph b units horizontally. If b > 0, the graph is translated to the right, and if b < 0, the graph is translated left. For example, the graph of the equation 1 results from translating a basic truncus 3 units y= ( x − 3)2 to the right. The vertical asymptote is x = 3
y b = −2 y=0
−2
b=3 0
3
x
1 y = ——— 2 (x − b)
Chapter 2
Functions and transformations
79
and the domain is R \{3}. If a basic truncus is translated 2 units to the left, it becomes 1 y= , where the vertical asymptote is x = −2 and the domain is R \{−2}. Hence, the ( x + 2)2 equation of the vertical asymptote is x = b and the domain is R \{b}. The range is still R+ and the equation of the horizontal asymptote is y = 0. Vertical translation y 1 The value c translates the graph c units vertically. If c > 0 the graph y = –– +c x2 is translated upward, and if c < 0, the graph is translated c units 1 c=1 downward. For example, the graph with equation y = 2 + 1 results x y=1 1 when a basic truncus is translated 1 unit upward. The horizontal 0 c = −1 x asymptote is y = 1 and the range is (1, ∞). If a basic truncus is 1 −1 y = −1 translated 1 unit down, it becomes y = 2 − 1, with y = −1 as x=0 x the horizontal asymptote and (−1, ∞) as the range. Hence the equation of the horizontal asymptote is y = c and the range is (c, ∞). Note: If a is positive (see graph below), the whole graph of the truncus is above the line y = c (the horizontal asymptote) and hence its range is y > c, (c, ∞). If a is negative, the whole graph is below its horizontal asymptote and therefore the range is y < 0, or (−∞, c). y
y y=3 y=3
3 0
The graph of y =
3
x x
a + c shows the combination of these ( x − b) 2
y
y=
a + c (x − b)2
transformations. y=c
c 0
b
x
x=b WORKED EXAMPLE 11
State the transformations required to change the graph of y = THINK
80
−1 1 − 1. into the graph of y = 2 ( x − 2 )2 x
WRITE
a +c ( x − b) 2
1
Write the general formula for the truncus.
y=
2
Identify the value of a.
a = −1
Maths Quest 12 Mathematical Methods CAS for the TINspire
1
3
State the effect of a on the graph.
The graph of y =
4
Identify the value of b.
b=2
5
State the effect of b on the graph.
The graph is translated 2 units to the right.
6
Identify the value of c.
c = −1
7
State the effect of c on the graph.
The graph is translated 1 unit down.
x2
is reflected in the xaxis.
WORKED EXAMPLE 12
2 − 4 , state: ( 3 + x) 2 a the equations of the asymptotes For the function y =
b the domain
THINK a
b c
c the range. WRITE
a y=
a +c ( x − b) 2
1
Write the general formula for the truncus.
2
Write the general equations of the asymptotes.
Vertical asymptote: x = b Horizontal asymptote: y = c
3
Identify the values of b and c.
b = −3, c = −4
4
State the equations of the asymptotes by substituting the values of b and c into corresponding formulas.
Asymptotes: x = −3 and y = −4
Write the domain of the truncus, which is R\{b}. 1
Check whether a is positive or negative.
2
Write the range (which for a > 0 is y > c).
b Domain: R\{−3} c a>0
Range: y > −4
a + c, then compare the given ( x − b) 2 equation with the general formula to see what changes should be made to the basic curve (the 1 graph of y = ) to transform it to the one you want. This should give you an idea of how the x2 graph will look. To sketch the graph of a truncus, first put it in the form y =
The following algorithm can then be used: 1. Find the position of the asymptotes. 2. Find the intercepts with the axes. 3. On the set of axes, draw the asymptotes (using dotted lines), label with the equation and mark the x and yintercepts. a>0 4. Treating the asymptotes as the new set of axes, sketch the basic truncus curve. 5. Make sure the curve passes through the points marked on the axes.
Chapter 2
a<0
Functions and transformations
81
WORKED EXAMPLE 13
1 , clearly showing the position ( x + 1)2 of the asymptotes and the intercepts with the axes (correct to 1 decimal place). Sketch the graph of y = 2 −
THINK
WRITE/DRAW
a +c ( x − b) 2
1
Write the general formula for the truncus.
y=
2
Identify the values of a, b and c.
a = −1,
3
Write a short statement about the transformations 1 the graph of y = 2 should undergo in order to be x changed into the one in question.
The graph of y =
4
Write the equations of the asymptotes (y = c and x = b).
Asymptotes: x = −1 and y = 2
5
Find the xintercept (round off to 1 decimal place).
xintercept: y = 0
b = −1,
1 is reflected in the x2 xaxis, translated 1 unit to the left and 2 units up.
0 = 2− 1 = 2 ( x + 1)2 ( x + 1)2 = 12 x = ± 1 2
x = .
Find the yintercept.
7
To sketch the graph: draw the set of axes and label them; use dotted lines to draw asymptotes; mark the x and yintercepts; treating the asymptotes as the new set of axes, draw the basic truncus curve upside down (since a is negative); make sure it intersects the axes in the right places.
≈
1 ( x + 1)2
1 2 1 2
x +1 = ±
6
c=2
−1
− 1 or x =
− 0.3
.
≈
−
1 2
−1
− 1.7
yintercept: x = 0 1 y = 2− (0 + 1)2 = 2 −1 =1 y
y=2
(−1.3, 1.3, 0) (0, 1) 0
(−1.7, 0)
x
x = −1
In the above example we have considered sketching the graph from the given equation. Sometimes the opposite task is required; that is, the equation of the function should be established from its graph.
82
Maths Quest 12 Mathematical Methods CAS for the TINspire
WORKED EXAMPLE 14
Find the equation of the curve shown in this diagram.
x=2
y
2
x
− 1–2
THINK
WRITE
y=
a +c ( x − b) 2
1
Write the general equation of the truncus.
2
Compared to y =
b = 2 and c = 0
3
Substitute the values of b and c into the formula.
y=
4
Substitute the coordinates of the yintercept into the formula.
Using (0, 2 ) :
1 , the graph is shifted 2 units to x2 the right. (There is no shift along the yaxis.)
a +0 ( x − 2)2 a = ( x − 2)2 −
1
−
1 a = 2 (0 − 2)2 a = 4 −1 a = 4 × ( 2)
5
Solve for a.
a = −2
6
Substitute for a in the equation.
y=
−2
( x − 2)2
REMEMBER
1. The graph of y = 2.
3. 4. 5. 6.
1
x2
is called a truncus.
a + c is the basic truncus curve, The graph of y = ( x − b) 2 dilated by a factor of a from the xaxis and translated b units along the xaxis (to the right if b > 0 or to the left if b < 0) and c units along the yaxis (up if c > 0 or down if c < 0). If a is negative, the graph is reflected in the xaxis. The vertical asymptote is x = b. The horizontal asymptote is y = c. The domain is R\{b}. The range is y > c if a > 0, or y < c if a < 0.
Chapter 2
y
y=
a (x − b)2 + c
c 0
b
x
Functions and transformations
83
EXERCISE
2D
The power function (the truncus) 1 1 WE11 State the transformations required to change the graph of y = 2 into the graph of x each of the following: − 1 2 a y= b y= 3 c y= 2 2 ( x + 2)2 x x d y=
2 ( x − 3)2
e y=
−5
(4 +
x )2
4 +1 ( x − 3)2 1 2 MC To obtain the graph shown, the graph of y = 2 was: x A reflected in the xaxis and translated 2 units down B translated 2 units to the left C reflected in the xaxis and translated 2 units to the left D reflected in the xaxis and translated 2 units to the right E reflected in the xaxis and translated 2 units up g y = 3− 1 x2
h y=
2 +6 x2
f
y=
i
y = 5−
1 ( x + 2)2 y
0
x
x=2
3 MC Which of the following translations took place, so that the 1 graph of y = 2 was changed into the one shown at right? x A m units to the left and p units up B m units to the right and p units up C m units to the left and n units up D m units to the right and n units up E m units to the left and n units down
y p y=n x x=m
4 WE12 For each of the following state: i the equations of the asymptotes ii the domain iii the range. − a y= 2 b y= 4 x2 3x 2 −5 2 d y= e y = ( x + 1)2 (4 + x ) 2 4 1 1 3 g y= + 2 h y= − 2 5 x 2 x
c y= f i
1 ( x − 2)2
2 −3 x2 2 y= +4 ( x − 1)2 y=
Questions 5 tο 7 refer to the following diagrams. i
ii
y
iii
y
y
−3 0
84
3
x
−3
Maths Quest 12 Mathematical Methods CAS for the TINspire
0
x
0
−3
x
iv
v
y
Digital doc Function grapher
x
3
0 −3
−3
0
Spreadsheet 236
y
x
0
3
eBoo k plus eBook
vi
y
x
3
5 MC Which of the above functions has the domain R\{−3}? A i i only B i i i only D i i , i i i and v i E v and vi
C
i i and i i i
6 MC Which of the above functions has the range y < −3? A i , i i and i v B i i i , v and vi E i i i and vi D v only
C
i v only
7 MC Which of the graphs has asymptotes y = 0 and x = −3? A i B ii D iv E v
C
iii
8 WE13 Sketch each of the following, clearly showing the position of the asymptotes and the intercepts with the axes (correct to 1 decimal place where appropriate). − 1 2 a y= 2 b y= c y= 2 2 ( x − 3) (4 + x ) 2 5x d y=
−1
( x − 1)2
2 −2 ( x − 1)2 3 1 y= − 2 4 4( x + 1)
g y= j
e y = 1− 4 x2
f
2 (3 + x )2 1 +3 k y= (2 − x ) 2
i
h y = 4−
9 Find the equations for each of the following curves, if it is known that all of them are a of the form y = + c and a is ( x − b) 2 + − either 1 or 1. a
b
y
l
1 −3 2x2 2 1 y= + 3 ( x − 2)2 4 y= −1 (2 x − 4) 2 y=
EXAM TIP Always check you have met the requirements of the answer format, for example, ensure that you have rounded correctly and show the correct number of decimal places. [Assessment report 1 2006 © VCAA]
c
y
y
p −q
n m
0
d
x
0
r
0
x
x
e
y
f
y
−cc −d −a 0
t 0
−s
x
x
−b
Chapter 2
y
−f −f
g
h
x
−e
Functions and transformations
85
g
h
y
y x
0 −l −i
j
0
x
−k [© VCAA 2006]
10 WE14 Find the equation for each of the following. a b y y
−1
1
0
x
−2
y=
− 3–4
−2
d
x y=0
e
f
y
y=4 x
0
−2.5
x
4
y = −3
−3
−1
0
y 0 −2
4
11
−5
y=1 x
x = −2
7
x=
−2
x=2
y
−1 0
y 1
2
0
x=0
c
x=4
1
x
y = −2
−7 x=1
The domain of a truncus is R\{−2}; its range is y > −3 and its graph cuts the xaxis at 1 and x = −3. Find the equation of the function.
12 The domain of a truncus is R\{1}; its range is (2, ∞) and its graph cuts the yaxis at y = 5. Find the equation of the function.
2E
The square root function in power form The square root function is given by y = x (or y = x ). y Power functions are functions of the form f (x) = xn, n ∈ R. The value of the power, n, determines the type of function. We saw earlier that when n = 1, f (x) = x and the function is linear. When n = 2, f (x) = x2 and the function is quadratic. When n = 3, f (x) = x3 and the x 0 function is cubic. When n = 4, f (x) = x4 and the function is quartic. When n = −1, f (x) = x−1 and the power function produces the graph of a hyperbola. When n = −2, f (x) = x−2 and the power function produces the graph of a truncus. 1 The power function that produces the graph of the square root function has a value of n = . 2 1 2
Thus, the function f ( x ) = x can also be expressed as the power function f ( x ) = f ( x ) = x . The function is defined for x ≥ 0; that is, the domain is R+ ∪ {0}, or [0, ∞). 1 2
86
Maths Quest 12 Mathematical Methods CAS for the TINspire
As can be seen from the graph, the range of the square root function is also R+ ∪ {0}, or [0, ∞). Throughout this section we will refer to the graph of y = x as ‘the basic square root curve’. Let us now investigate the effects of various transformations on the basic square root curve. a( x − b) + c. Consider the function y = a x − b + c, or y = a( 1 2
Dilation
a=3
y
a=2
The value a is a dilation factor; it dilates the graph from the xaxis. The domain is still [0, ∞).
a=1 a = 1–2 0
y=a x
x
Reflection If a is negative, the graph of a basic square root curve is reflected in the xaxis. The range becomes (−∞, 0]. The domain is still [0, ∞).
y y=√ √x (1, 1)
(0, 0)
x
(1, −1) y = −√ √x
If x is replaced with −x, the graph is reflected in the yaxis. For example, the graphs with equations y = x and y = − x are reflected across the yaxis. The domain becomes (−∞, 0] and the range is [0, ∞).
y y = √−x
y=√ √x
(−1, 1)
(1, 1) x
(0, 0)
Translation Horizontal translation The value b translates the graph horizontally. If b > 0, the graph is translated to the right, and if b < 0, the graph is translated to the left. The graph with the equation y = x − 3 results when the basic curve is translated 3 units to the right. This translated graph has domain [3, ∞) and range [0, ∞). If the basic curve is translated 2 units to the left, it becomes y = x + 2 and has domain [−2, ∞) and range [0, ∞). The domain of a square root function after a translation is given by [b, ∞). Vertical translation The value c translates the graph vertically. If c > 0, the graph is translated vertically up, and if c < 0, the graph is translated vertically down. If y = x is translated 2 units vertically up, the graph obtained is y = x + 2, with domain [0, ∞) and range [2, ∞). If the basic curve is translated 4 units down, it becomes y = x − 4, with domain [0, ∞) and range [−4, ∞). The range of the square root function is [c, ∞) for a > 0.
Chapter 2
y b = −2 (−2, 0) 0
(3, 0)
b=3 x
y= x−b
y (0, 2) 0 (0, −4)
c=2
c = −4
x
y= x+c
Functions and transformations
87
Combination of transformations
y
The graph of y = a x − b + c shows the combination of these transformations. The point (b, c) is the end point of the square root curve. For example, the end point of y = x − 2 + 1 is (2, 1). It is always good practice to label the end point with its coordinates. Make sure it is an open circle if the xvalue is not in the required domain and a closed circle if its xvalue is within the function’s domain. Consider the function y = a b − x + c. The graph of y = a x + b + c has (−b, c) as its end point. If this function is reflected in the yaxis, it becomes y = a − x + b + c with end point (b, c). The equation y=a
−x
y=a x−b+c
(b, c) x
0
y
y=a b−x +c y=a x+b +c
+ b + c can then be rewritten as (−b, c)
y = a b − x + c.
(b, c) x
For example, the graph of y = 2 − x + 1 can be −x
rewritten as y =
+ 2 + 1, which has an
end point of (2, 1) and bends to the left. The domain is (−∞, 2] and the range is [1, ∞). The equation y = x + 2 + 1 results in y = changes from
[−2,
∞) to
(−∞,
be rewritten as y = 2( x +
3 ) 2
−x
+ 2 + 1 when it is reflected in the yaxis. The domain
2] and the range remains [1, ∞). The equation y = 2 x + 3 − 1 can −
3
− 1; 1 the domain is [ 2 , ∞) and the range is [−1, ∞).
WORKED EXAMPLE 15
State the transformations required to change y = x to y = − 3 x + 5 + 3. THINK
WRITE
1
Write the general formula for the square root curve.
y=a x−b +c
2
Identify the value of a.
a = −3
3
State the effect of a on the graph.
The graph is dilated by a factor of 3 from the xaxis and reflected in the xaxis.
4
Identify the value of b.
b = −5
5
State the effect of b on the graph.
The graph is translated 5 units to the left.
6
Identify the value of c.
c=3
7
State the effect of c on the graph.
The graph is translated 3 units up.
WORKED EXAMPLE 16
eBoo k plus eBook
For each of the following functions find the domain and range.
88
a
y = 2 x − 3 +1
c
y= 4− x +2
b
y = −4 3x + 2 − 4
Maths Quest 12 Mathematical Methods CAS for the TINspire
Tutorial
int0523 Worked example 16
THINK
WRITE
Write the general formula.
y=a x−b +c
a
a y = 2 x − 3 +1
b
c
1
Write the question.
2
Identify the values of b and c.
b = 3, c = 1
3
State the domain x ≥ b.
The domain is [3, ∞).
4
State the range (y ≥ c for a > 0).
The range is [1, ∞). b y = − 4 3x + 2 − 4
1
Write the question.
2
Factorise the expression under the square root sign.
y = − 4 3( x + 23 ) − 4
3
State the domain.
∞). The domain is [ 3 , ∞)
4
Identify the value of c and check whether a is positive or negative.
c = −4, a < 0
5
State the range.
The range is (−∞, − 4].
1
Write the question.
2 3
−
c y=
2
4− x +2
Identify the values of b and c.
b = 4, c = 2
Since the function is of the form
The domain is (−∞, 4].
y = a b − x + c, the domain is x ≤ b. 4
State the range (y ≥ c).
The range is [2, ∞).
To sketch the graph of the square root function, we a>0 a>0 need to compare the given formula with y=a b−x+c y=a x−b+c y = a x − b + c. This will give us an idea of the (b, c) a<0 a<0 changes required to transform the basic square root curve into the one we want. It will also let us know the way the curve will look. The diagram above illustrates the idea. Once the coordinates of the end point and the direction of the curve are known, the intercepts with the axes (if any) should be found before sketching.
WORKED EXAMPLE 17
Sketch the graph of y = − 2 x − 3 + 1 , clearly marking intercepts and the end points. THINK
WRITE/DRAW
1
Write the equation.
y = −2 x − 3 + 1
2
Write the coordinates of the end point.
End point: (3, 1)
3
State the shape of the graph.
Shape:
Chapter 2
Functions and transformations
89
4
Find the xintercept by letting y = 0.
xintercept: y = 0 0 = − 2 x − 3 +1 2 x−3 =1 1 x−3 = 2 1
x − 3 = ( 2 )2 =
1 4
1
x =34 5 6
Find the yintercept if there is one. Sketch the graph by plotting the end point, marking the xintercept, and drawing the curve so that it starts at the end point and passes through the xintercept.
There is no yintercept. y 1 0
(3, 1) (3 14– , 0) x 3
WORKED EXAMPLE 18
Given f : [0, ∞) → R, where f ( x ) = x and g(x ( ) = af (x (x (x) + b, where a and b are positive real constants, consider the effect on g(x ( ) as a and b increase individually. (x THINK 1
WRITE
On a Graphs page, complete the function entry line as: f 1( x ) = v1 x + v 2 Insert a slider by pressing: • MENU b • 1:Actions 1 • A:Insert Slider A Then press ENTER ·. Repeat to insert a second slider. Grab the slider for each variable and move it back and forth taking note of the effect of each variable increasing.
2
Write your description in words.
As a increases, the graph is dilated away from the xaxis, with the graph stretched further from the xaxis. As b increases, the graph is translated up parallel to the yaxis.
REMEMBER
1. The graph of the function y = a x − b + c is the graph
2. 3. 4. 5.
of y = x, x dilated by the factor of a from the xaxis and translated b units along the xaxis and c units along the yaxis. If a < 0, the basic graph is reflected in the xaxis. The end point of the graph is (b, c). The domain is x ≥ b. The range is y ≥ c for a > 0, or y ≤ c for a < 0.
y
y=a x−b+c
(b, c) 0
6. If y = a b − x + c, the domain is x ≤ b; the graph of y = a x is reflected in the yaxis.
90
Maths Quest 12 Mathematical Methods CAS for the TINspire
x
EXERCISE
2E
The square root function in power form State the transformations required to change y = x to each of the following.
1 WE15 a y=
b y=
−1 3
x
c
y = 3 x −1
d y = −2 4 + x
e y = x −1
f
y = 2−3 x
g y = x − 4 +3
h y = 6−2 3+ x
i
y=
x
−1 2
2
2− x + 3
2 For each of the functions in question 1 write the coordinates of the end point. 3 MC The graph shown below was obtained by translating the graph of y = A 3 units up and 9 units to the right y B 3 units down and 9 units to the right C 9 units up and 3 units to the right 3 D 3 units down and 9 units to the right E none of the above
−
x:
0
y
(2, 4)
e y = 5− x
f y = x −1+3
g y = 2 + x −1
h y = 4 − 2 2x + 1
i
j y= 3− x −7
k y = 6 + 4 − 2x
l y = 1− 2 − x
Questions 6 tο 7 refer to the diagram at right. 6 MC The equation of the graph is of the form:
−3 5
x
0
d y = 4+2 x
y=
x
9
4 MC To obtain the graph in the diagram at right, the graph of y = x was: A translated 2 units to the right and 4 units up B translated 4 units to the right and 2 units up C translated 2 units to the right, 4 units up and reflected in the xaxis D reflected in the yaxis, translated 4 units up and 2 units to the right E reflected in the xaxis, translated 4 units up and 2 units to the left. 5 WE16 Find the domain and range for each of the following functions. a y = x +1 b y= x−3 c y = x −3
3x − 4 + 2
y
A y = a x − b + c, a > 0
B y = a x − b + c, a < 0
C y = a b − x + c, a > 0
D y = a b − x + c, a < 0
(−2, 2) 0
x
E could be either B or C 7 MC The domain and range (in that order) of the function are: A (−∞, −2] and (−∞, −2] B (−∞, −2) and (−∞, 2) C (−∞, 2} and {−2, +∞) − − − − − − D ( ∞, 2] and ( ∞, 2] E ( ∞, 2] and ( ∞, 2] 8 WE17 Sketch the graph of each of the following, clearly marking intercepts and end points. a y= x+2 b y = 13 x + 3 c y = 2− x
d y = x − 6 +1
e y= 3+ x +2
f y = 12 − 4 + x
g y = 2x − 3
h y = 6 + 3x + 2
i y = 2 − x −1
EXAM TIP Graphs should be drawn showing correct features, such as smoothness and end points. [Assessment report 2 2007]
[© VCAA 2007]
Chapter 2
Functions and transformations
91
eBoo k plus eBook
9 WE18 A
y = 2− x −1
B
y = 2− 2 1− x
Spreadsheet 236
C
y = x −1−2
D
y = x − 2 +1
Function grapher
E
y = 2−2 x −1
Digital doc
y
MC The equation of the graph shown at right is:
0
(1, 2) x
2
The graph of y = x was dilated by the factor of 2 from the xaxis and translated m units to the right and 4 units down. It intersects with the xaxis at x = 5. Find: a the value of m b the equation of the curve. 11 The end point of the square root curve is at (4, 3) and its yintercept is 9. Sketch the graph of the curve and hence establish its equation. 10
12 The graph of y = x was dilated by the factor of 4 from the xaxis, reflected in the xaxis, translated 1 unit to the left and p units up. Find: a the value of p, if the graph cuts the yaxis at y = 4 b the equation of the curve c the xintercept d the domain e the range. f Hence, sketch the graph, showing the coordinates of the end points and the intercepts with the axes.
2F
The absolute value function The function f (x) =  x  is called an absolute value function or modulus function. The domain of this function is R and its range is R+ ∪ {0}. Its graph is symmetrical in the yaxis and has a cusp (a sharp point) at the origin. The symbol  x  represents the magnitude of x, (that is, the size of x), regardless of its sign. x, if x ≥ 0 Therefore, x = − x, if x < 0 Compare the graphs of y = x and y =  x . For x ≥ 0, the graphs of the two functions are identical, while for x < 0 the graph of y =  x  is the reflection of y = x in the xaxis. In general, any graph of the form y = ff (x) is called an absolute value function. To sketch the graph of y = ff (x), we need to sketch the graph of y = f (x) first and then reflect in the xaxis the portion of the graph which is below the xaxis.
y
y
x
0 y = x y
x
0 y=x
x
0 y = x
WORKED EXAMPLE 19
Sketch the graph of y =  x2 − 1. THINK 1
92
We first need to sketch the graph of y = x2 − 1. State the shape of this graph.
WRITE/DRAW
Let y = x2 − 1 Shape: positive parabola, translated 1 unit down
Maths Quest 12 Mathematical Methods CAS for the TINspire
2
State the coordinates of the turning point (note that it is also the yintercept).
Turning point: (0, −1)
3
Find the xintercept by letting y = 0.
xintercept: y = 0 x2 − 1 = 0 x2 = 1 x=± 1 x = +1 or −1
4 5
Sketch the graph of the parabola (Figure A).
y
y
−1
Reflect the portion of the parabola for < x < 1 in the xaxis; mark the new yintercept (Figure B).
(0, 1) (−1, 1, 0)
0
(1, 0) x
(−1, 0) 0 (1, 0)
(0, −1)
x
y = x2 − 1
Figure A
Figure B
Similar to the graphs discussed in the previous sections, the graph of the absolute value function can be transformed through dilations, translations and reflections. If y = a f (x) + c, a is the dilation factor. It dilates the graph from the xaxis. The larger a is, the thinner the graph. If a < 0, the graph is reflected in the xaxis.
y
a=2 a=1
0
The value c translates the graph along the yaxis. If c > 0, the graph is moved c units up and if c < 0, it is moved c units down.
y
a = −1
x
c=1 c = −2
1 0 −2
WORKED EXAMPLE 20
y = x + c
eBoo k plus eBook
Sketch the graph of y = x − 2 + 1. THINK
x
Tutorial
WRITE/DRAW
int0524 Worked example 20
1
Compare the given function with y = aff (x) + c, and write a short comment.
The graph of y = x − 2 is translated 1 unit up.
2
To sketch the required shape we first need to sketch y = x − 2 (a straight line). Find the yintercept by letting x = 0.
Let y = x − 2. yintercept: x = 0 y=0−2 = −2
3
Find the xintercept by letting y = 0.
xintercept: y = 0 x−2=0 x=2
Chapter 2
Functions and transformations
93
4
y
Sketch the line.
(2, 0) x
0 (0, −2)
5
Reflect the portion below the xaxis in the xaxis; mark the new yintercept.
y (0, 2) 0
6
Move the graph 1 unit up; mark the new yintercept and the coordinates of the cusp.
(2, 0)
x
y (0, 3) (2, 1) 0
2
x
Absolute value functions as hybrid functions An absolute value expression can be thought of as two separate expressions, depending on whether it is negative or positive. 2x 2 + 3 can be written as (2 2x (2x + 3) or −(2x (2 + 3) depending upon the value that x takes. To determine these particular values of x, we need to solve the two inequalities −3 −3 2 + 3 > 0 and 22x + 3 < 0, giving x > 2x and x < , respectively. 2 2 This gives us a domain for the two expressions above, so we can write a representation for the absolute value expression as: −3 2 x + 3, wh ere x ≥ 2 2x 2 + 3 = 2x − − (2 x + 3), wh ere x < 3 2 This is a useful process when used to rewrite an absolute value function that is to be graphed, because it gives us a rule for each part of the graph in the form of a hybrid function. It is also important when needing to differentiate a function of this type in a later chapter. WORKED EXAMPLE 21
k plus eBook eBoo
Express f (x) = 5x 5 − 4 as a hybrid function, defining the domain of each part 5x and graphing the function. THINK 1
Break the function into two parts: a negative and positive part.
2
Simplify the domain and function for each.
Tutorial
int0525 Worked example 21
WRITE/DRAW
5 x − 4, wh ere 5 x − 4 ≥ 0 f ( x) =  5x − 4  = − (5 x − 4), wh ere 5 x − 4 < 0 First function: 5x − 4 First domain: 5x − 4 ≥ 0 4
x≥ 5 − Second function: (5x − 4) = −5x + 4 Second domain: 5 x − 4 < 0 ∴x<
94
Maths Quest 12 Mathematical Methods CAS for the TINspire
4 5
3
Rewrite the function in hybrid form with the two rules with their respective domains.
4
Graph the two functions for the specific domains.
4 5 x − 4, wh ere x ≥ 5 f ( x) = − 5 x + 4, wh ere x < 4 5 y
5 4 (0, 4) 3 2 1 −2
5
−1
0
f x) = 5x − 4 f(
1 ( –45 , 0)
x
2
The absolute value function can easily be drawn using a CAS calculator. On a Graphs page, complete the function entry line as: f 1(x) = 5x − 4 Then press ENTER ·.
REMEMBER y
1. The symbol  x  denotes the magnitude of x. 2.  x  = x, if x ≥ 0  x  = −x, if x < 0 3. To sketch the graph of y = ff (x): (a) sketch the graph of y = f (x) (b) reflect the portion of the graph which is below the xaxis in the xaxis. EXERCISE
2F
x
0 y = x
The absolute value function 1
Sketch the graph of each of the following, showing exact values of intercepts
WE19
of axes. a y = 2x 2  2x d y=
x2
b y = x − 1
− 6
e y = 4 −
g y = 3x3 h i
y = (x + y=
2)3
2 x −1
c
x2
y = 3 − 6x
f y = (x − 3)2 − 4 EXAM TIP Always give answers in exact form unless the question asks for a numerical approximation (for example, ‘give your answer to 2 decimal places’).
− 1
[Assessment report 2 2007]
[© VCAA 2007]
Chapter 2
Functions and transformations
95
2
y
MC Which of the following functions best describes this graph?
A y = (x − 1)3
B y = (x + 1)3
C y = x3 + 1
D y = x3 − 1
E y = (x +
1)3
1
+ 1
–1 0
3 For each of the following functions state the domain and range.
eBoo k plus eBook
a y = 2x
b y = x + 1
d y = x2 − 3 − 2
e
4
Spreadsheet 002
d y=
Absolute value function
1 +1 x +1
y = 4 − 3x
f
y= 2−
1 x2
WE20 Sketch the graphs of each of the following.
a y = −2x
Digital doc
y=
c
x
x2
b y = x + 5 − 6
− 1 + 1
1 3 − x 4
g
y=
j
y=
l
y=2−
−
1 −1 x2
c
y = 23 − x + 1
− 2
f
y = (x + 1)2 − 1 − 2
2 +3 6− x
i
y=
e y=2− h
y=
k
y=
x2
1 1 −4 4 x2
2− x −2 + 3
x +1 − 8
EXAM TIP
Use a ruler for linear and absolute
value graphs. [Assessment report 2 2007 © VCAA]
5
WE21 Given the function f (x (x) =  3x − 1:
a rewrite the function as a hybrid function with appropriate domains b find f (0) and f (2) c sketch the graph, labelling any significant points.
6 Given the function f (x (x) =  x2 − 3x + 2: a rewrite the function as a hybrid function with appropriate domains b find f (−1) and f (2) c sketch the graph, labelling any significant points.
eBoo k plus eBook Digital doc
WorkSHEET 2.2
2G eBoo eBook k plus Interactivity
int0247 Transformations with matrices
96
7 The design shown at right is to be embroidered on the outer side of a pair of children’s socks. The total length of the design is 12 cm and its width is 8 cm. y 6 If we draw the set of axes through the centre of the design, the red section can be thought of as the absolute value function on a restricted domain. a Find the rule for the red section and specify the domain. b Using your knowledge of the transformations, and the rule for 4 4x the red section, find the rules for the blue, green and yellow sections of the design. c Using a graphics calculator, sketch the 4 functions that were obtained in a and b . Have you obtained the right design? −6
Transformations with matrices Transformations of graphs can be described using matrices as an alternative to function notation. The transformations that have been considered so far (dilations, reflections and translations) can be represented in matrix form. This describes how a particular point on a graph will be moved (or mapped) to a resultant location by the application of a dilation, reflection or translation, or a combination of the three. Remember the definition of a transformation is a rule that ‘links’ each point in the Cartesian plane to another point. So the matrix can be used for any
Maths Quest 12 Mathematical Methods CAS for the TINspire
point on a curve, or in fact used to determine the new rule for a function that has undergone one or a series of transformations. The new points or rules are termed ‘images’ of the original.
Reflections and dilations
x We can summarise the use of matrices to map these transformations, T , of points on a y curve as follows: (Let x' be the transformed value of x, and y' be the transformed value of y.) x x' − 1 0 x − x T = = = represents a reflection in the yaxis. y y' 0 1 y y x x' 1 T = = y y' 0
0 x
x = − represents a reflection in the xaxis. y
−1 y
x x' a 0 x ax T = = = represents a dilation of a factor of a from the yaxis. y y' 0 1 y y x x' 1 0 x x T = = = represents a dilation of a factor of a from the xaxis. y y' 0 a y ay These operations can be combined to represent more than one transformation, for example x x' − 2 0 x − 2 x T = = = y y' 0 3 y 3 y represents a reflection in the yaxis, a dilation of a factor of 2 from the yaxis, and a dilation of a factor of 3 from the xaxis. WORKED EXAMPLE 22
Using matrices, find the location of the point ((x′, y′) under the following transformations of the point (1, 3): • dilation by a factor of 2 from the yaxis • reﬂection in the xaxis. THINK 1
Construct the correct matrix that represents the transformations described. x x' 2 T = = y y' 0
2
WRITE/DISPLAY
0 x × − 1 y
2 0
0
−1
On a Calculator page, press: • Ctrl / • ×r Select the appropriate matrix templates∗ to construct and solve the matrix equation using the point (x, y) = (1, 3): x x' 2 T = = y y' 0
0 1 × 3
−1
Chapter 2
Functions and transformations
97
3
Write the matrix equation and interpret this to answer the question.
2 0
0 1
2 =− 3
−1 3
The image of the point (1, 3) is at (2, −3). ∗
Matrix operations can be done using a CAS calculator, but as the matrix multiplication required here is simple, it is recommended it be done by hand.
The difficult part is to correctly identify the transformation matrix. Once you have done that it is a matter of performing a matrix multiplication.
Translations Translations require a slightly different process. The transformation matrix is a 2 × 1 matrix, and finding the new image requires addition of the matrices rather than multiplication. b Matrices describing translations are of the form . c This represents: • a translation of b units in the positive direction of the xaxis and • a translation of c units in the positive direction of the yaxis. Note b and c > 0: • If either of the terms is negative, the translation is in the negative direction. • A zero entry indicates there is no translation in a particular direction. So a translation of a point (x, y) can be described as follows: x x' x b T = = + y y' y c x + b = y + c WORKED EXAMPLE 23
Find the location of the point ((x', y') under the following transformations of the point (−2, 4): • translation of 3 units in the x direction • translation of −5 units in the y direction. THINK 1
WRITE
Construct the correct matrix that represents the transformations described. x x' x 3 T = = + − y y' y 5
2
Construct and solve the appropriate matrix equation.∗ x x' 3 T = = + − y y ' 4 5 − 2
3
Interpret this to answer the question.
3 − 5
− 2 3 1 +− = − 4 5 1 − 2 3 1 + − = − 4 5 1 The image of the point (−2, 4) is at (1, −1).
∗
Matrix operations can be done using a CAS calculator, but as the matrix addition required here is simple, it is recommended it be done by hand.
98
Maths Quest 12 Mathematical Methods CAS for the TINspire
Note it is very important to use the correct mathematical language, both for the size and the direction of the transformation. For example, dilations are ‘by a factor of’, in other words a multiple of the original value, as distinct from translations, which are described as ‘of n units’, which is a set distance. In terms of direction, the expression ‘from the yaxis’ can also be expressed as ‘parallel to the xaxis’, ‘in the x direction’ or even ‘horizontally’. The same applies for ‘from the xaxis’.
Putting it all together The formal notation often used to describe a transformation begins as T : R2 → R2, which is saying ‘the transformation that maps a point (x, y) to another point (x, y) is ….’ and then the transformation is described. For example, a transformation involving a dilation by a factor of 3 from the xaxis followed by a translation of −2 in the x direction and 1 in the y direction could be defined as T : R2 → R2, T( T x, y) = (x − 2, 3y + 1), or simply (x, y) → (x − 2, 3y + 1). When more than one transformation is described, it is known as a composition of the transformations. When a series of transformations are described, they need to be done in the correct order as stated in the question. WORKED EXAMPLE 24
Find the location of the point ((x', y') under the following transformation of the point (3,−2): 1 from the yaxis 2 • dilation by a factor of 3 from the xaxis • reﬂection in the xaxis • translation of 3 units in the x direction. • dilation by a factor of
THINK
WRITE
1
Perform the first three transformations together. Construct the 1 correct matrix that represents these transformations described. 2 0 x x' 1 0 x T = = 2 × y y' 0 − 3 y
0 −3
2
Construct and solve the appropriate matrix equation. x x' 1 0 3 T = = 2 × y y' 0 − 3 − 2
1 2 0
0 3 3 = 2 −3 − 2 6
3
Now perform the translation. Construct and solve the appropriate matrix equation. Note we are using the product matrix from step 2.
3 3 9 2 + = 2 6 0 6
x x' 1 T = = 2 y y' 0 4
0 3 3 × − + − 3 2 0
Interpret this to answer the question.
The image of the point (3, −2) is at
( , 6) 9 2
Remember that a transformation maps any point on a curve to another by the same rule. Rather than mapping a series of individual points on the same curve, we can simply find a new rule under a transformation (or a series of them) and use this new rule to determine the location of any points from the curve described by the original rule.
Chapter 2
Functions and transformations
99
WORKED EXAMPLE 25
k plus eBook eBoo
Write the resultant equation from the following transformations of the curve described by y = x3: • dilation by a factor of 2 from the yaxis • reﬂection in the xaxis • translation of 2 units in the negative y direction. THINK 1
3
2 0
−1
Construct and solve the appropriate matrix equation.
2 0
−1 y
Now perform the translation. Construct the correct matrix that represents this transformation described.
0 − 2
Perform the first two transformations together. Construct the correct matrix that represents these transformations described.
0
0 x × y
−1
5
x x' 2 0 x 0 T = = − × + − y y' 0 1 y 2 Construct and solve the appropriate matrix equation. Note we are using the product matrix from our first equation in step 2. The transformed values of x and y are 22xx and −y − 2.
6
Express x' and y' in terms of x and y.
7
State the resultant equation.
4
int0526 Worked example 25
WRITE
x x' 2 T = = y y' 0 2
Tutorial
0 x
2x = − y
x' 2 x 0 2 x = − + − = − y' y 2 y − 2 Therefore x' = 2x 2x and y' = −y − 2. x' and 2 −y = y' + 2 y = −y' − 2 x=
3
− x y= −2 2
So for any point on the graph of the original function, y = x3, we can map the corresponding point under the transformations above by substituting the values into this transformed equation. Let’s have a look at another example. This time we will complete reflections/dilations and translations in the one step. WORKED EXAMPLE 26
Find the image of the curve with equation y = x after a reflection in the xaxis, followed by a dilation of a factor of 2 from the xaxis, and then a translation by +3 in the x direction. THINK 1
100
Construct the correct matrices that represent the transformations described. x x' 1 0 x 3 T = = − × + y y' 0 2 y 0
Maths Quest 12 Mathematical Methods CAS for the TINspire
WRITE
1 0
0
3 and 0
−2
2
Construct and solve the appropriate matrix equation.
x' 1 y' 0
0 x 3 x + 3 + = − y 0 2y
−2
3
The transformed values of x and y are x + 3 and −2y.
So x' = x + 3 and y' = −2y.
4
Express x' and y' in terms of x and y.
− x = x' − 3 and y = y' 2
5
Substitute new y and xterms for y = x, x and drop the primes.
−y
2 −y
= x−3 = 2 x−3
y = −2 x − 3 6
State the resultant equation showing all transformations.
y = −2 x − 3
REMEMBER
a 0 1. represents a dilation of a from the yaxis and a dilation of b from the xaxis. 0 b −1 0 1 2. represents a reflection in the yaxis, and 0 0 1 the xaxis.
0
−1
represents a reflection in
b 3. represents a horizontal translation of b and a vertical translation of c. c EXERCISE
2G
Transformations with matrices 1
Identify the dilations and/or reflections described by the following matrices. i
−1 0 0 2
ii
1 2 0
0 −4
iii
−2 0 0 3
iv
1 0
−1
0 2
2 WE22 WE24 Find the image of the point (−3, 5) under the above transformations in question 1 . 3 Find the image of the graphs of the following equations under the transformations in 1 i and 1 ii. a y=
1 x2
b y = x3 − 5
c
y= x
4 Identify the translations described by the following matrices. i
3 2
ii
2 − 2
iii
Chapter 2
−1 5 0
Functions and transformations
101
5 WE23 Find the image of the point (1, −2) under the transformations given in question 4 . 6 Find the image of the following equations under each of the transformations defined in questions 4 i and 4 ii. a y = x b y = x2 − 3x 7 The transformation T : R2 → R2 which maps the curve with the equation y = x3 to the curve with the equation y = (3x − 6)3 + 1, could have: A
x 1 0 x − 6 T = + y 0 3 y 1
B
x 1 0 x 6 T = 3 + y 0 1 y 1
C
x 1 0 x 2 T = 3 + y 0 1 y 1
D
x 1 0 x − 3 T = + y 0 3 y 1
E
x 3 0 x 6 T = + y 0 1 y 1
For the following transformations, where T : R2 → R2, state what the transformation T 1 represents and determine the image of the equation f ( x ) = . x x − 1 0 x 6 x 2 0 x 3 a T = b T = − 1 + − + y 0 2 y 1 y 0 2 y 1 x 1 c T = 3 y 0
0 x 1 + − −1 y 2
9 A function g(x ( ) is mapped to the curve h(x (x ( ) = −g(4(x (x (4( + 1)) + 3. Create a matrix equation that (4(x will map g(x ( ) to h(x (x ( ). (x 10 WE25 The following transformations are applied, in order, to the graph of y = x3 − 4x 4 : • dilation by a factor of 2 from the xaxis • reﬂection in the yaxis • translation of −1 unit in the y direction. a Use matrices to determine the image equation under these transformations. b Find the image of the point (2, 0) and check whether this point lies on the curve of the equation from a. −3 1 + 1 , describe, in order, the transformations and g( x ) = 2 ( x − 2)2 x performed to the graph of f (x) to give g(x) and create a matrix equation which would map f (x) to g(x).
11 WE26
If f ( x ) =
12 If f (x (x) = −g(2(x (2( + 1)) + 1 and g( x ) = x , find f (x (2(x (x) in terms of x only, using: a an algebraic method without the use of a CAS calculator b matrices and a CAS calculator. 13 If f (x (x) = 2g(x ( − 1) − 2 and g(x (x ( ) = x2 − 3x, find f (x (x (x) in terms of x only, using: a an algebraic method without the use of a CAS calculator b matrices and a CAS calculator. 14 If
102
−
− x3 1 1 ( ) using matrix methods. (x h( x + 2) + 1 = − 3 x 2 − 6 x − , find h(x 2 2 2
Maths Quest 12 Mathematical Methods CAS for the TINspire
2H
Sum, difference and product functions Sum functions
y
A sum function is of the form y = f (x) + g(x), or alternatively y = x2 4 y = (f ( + g)(x). 3 Many functions include two or more terms added (or 2 y = 1x 1 1 2 subtracted) together. For example the function y = x + can be x −4 −3 −2 −1−10 1 2 3 4 x 1 2 thought of as the sum of the functions y = x and y = . −2 x These graphs can be drawn by sketching the two individual functions on the same set of axes then adding the yvalues (ordinates) for each xvalue and plotting the resulting points. This is a useful method when we know the basic shape of the individual functions but do not recognise the whole function. We would not use this method for a familiar function such as y = x2 + 3x, as we have learnt ways of sketching this without breaking it up into parts. 1 y Using the example in the first paragraph, y = x 2 + , we 4 y = x2 x do not recognise the shape of this function, but we know the 3 2 two individual functions are the basic positive parabola and y = x2 + 1x 1 y = 1x the hyperbola. We could therefore sketch the graph of the parabola and the hyperbola and add the yvalues together for −44 −3 −22 −11−10 1 2 3 4 x corresponding xvalues to obtain points on the curve of the sum −2 function — which can be joined together to obtain the graph of the sum function. Note that the domain of the hyperbola is restricted to R \ {0} so the yvalue at x = 0 is undefined. As you cannot add an undefined number, this xvalue is also undefined for the sum function. A general rule is that the sum function is only defined for the domain over which both of the individual functions are defined. The domain of the sum function is, therefore, the intersection of the domains of the individual functions. If h(x) = f (x) + g(x), then the domain h(x) = domain f (x) ∩ domain g(x).
Difference functions A difference function is of the form y = f (x) − g(x), or alternatively y = (f ( − g)(x). It is essentially the same as a sum function except that one of the individual functions is subtracted from the 1 other. So y = x 2 − could be sketched by the same method as described above but instead of x adding the yordinates, we would subtract one from the other. The domain of a difference function is determined in the same way as a sum function. We could extend our rule above to include difference functions. y If h(x) = f (x) ± g(x), then the domain h(x) = 4 y = x2 domain f (x) ∩ domain g(x). 3 y = x2 − 1x 2 We can also think of a difference function as ‘adding a y = 1x 1 −1 negative’ and it could be written y = x 2 + . With this x −4 −3 −2 −1−10 1 2 3 4 x in mind, an alternative method of sketching the graph of a −2 difference function is to reflect the graph of the second function 1 (in this example, ) in the xaxis and then add the ordinates as for a sum function. x
Chapter 2
Functions and transformations
103
When sketching graphs of sum/difference functions, there are key points that can be found on either individual function to easily identify the value of the ordinate of the sum or difference function. These are the xintercepts and any point of intersection of the individual functions. The xintercept is where the ordinate of that particular function is zero, so the graph of the sum or difference function is actually the ordinate of y the other function for that value of x. 2.5 At the intersection, the ordinate of the sum function 2 will be double that of the two individual functions. 1.5 For a difference function, an intersection of the two individual functions corresponds to on xintercept 1 (y = 0) of the difference function. 0.5 Another useful yvalue to look for is where the graphs of individual functions have yvalues that are −1.5 −1 −0.5 0 0.5 1 1.5 x of the same magnitude but one is positive and one −0.5 is negative. This point is an xintercept of the sum function.
WORKED EXAMPLE 27
eBoo k plus eBook
Using addition of ordinates, sketch the graph of
Tutorial
f ( x ) = log log e ( x + 2 ) + x , x ∈[ ∈[ − 1, 2 ].
int0527 Worked example 27
THINK 1
WRITE/DRAW
Sketch the graphs of y = loge (x + 2) and y = x on the same set of axes over the required domain x ∈ [−1, 2].
y 3 2 y = loge (x + 2) (−1, 1)
1
−1.5 −1 −.5 0 2
104
Moving from left to right, add the ycoordinates of the two graphs for the key points and plot the resultant points. The key points are the: • end points • yintercepts • points of intersection. The new points on the graph are marked by an asterisk.
y =x (2, loge (4)) 0.5 1 1.5 2 2.5 x
y
∗
3 (−1, 1)
∗ ∗
2 y = log (x∗ + 2) e 1
−1.5 −1 −0.5 0
(2, 2)
(2, 2)
y =x (2, loge (4)) 0.5 1 1.5 2 2.5 x
Left end points (−1, 0) and (−1, 1), so the new point will be at (−1, 1). Right end points (2, loge (4)) and (2, 2), so the new point will be at (2, 2 + loge (4)). yintercepts (0, 0) and (0, loge (2)). Points of intersection (−0.44, 0.44) and (1.15, 1.15), so the new points will be (−0.44, 0.88) and (1.15, 2.30).
Maths Quest 12 Mathematical Methods CAS for the TINspire
3
Join these points with a smooth curve to create f (x).
y f( f x) = loge (x + 2) +x 3 (2, 2) y = x (−1, 1) 1 y = loge (x + 2) (2, loge (4)) −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 x 2
4
Remove the two individual graphs to leave the sum function.
(2, 2 + loge (4))
y 3 2 (−1, 1)
f x) = loge (x + 2) +x f(
1
−1.5 −1 −0.5 0
0.5
1 1.5
2 2.5 x
Product functions A product function is of the form y = f (x) g(x), or alternatively y = (fg ( )(x). When graphing product functions, it is useful to graph the individual functions, f and g, and for any relevant values of x, to identify the yvalues, or ordinates and multiply these together to obtain the yvalue of the product function. If the yvalue is undefined at a particular value of x for either of the individual functions, then the product function is undefined for that value. We cannot multiply by an undefined number. If h(x) = f (x)g(x), then the domain h(x) = domain f (x) ∩ domain g(x). When examining the graph of the two individual functions, it is useful to look at xintercepts and points where the value of either function is ±1. The product function will also have an xintercept at a point where either individual function has an intercept (as multiplying by zero gives zero). At a point where a function = ±1, the product function will have a value equal to the value of the other function, or its negative. It is also useful to observe that where the individual functions are both above the xaxis, or both below the xaxis, the value of the product function will be positive, that is, above the xaxis. This is because the product of two positive numbers or two negative numbers is positive. Alternatively, where one function is above and one below the xaxis, the value of the product function will be negative, that is, below the xaxis. WORKED EXAMPLE 28
If f (x (x) = 2x 2 and g( x ) = x + 1 , sketch the graph of f ( x ) g( x ) = 2 x x + 1 . THINK 1
Sketch the graphs of f (x) and g(x).
WRITE/DRAW y
y = 2x 2 y= x+1
(0, 1) (−1, 0) 0
Chapter 2
x
Functions and transformations
105
2
Find the domain of f (x) and the domain of g(x).
Dom f = R and dom g = [−1, ∞)
3
Find the domain of f (x) g(x).
Dom fg = [−1, ∞)
4
Find the xintercepts of both f and g and hence find the xintercepts of the product fg.
xintercept for f (x) is when x = 0 and f (x) = 0 xintercept for g(x) is when x = −1 and g(x) = 0 Hence, the xintercepts for the product are when x = 0 and x = −1.
5
Find the values of x for which the product is negative.
f (x) is negative and g(x) is positive for x ∈ (−1, 0), so fg is negative for x ∈ (−1, 0).
6
Find the values of x for which the product is positive.
7
Find the turning point using a CAS calculator. Round the answer to 2 decimal places as appropriate.
f (x) and g(x) are both positive for x ∈ (0, ∞), so fg is positive for x ∈ (0, ∞). The turning point is ( 32 , −0.77).
8
Sketch the graph of the product.
−
y
y = 22xx x + 1
(−1, 0) (0, 0)
x
2 (−— , −0.77) 3
REMEMBER
1. For the sum/difference function, dom( dom(f (x) ± g(x)) = dom f (x) ∩ dom g (x). The graph of the sum/difference function can be obtained by using the addition of ordinates method. 2. For the product function, dom ((ff (x) g(x)) = dom f (x) ∩ dom g(x). Some features of the graph of the product function are as follows: (a) the xintercepts of f (x) g(x) occur where either f (x) or g(x) have their xintercepts (b) f (x) g(x) is above the xaxis where f (x) and g(x) are either both positive or both negative (c) f (x) g(x) is below the xaxis where one of the functions f (x) or g(x) is positive and the other is negative. EXERCISE
2H
Sum, difference and product functions 1
106
Sketch the graphs of f (x (x) = g(x ( ) + h(x (x ( ) using addition of ordinates, given the following (x functions g(x) and h(x). State the domain of f (x) in each case. a g(x) = x2, h(x) = x x3 b g( x ) = , h( x ) = x 2 c g(x) = 3x2, h(x) =  x  1 x) = x + 2 d g( x ) = , h( x) x
Maths Quest 12 Mathematical Methods CAS for the TINspire
Determine the equation of g(x ( ) − h(x (x ( ) in each of the following cases then, using addition (x of ordinates, sketch the graph of g(x) − h(x). − x + 1, h( x ) = x a g( x ) = b g(x) =  x , h(x) =  x + 1 − 2 3 For each of the following, find the domain of f (x (x)g(x ( ). (x 2 a f ( x ) = x , g( x ) = 3 − x b f (x) = −x + 2, g(x) = 2x 2 + 1 2x c f ( x ) = x , g( x ) = 1 − x d f (x) =  x , g(x) = x2 − 1 e f ( x ) = x 3 , g( x ) = − x + 2 4
WE27 Sketch the graph of f ( x ) = x +
the general shape and any asymptotes.
2 for −4 ≤ x ≤ 4, by the addition of ordinates, showing x
5 Two functions are defined as f (x ( ) = −x2 and g( x ) = x . a Sketch the graph of each on the same set of axes for −2 ≤ x ≤ 2. b Find the smallest possible value of a given that the domain of the function h, where h(x) = (f ( + g)(x), is a ≤ x ≤ 2. c Find f (0) and g(0), and hence find h(0). d Find f (1) and g (1), and hence find h (1). e Find f (2) and g (2), and hence find h (2). f Using this information, sketch the graph of h(x ( ) (on the same set of axes as in a). (x Given the functions f (x ( ) = −x3, g(x ( ) =  x  and h(x (x ( ) = f (x (x ( ) + g(x ( ): (x − a Sketch the graph of each on the same set of axes for 2 ≤ x ≤ 2. b Find f (−2)and g (−2), and hence find h (−2). c Find f (0) and g (0), and hence find h (0). d Find f (1) and g (1), and hence find h (1). e Find f (2) and g (2), and hence find h (2). f What is the range of the function h (in exact form). g Using this information, sketch the graph of h(x) (on the same set of axes as in a). 7 WE28 Two functions are defined as f (x ( ) = x − 3 and g( x ) = x . Let h( x ) = f ( x ) g( x ). a Find the domain of h. b Sketch the graph of each on the same set of axes. c Find f (0) and g (0), and hence find h (0). d Find f (1) and g (1), and hence find h (1). e Find f (2) and g (2), and hence find h (2). f What is the range of the function h (in exact form)? g Using this information, sketch the graph of h(x) (on the same set of axes as in a). 6
8 Sketch the graphs of the functions f ( x ) = x + 5 and g( x ) = 8 − x and use these to find the domain of the function h(x ( ) = f (x (x ( ) + g (x ( ). On the same axes, sketch the graph of h(x ( ), including (x the coordinates of any end points. x2 − 4, for 9 Use a CAS calculator to view and sketch the graphs of f (x ( ) =  x + 2  and g( x ) = 2 −2.5 ≤ x ≤ 2.5. Then, without using the calculator, use these graphs to sketch the graph of h, if h(x) = (f ( − g)(x) on the same set of axes. Using the calculator, check the shape of the graph you have drawn and use it to identify any significant points such as intercepts and cusp points to 2 decimal places. (You may need to adjust your window settings in order to clearly identify these points.)
Chapter 2
Functions and transformations
107
2I
Composite functions and functional equations Composite functions A composite function is formed from two functions in the following way. If f (x) = x + 5 and g(x) = 2x 2x are two functions, then we combine the two functions to form the composite function g(f ( (x)) = 2f (f 2 (x) = 2(x + 5). That is, f (x) replaces x in the function g(x). The composite function reads g of f and can be written g ° f. f Another composite function is f (g(x)) = g(x) + 5 = 2x 2 + 5. In this case, g(x) replaces x in f (x). This composite function reads f of g and can be written f ° g. For the composition function f (g(x)) to be defined, the range of g must be a subset of (or equal to) the domain of ff, that is ran g ⊆ dom ff. It is easiest to list the domain and function of both f (x) and g(x) first when dealing with composite function problems. For example: f (x) = x2 and g( x ) = x :
Domain Range
f (x ( )
g(x ( ) (x
R
[0, ∞)
[0, ∞)
[0, ∞)
Composite functions can be rather complex to graph by hand, so a CAS calculator can be used for assistance when sketching.
WORKED EXAMPLE 29
For the pair of functions f ( x ) = a show that f (g ( (x ( )) is defined c state its domain.
1 and g( x ) = x : x+2 b find f (g ( (x ( ))
THINK a
1
2
b
WRITE
Create a table showing the domain and range of both functions.
For f (g(x)) to exist the range of g must be a subset of f. f
Form the composition function f (g(x)) by substituting g(x) into f (x).
f (x (x)
g(x ( ) (x
Domain
R \{−2}
R+
Range
R \{0}
R+
ran g(x) ⊆ dom f (x) ∴ f (g(x)) is defined. f ( g( x )) = f ( x ) f ( g( x )) =
c
108
1 x +2
The domain of f (g(x)) must be the same as Domain of f (g(x)) = R+ the domain of g(x). Since the domain of g(x) is R+, it is the domain of f (g(x)).
Maths Quest 12 Mathematical Methods CAS for the TINspire
Functional equations Sometimes we are required to solve or analyse equations that are in terms of unknown functions, for example, f (x) or f (y), rather than being in terms of unknown variables, for example x or y. An example of the type of problem you might encounter is to find a function that satisfies f (x + y) = f (x) + f (y). Alternatively, you might be required to determine if a particular function satisfies the rule f (2x (2x) = 2f 2f (x). Equations such as f (x + y) = f (x) + f (y) are called functional equations. There are generally two ways to solve these types of problems: algebraically or using a CAS calculator. WORKED EXAMPLE 30
Determine if f (x ( ) = 3x satisfies the equation f (x ( + y) = f (x ( ) × f (y ( ). THINK 1
WRITE
Substitute the function into the LHS and RHS of the equation separately. Simplify the LHS of the equation to determine if it equals the RHS of the equation. Answer the question.
2 3
LHS = f (x + y) = 3x + y RHS = f (x) × f (y) = 3x × 3y LHS = f (x + y) = 3x + y LHS = f (x + y) = 3x × 3y ∴LHS = RHS x ∴ff (x) = 3 satisfies the equation f (x + y) = f (x) × f (y)
WORKED EXAMPLE 31
Determine if g(x ( ) = 10x (x 10 satisfies the equation g(2x (2 ) = 2g (2x 2 (x ( ). THINK 1
On a Calculator page, press: • MENU b • 1:Actions 1 • 1:Deﬁne 1 Complete the entry line as: Define g(x) = 10x Then press ENTER ·.
2
To calculate g(2x) and 2g(x), complete the entry lines as: g(2x) 2g(x) Press ENTER · after each entry. Examine the solutions to determine if the expressions are equal.
WRITE/DISPLAY
LHS = g(2x (2 ) = 20x (2x 20
Chapter 2
RHS = 2g(x) = 20x 20
Functions and transformations
109
3
Check that the equation holds true for all values of x by completing the entry line as: g(2x) = 2g(x) Then press ENTER ·. As the statement is true the equation must hold for all values of x.
4
Write the solutions for g(2x (2 ) and 2g(x) (2x and answer the question.
g(2x (2 ) = 20x (2x 20 2g(x) = 20x 20 ∴g(2x (2 ) = 2g(x) (2x When g(x) = 10x 10 it satisfies the equation g(2x (2 ) = 2g(x). (2x
If we consider the same equation f (2x (2x) = 2f 2f (x (x) for a different function, for example, f ( x ) = x , we obtain two different equations, f (2 x ) = 2 x and 2 f ( x ) = 2 x , which are not equal. However, if we define this function on a CAS calculator and enter the statement f (2x (2x) = 2f 2f (x), the result is x = 0. This means this equation holds true when x = 0 but not for any other values of x. REMEMBER
1. For the composite function f (g(x)) to be defined, the range of g must be a subset of the domain of ff. Furthermore, if f (g(x)) is defined, the domain of f (g(x)) equals the domain of g(x). 2. For equations involving algebra of functions, determine if the equation is true for the particular function by considering the LHS and RHS of the equation separately to test if the equation holds true for all values of x. EXERCISE
2I
Composite functions and functional equations 1
WE29 For each of the following pairs of functions:
i show that f (g(x)) is defined
i i find f (g(x)) and state its domain.
a f (x) = 2x 2 − 1 and g( x ) = x + 3 c f (x) = 3(x − 2)3 and g(x) = x2 e f (x) = (x + 1)(x + 3) and g(x) = x2 2 Show that the function f (x ( ) = 3xx satisfies x + y f ( x) + f ( y) = the equation f . 3 3 3
WE30 Show that f ( x ) =
equation
110
[© VCAA 2007]
2 satisfies the x
f ( x) + f ( y) = x + y. f ( xxyy )
Maths Quest 12 Mathematical Methods CAS for the TINspire
1 and g(x) =  x  + 1 x+2 d f (x) =  x  and g(x) = x3 b f ( x) =
EXAM TIP ‘Show that’ questions require detailed working out to be given, so provide as many steps as possible to attain full marks. Write ‘as required’ at the end of your solution once you have shown the problem is correct. [Assessment report 2 2007]
4 WE31 Determine which of i–v –v hold for the following functions: –v f ( x) ii f ( x − y ) = i f (x − y) = f (x) − f (y) f ( y) x f ( x) f = iii f (x) + f (y) = (x2 + y2) × f (xy) iv y f ( y ) v f (xy) = f (x) × f (y)
5 6 7 8
9
10
b f (x) =  x  a f ( x) = x 1 1 c f ( x) = d f ( x) = 2 x x e f (x) = x2 f f (x) = 2x 1 f ( x) = and g( x ) = x , determine the values of a such that f (g(x ( )) exists. (x ( x + a) 2 1 + 2, determine if f ° g If f: f R→R, where f ( x ) = x − 2 and g:R {−1} → R, where g( x ) = x +1 and g ° f exist and, if so, find the composition functions. If f: f R→R, where f ( x ) = 3 − x and g:R→R, where g(x ( ) = x2 − 1, show that f ° g is not defined. (x By restricting the domain of g, find a function h such that f ° h is defined. Given w(x ( ) = x + 3, x > −3 and v(x (x ( ) =  x  − 2, x ∈ R+, state the domain and range of each (x function. Hence, find if w ° v and v ° w exist and, if so, state their rules including their domains. Show that the equation g(x ( ) = x3 satisfies the equation g(−x) = −g(x (x ( ). Show that (x n this statement is true for all functions of the form g(x) = x , where n is an odd natural number. Show that g(x ( ) = x4 satisfies the equation g(xy (x ( ) = g(x (xy ( )g(y (x ( ). Show that this equation is true (y for all functions of the form g(x) = xn, where n is a natural number.
11 Given the two functions: 1 f : [0, 6] → R R,, f ( x) x ) = 4 x ( x − 6)2 g: [0, 6] → R, g (x) = −x2 + 6x find the maximum value of  f (x) − g(x) for 0 ≤ x ≤ 6, correct to 2 decimal places, and determine the values of x in this interval for which the maximum occurs. [© VCAA 2007]
EXAM TIP Be careful to answer all aspects of the question (this question requires the xvalue and the maximum). Always reread the question to check you have met all requirements before moving on to the next question. [Assessment report 2 2007]
12 Consider f : [4, ∞] → R ( ) = 1 − x. What transformations are (x R,, f ( x) x ) = x − 4 and g: R → R, g(x required to obtain f (g(x ( )) from f (x (x ( )?
2J
Modelling People such as scientists, financial advisers, business analysts, economists, statisticians and others often have to deal with large and small sets of data. Once the data are collected, we are often interested in finding the rules that link features of the data. The process of finding such a rule is called modelling and the rule itself is known as the mathematical model. When finding the model, the best way to start is to plot the data, as the shape of the graph might suggest the type of relationship between the variables.
Types of graphs By recognising the shape of a graph, it is possible to find the rule or mathematical model that describes it. Throughout this chapter, several types of graphs have been investigated.
Chapter 2
Functions and transformations
111
The parabola: y = x2
The graph of a cubic function: y = x3
y
The hyperbola: y =
y
y
y=0
x
0
1 x
x
0
x
0
x=0
The truncus: y = y
y=0
The graph of a square root function:
1 x2
y= x y
x
0
x
0 x=0
Reflections and translations can be applied to each of these graphs, but the basic shape of each graph remains the same.
WORKED EXAMPLE 32
Match each of the following graphs with the appropriate model. i y = ax2 a
ii y = ax3 b
y x
iii y = c
y
a x
iv y = d
y
x
x
THINK
a x2
v e
y
y= a x y
x
x
WRITE
Match the graphs using the information in the summary above.
i is a parabola; it matches graph b . i i is a cubic; it matches graph e . i i i is a hyperbola (the graph is in opposite quadrants); it matches graph c . iv is a truncus (the graph is in adjacent quadrants); it matches graph a . v
112
is a square root function; it matches graph d .
Maths Quest 12 Mathematical Methods CAS for the TINspire
WORKED EXAMPLE 33
x
0
1
2
3
4
5
y
0
2.5
3.54
4.33
5
5.59
The data in the above table exactly fit one of these rules: y = ax 2 , y = ax 3 , y = a Plot the values of y against x.
b Select the appropriate rule and state the value of a.
THINK
WRITE/DRAW
a Plot the values of y against x.
a
y 5 4 3 2 1 0
b
a a , y= or y = a x . x x2
1 2 3 4 5
x
b Assume that y = a x .
1
Study the graph. It appears to be a square root curve. Write the appropriate rule.
2
To find the value of a: select any pair of corresponding values of x and y. (Since we need to take a square root, the best to choose is the one where x is a perfect square.)
Using (1, 2.5):
3
Substitute selected values into the rule and solve for a.
2.5 = a 1 =a×1 a = 2.5
4
We need to make sure that the selected rule is the right one. Replace a with 2.5 in the rule. Substitute the values of x from the table into the formula and check if you will obtain the correct values of y.
Verifying: y = 2.5 x
As the values of y obtained by using the rule match those in the table, the choice of model is correct.
The rule that fits the data is y = a x , where a = 2.5.
5
6
y = 2.5 0 =0 (2, 3.54): y = 2.5 2 = 3.54 (3, 4.33): y = 2.5 3 = 4.33 y = 2.5 4 (4, 5): =5 (5, 5.59): y = 2.5 5 = 5.59 (0, 0):
Chapter 2
Functions and transformations
113
The process of fitting a straight line to a set of points is often referred to as regression. Statistical data is easiest to deal with in linear form. If the data is not linear, then a linear relationship can still be found by transforming the x scale. A regression line can then be fitted. m For example, y = x + c is a hyperbola. However, if we substitute X for 1 , the rule becomes x linear: y = mX + c. The graph of y versus X will be a straight line with a gradient of m and a yintercept of c. These values (m and c) can then be established from the graph and thus the hyperbolic model can be determined. Note: In a quadratic relationship, X is substituted for x2; in a cubic relationship, X is substituted for x3. WORKED EXAMPLE 34
It is believed that, for the data in the table below, the relationship between x and y can be modelled by y = aaxx 2 + bbxx + c. x
0
1
2
y
4
5.3
8.6
3
4
5
14.8
23
34.4
a Plot the values of y against x.
THINK a
114
b Calculate the values of a, b and c (correct to 3 decimal places) and write the equations. WRITE/DISPLAY
1
On a Lists & Spreadsheet page, enter xvalues into column A and yvalues into column B. Label each column x and y respectively.
2
On a Data & Statistics page, move the pointer to the horizontal and vertical axes and select the xvalues variable and yvalues variable respectively.
Maths Quest 12 Mathematical Methods CAS for the TINspire
b
1
Return to the Lists & Spreadsheet page, then press: • MENU b • 4:Statistics 4 • 1:Stat Calculations 1 • 6:Quadratic Regression 6 Change the X List to x, and the Y List to y. Select OK and press ENTER ··
2
Interpret the variables given on the screen.
a = 1.252 b = −0.222 c = 4.096 ∴ y = 1.252x 1.252 2 − 0.222x 0.222 + 4.096 Correct to 3 decimal places.
If the relationship between the variables is not given, we have to make an assumption of a model from the graph of the data. We then have to transform the data according to our assumption. If the assumption was correct, the transformed data, when plotted, will produce a perfectly straight, or nearly straight, line. Note: In this section we will consider only the rules of the type y = ax2 + b, y = ax3 + b, a y = + b and so on (we will not allow for a horizontal translation), so that the appropriate x substitution can be made. WORKED EXAMPLE 35
x
1
2
3
4
5
6
y
35
21
16
12
11
10
Establish the rule connecting x and y that fits these data. THINK 1
Using either graph paper or a CAS calculator, plot y against x.
WRITE/DRAW y 35 30 25 20 15 10 5 0
2
The scatterplot appears to be a hyperbola. Write the appropriate formula (remember that we do not consider horizontal translations in this section).
1 2 3 4 5 6
Assumption: y =
x
a +b x
Chapter 2
Functions and transformations
115
3
4
Check your assumption: prepare a new 1 table by replacing values of x with x (leave the values of y unchanged). 1 Plot y against . x
1 x
1
y
35
116
Comment on the shape of the graph.
6
If we replace
1 with X, the rule x becomes y = aX + b, which is the equation of the straight line, where a is the gradient and b is the yintercept. These (a and b) can be found from the graph as follows: draw in the line of best fit.
21
0.33
0.25
16
12
0.2 11
0.17 10
y 35 30 25 20 15 10 5 0
5
0.5
0.2 0.4 0.6 0.8
1
1 x–
The graph is very close to a straight line, therefore the assumption of a hyperbolic model is correct. y 35 30 25 20 15 10 5 0
0.2 0.4 0.6 0.8
1
X
y2 − y1 x2 − x1
7
Write the formula for the gradient.
m=
8
Select any 2 points on the line.
Using (0.17, 10) and (1, 35):
9
Substitute the coordinates of the points into the formula and evaluate.
m=
10
Write the value of a.
Since a is the gradient, a = m = 30.12.
11
Write the general equation of the straight line.
y = mx + c
12
Substitute the value of m and the coordinates of any of the 2 points, say (1, 35) into the equation.
35 = 30.12 × 1 + c
13
Solve for c. (Alternatively, read the yintercept directly from the graph.)
14
State the value of b.
35 = 30.12 + c c = 35 − 30.12 = 4.88 Since b is the yintercept, b = c = 4.88 88.
15
Substitute the values of a and b into a y = + b to obtain the rule that fits x the given data.
35 − 10 1 − 0.17 25 = 0.883 = 30.12
The rule for the given data is: y=
30.12 + 4.888 x
Maths Quest 12 Mathematical Methods CAS for the TINspire
REMEMBER
1. 2. 3. 4.
Modelling is the process of finding the rule that fits the given data. The rule itself is called a mathematical model. The best way to start modelling is to produce a scatterplot of the original data. Use the scatterplot of the data to make an assumption of the model of the relationship. a It should be of the type y = ax2 + b, y = ax3 + b, y = + b and so on. To test the x assumption, transform the data accordingly. If the assumption is correct, the transformed data when plotted will produce a straight, or nearly straight, line. 5. To find the values of a and b in the model, draw a line of best fit; a is the gradient of the line and b is the yintercept. EXERCISE
2J
Modelling 1
WE32 Match each of the graphs with the appropriate model:
i y = ax2 + b a iii y = + b x
ii y = ax3 + b a iv y = 2 + b x
v y=a x +b a
b
y
c
y
x
x
d
y
e
y
x
y
x x
2 eBoo k plus eBook Digital doc
Spreadsheet 076 Modelling
WE33 The data in each of the tables below exactly fit one of these rules: y = ax2, y = ax3,
a a ,y = or y = a x . For each set of data, plot the values of y against x and draw the x x2 graph. Select the most appropriate rule, and find the value of a. y=
a
b
x
−3
−2
−1
0
1
2
3
y
−8.1
−2.4
−0.3
0
0.3
2.4
8.1
x
−2
−1
0
1
2
3
y
−24
−6
0
−6
−24
−54
Chapter 2
Functions and transformations
117
c
−5
x
d
e
f
−2
−1
1
2
5
2
2
0.5
0.08
y
0.08
0.5
x
0
0.5
1
1.5
2
y
0
1.13
1.6
1.96
2.26
x
1
2
4
5
y
5
2.5
1.25
1
x
−3
−2
−1
0
1
0
−1.5
y
40.5
12
1.5
10 0.5 2 −12
a + b? x2 iii y
3 MC Which of the graphs below could be modelled by y = i
ii
y
y
x
x
iv
v
y
y x
x
A i only D i , i i and i v 4
x
B i , i i and i i i E i , i v , and v
C i v and v
WE34 It is believed that for the data in the table below, the relationship between x and y can be modelled by y = ax2 + b.
x
0
1
y
−3.2
−1
2
3
4
5
4.9
14.5
29
46.8
a Plot the values of y against x. b Plot the values of y against x2 and draw the line of best fit. c Find the values of a and b and hence the equation describing the original data. 5 The table below shows the values of 2 variables, x and y. x
−4
y
−28
−2
0
−13.5
−12.5
2 −10
4
6
4.3
41
Establish the mathematical model of the relationship between the variables, if it is known that it is of the form y = ax3 + b. 6 The table below shows the results, obtained from an experiment, investigating the frequency of a sound, ff, and the length of the sound wave, λ.
λ f
118
0.3 1130
0.5 680
1
3
5
8
10
340
110
70
40
35
Maths Quest 12 Mathematical Methods CAS for the TINspire
a Plot f against λ. b From the following relationships select the one which you think is suggested by the plot: a f = aλ2, f = , f = a λ . λ 1 c Based on your choice in part b, plot f against either λ2, or λ , draw in the line of best λ fit and use it to find the rule that connects the 2 variables. 7 For her science assignment, Rachel had to find the relationship between the intensity of the light, II, and the distance between the observer and the source of light, d. From the experiments she obtained the following results. d
1
I
270
1.5
2
120
2.5
68
3
43
30
3.5 22
4 17
a Use a graphics calculator to plot the values of I against d. What form of relationship does the graph suggest? b Nathan (Rachel’s older brother) is a physics student. He tells Rachel that from his studies a he is certain that the relationship is of the type I = 2 . Use this information to help d Rachel to find the model for the required relationship. 8 WE35 The table below gives the values of 2 variables, x and y. Establish the rule, connecting x and y, that fits these data. x
0
1
3
5
7
9
y
4
7
9
11
12
13
9 Joseph is a financial adviser. He is studying the prices of shares of a particular company over the last 10 months. Months
1
2
3
4
5
6
7
8
Price, $
6.00
6.80
7.45
8.00
8.50
8.90
9.30
9.65
9 10.00
10 10.30
a Represent the information graphically. b Establish a suitable mathematical model, which relates the share price, P, and the number of the month, m. c Use your model to help Joseph predict the share price for the next 2 months. eBoo k plus eBook Digital doc
Investigation Goal accuracy
Chapter 2
Functions and transformations
119
SUMMARY Graphs of the power functions
Name Parabola
Equation y = a(x − b)2 + c
Basic shape y
(b, c) 0
Cubic
y = a(x − b)3 + c
y=
a +c x−b
or y = a( x − b) − 1 + c Truncus
a +c ( x − b) 2 or y = a( x − b) − 2 + c
If a > 0 y≥c If a < 0 y≤c
Turning point at (b, c)
R
R
Stationary point of inflection at (b, c)
R \{b}
R \{c}
Horizontal asymptote y = c, vertical asymptote x=b
R \{b}
If a > 0 y>c If a < 0 y
Horizontal asymptote y = c, vertical asymptote x=b
x≥b
If a > 0 y≥c If a < 0 y≤c
End point at (b, c)
c x
0 b y
y=
y=a x−b +c or y = a( x − b) + c
Special feature
(b, b, c) c x
y
c 0
Square root
Range
R
x
y 0
Hyperbola
Domain
b
x
y
1 2
0
(b, c)
x
• The equation for any graph y = f (x) above can be written in the general form: y = af (x − b) + c. This form can be used to describe transformations of all of the functions considered. • For all of the above functions: 1. a is the dilation factor: it dilates the graph from the xaxis. 2. When an equation for these types of graphs is put into its general form of y = af (x − b) + c, the horizontal dilation can be described in terms of a vertical dilation. 3. If a < 0, the basic graph is reflected in the xaxis. 4. f (b − x) or f (−x + b) is the reflection of f (x + b) in the yaxis. 5. b translates the graph b units along the xaxis (to the right if b > 0, or to the left if b < 0). 6. c translates the graph c units along the yaxis (up if c > 0, or down if c < 0).
120
Maths Quest 12 Mathematical Methods CAS for the TINspire
• To put equations into general form: 7. If the coefficient of x is a number other than 1, to find the value of b and a, the equation should be transposed to make the coefficient of x equal to 1. For example, y = (3x + 5)2 + 4 = [3( x + 35 )]2 + 4 = 32 ( x + 35 )2 + 4 = 9( x + 35 )2 + 4 Hence, a = 9, b =
−
5 . 3
The absolute value function
• y = x means •
y = x, if x ≥ 0 and y = −x, if x < 0 y y = x x
Domain: R Range: R+ ∪ {0} • To sketch y =  f (x): 1. Sketch the graph of y = f (x). 2. Reflect the portion of the graph that is below the xaxis in the xaxis. Or: 1. Express the function in hybrid form with specific domains where the absolute value expression is positive and negative. 2. Sketch each rule for the specified domain. • For functions of the form y = a f (x) + c, a and c have the same impact on the graph of the absolute value function, as on the graphs of all other functions discussed in this section. Transformations with matrices
• The use of matrices to map transformations of points and equations can be summarised as follows, where (x′, y′) is the image of the point (x, y) under the transformation. x T y x T y x T y x T y
x' −1 = = y' 0 x' 1 = = y' 0
0 x −x = represents a reflection in the yaxis. 1 y y 0 x
x = − represents a reflection in the xaxis. y x' a 0 x ax = = = represents a dilation of a factor of a from the yaxis. y' 0 1 y y x' 1 0 x x = = = represents a dilation of a factor of a from the xaxis. y' 0 a y ay −1 y
• Transformations can be combined to represent more than one transformation. For example, 4x + 2 x ' 4 0 x 2 = + = 0 1 −y y' y 3 + 3 2 2 −
∴ x′ = 4x + 2 −y y′ = +3 2
Chapter 2
Functions and transformations
121
1
describes the following: dilation by a factor of 4 from the yaxis, a dilation by a factor of 2 from the xaxis, reflection in the xaxis, a horizontal translation of +2 and a vertical translation of +3. Sum and difference of functions
• For the graph of the sum/difference function, dom ((f (x) ± g(x)) = dom f (x) ∩ dom g(x). The graph of the sum/difference function can be obtained by using the addition of ordinates method. • For the product function, dom ((ff (x)g(x)) = dom f (x) ∩ dom g(x). Some features of the graph of the product function are as follows: ° the xintercepts of f (x)g(x) occur where either f (x) or g(x) have their xintercepts ° f (x)g(x) is above the xaxis where f (x) and g(x) are either both positive or both negative ° f (x)g(x) is below the xaxis where one of the functions f (x) or g(x) is positive and the other is negative. Composite functions and functional equations
• For the composite function f (g(x)) to be defined, the range of g must be a subset of the domain of f. f Furthermore, if f (g(x)) is defined, the domain of f (g(x)) equals the domain of g(x). • Equations involving algebra of functions, for example f (2x (2 ) = 2f 2 (x), are generally tested to determine if they are true for particular functions. ° To determine if an equation is true for a particular function, consider the LHS and RHS of the equation separately to determine if the equation holds true for all values of x. ° Alternatively, you may find a particular xvalue for which the equation does not work; that is, a counterexample. ° These types of equations can be investigated by defining the functions on the CAS calculator and then testing the algebraic function equation. Modelling
• Modelling is the process of finding the rule (mathematical model) that fits the given data. • To model: 1. Plot the original data on graph paper or use a CAS calculator. 2. Make an assumption of the model. 3. Transform the data in accordance with your assumption. 4. Check the assumption by plotting the transformed data (if correct, the graph will be a straight or nearly straight line). 5. Draw in a line of best fit. 6. Find the equation of the line (y = mx + c). 1 7. Replace x in the equation with the transformed variable (for example, x2, ). x
122
Maths Quest 12 Mathematical Methods CAS for the TINspire
CHAPTER REVIEW SHORT ANSWER
1 For the function x − 3)2 − 4: a state the coordinates of the turning point b state the domain and range c sketch the graph. a 2 The graph of y = + c has a vertical x−b asymptote at x = 2 and a horizontal asymptote at y = −1. a Find the values of b and c. This graph then undergoes the following transformations: • reflection in the xaxis • dilation by a factor of 3 from the xaxis • horizontal shift of 2 units right. b If the intersection of the two graphs is at (m, 2), find the value of m. c Hence find the equation of the transformed graph. 6x − 5 3 Express the function f ( x ) = in the form 3x + 1 b a+ . Hence describe the transformations x+c required to produce this curve from the graph of 1 y= . x 4 The graph of a cubic function has a stationary point of inflection at (1, 1). It cuts the yaxis at y = 4. Find the equation of the graph. 1 5 The graph of y = was dilated by the factor of x 4 from the xaxis, reflected in the xaxis and then translated 2 units to the left and 1 unit down. a State the equation of the asymptotes. b State the domain and range. c State the equation of the new graph. d Sketch the graph. 6 a State the changes necessary to transform the 1 y graph of y = 2 into x the one shown. b Find the equation of the graph.
−2
−1 − –23
x
7 The domain of a truncus is R \{1}, the range is (−∞, 2) and the graph cuts the yaxis at y = −3. Find the equation of the function. 8 The basic square root curve was reflected in both axes and then translated so that its intercepts at the axes were (0, 1) and (−5, 0). Find the size and the direction of the translations; hence, find the equation of the new graph. 2 − 2, 9 a Sketch the graph of y = 2 − ( x + 2)2 clearly showing the coordinates of the cusps, the intercepts with the axes and the position of the asymptotes. b State the domain and range of the graph in a. 10 The graph of ff: [−5, 1] → R where f (x ( ) = x3 + 6x 6 2 + 9x is shown. y
−3
0
x
a On the same set of axes sketch the graph of y = /f (x)) /. b State the range of the function with the rule y = /f (x)) / and domain [−5, 1]. [© VCAA 2006]
11 The point (−1, 3) undergoes a translation given by a the matrix to (2, 0). Find a and b and describe b the transformations involved. 12 The point (1, 2) undergoes a series of a 0 transformations given by the matrices and 0 b 1 then to (−7, 4). 2 a Find the values of a and b. b Find the image under the transformations of: i y=2 x ii y = x3 + x 13 A point on a curve ((x, y) undergoes a a 0 transformation descibed by to (x', y'), 0 2 where a is a real constant such that a > 0.
Chapter 2
Functions and transformations
123
a Find the values of x and y in terms of a, x' and y'. b If the point is on the curve y = 2x 2 2 − x, find the image of the curve in terms of a under this transformation. c If the point (3, 6) is on the transformed curve, find the value(s) of a and hence the rule of this image. 14 The graph of the function f : (−2, 1) → R, f (x ( ) = x3 + 2x 2 2 is shown below. y
f x) = x3 + 2x f( 2 2
2 1
(−2, 0)
−2 −1.5 −1 −0.5 0 −1
0.5
1
x
Let g(x) = f (x) + 1, and sketch this graph on the same set of axes. Hence, sketch ((f + g)(x). 15 The graph of the function f:( f −2, 2), f (x ( ) = /x / is shown below. (−2, 2)
f x) =x y f( 2 (2, 2) 1
0 −2 −1 −1
1 2 x
−2
Let g(x) = −f (x) + 1 and sketch the graph of this function on the same set of axes. Hence, sketch f (g(x)). 16 Let f (x ( ) = x2 determine which of the following relationships are true. a f (x) − f (−y) = f (x) − f (y) b −f (x) − f (y) = f (x) + f (y) c f (−x) + f (−y) = f (x) + f (y) d f (x) − f (−y) = f (x) + f (y) 17 The data in the table below exactly fit one of these a models: y = ax3, y = 2 or y = a x . x x
2
y
25
4
5
10
6.25
4
1
20 0.25
25 0.16
a Plot the values of y against x and use the scatterplot to choose a suitable model. b Plot the values of y against either x3, 1 or x x2 (depending on your choice in part a ). Did you choose the right model? Explain your answer. c Find the value of a.
124
1 The equation of a parabola is given by y = m − 2(x 2( + 3)2, where m > 0. The increase in m will result in: A the graph being thinner B the graph being wider C the increase of the domain D the increase of the range E the graph being shifted further to the right 2 The coordinates of the turning point of the parabola y = 2(3x + 6)2 − 3 are: A (6, −3) B (−6, −3) C (2, −3) − − − D ( 2, 3) E ( 2, 3)
(1, 3)
3
MULTIPLE CHOICE
bx − 3)3 + 1 is dilated in the 3 The graph of y = 23 (bx y direction by the factor of: 3 C 2b B 23 b A 23 3 2 3 D b E 3b3 4 The graph of y = 2 − (3 + 4x 4 )3 has a stationary point of inflection at: A ( 43 , 2)
B (
−
3 − , 2) 4 − 3 , 2) 4
C (−3, 2) D ( E (−3, −2) 2 5 If f ( x ) = + 1, then f (x ( ) + 2 will have: x A the horizontal asymptote y = 2 B the horizontal asymptote y = 1 C the horizontal asymptote y = 3 D the vertical asymptote x = 2 E the vertical asymptote x = 1 6 The equation of the graph shown is likely to be: −2 y A y= −1 x−2 −2 B y =1 x+2 −2 x −2 −1 C y= −2 x +1 2 −1 D y= x+2 −2 E y= −1 x+2 7 If the graph of y = 1 is reflected in the yaxis, x translated 43 units to the right and 2 units up, the resulting graph would have the equation: 1 3 A y= + x−2 4 3 +2 B y= 4x
Maths Quest 12 Mathematical Methods CAS for the TINspire
1 3 − 4x −1 +2 D y= 4x − 3 1 +2 E y= 3 − 4x C y = 2−
8 Which of the following is not true for the graph at right? y A The vertical asymptote is x = 2. B The horizontal asymptote is y = −4. x 2 C The domain is R\{2}. D The range is R\{−4}. −4 E The value of the yintercept is greater than −4. 9 To obtain the graph shown, we need to: A translate the graph of y 1 y = 2 one unit to the left x and reflect in the xaxis x −1 B translate the graph of 1 y = 2 one unit to the left x and reflect in the yaxis 1 C translate the graph of y = one unit to the x2 left and reflect in both axes 1 D translate the graph of y = 2 one unit to the x left and dilate it in the x direction E none of the above 10 The equation of this graph is of the form: A y = a x − m + n, a > 0 B y = a m − x + n, a > 0
y (m, n) x
C y = a x − m + n, a < 0 D y = a m − x + n, a < 0 E y = a x + m + n, a < 0 11 The equation of this graph could be: A y=d− a−x
y d
B y= x−a +d C y=d− x−a D y = c− a − x E y= a− x −c
b a c
x
12 The domain of the function f ( x ) = 2 x − 1 + 3 is: A [1, ∞) B [3, ∞) C [0.5, ∞) D [2, ∞) E [−1, ∞) 1 13 The range of the function y = − 3 − − 2 is: x2 A (1, ∞) B (−∞, −5] C [3, ∞) D (−∞, 2) − − E ( ∞, 2] 14 The equation of the graph shown in the diagram below is best described by: y A y = x + 2 + 2 2 B y = 2 − x + 2 C y = 2 − x + 2 x −2 D y = 2 − 2 − x E y = x + 2 − 2 15 The value(s) of k for which 2k + 1 = k + 1 are: − −2 2 B A 0 only C 0 or 3 3 − 1 2 E or 0 D 0 or 2 3 [© VCAA 2006] 16 Under the transformation T : R2 → R2 of the plane x 2 0 x 1 defined by T = + , the y 0 1 y − 3 image of the curve y =  x  has the equation: A B C D E
y=
1 2
x −1 + 3
1 2
x −1 − 3
y =  22x + 1 − 3 y = 2  x + 1 − 3 y = 2  x − 1 + 3 y=
17 If g:[2, 4] → R, where g(x ( ) = x2(x (x ( − 3), and h:(0, 3] → R, where h(x ( ) = 3 − x, then the (x function f (x ( ) such that f (x ( ) = g(x ( ) × h(x (x ( ) is defined (x by the rule: A f: f R → R, where f (x) = −x2(x − 3)2 B ff: (0, 3] → R, where f (x) = −x2(x − 3)2 C ff: [2, 3] → R, where f (x) = −x2(x − 3)2 D ff: (0, 4] → R, where f (x) = −x2(x − 3)2 E ff: [2, 3] → R, where f (x) = −x2(x − 3) (3 + x) 18 Given the function f ( x ) = x − 3 then f ° h could exist if h(x ( ) was defined as: (x A h(x) = −x + 2 B h(x) = x2 − 3 C h(x) = x3 D h(x) = x2 + 3 E h( x ) = x − 2
Chapter 2
Functions and transformations
125
19 The data in the following table exactly fit one of these models: y = ax2, y = ax3 or y = a x . x
1
2
3
y
0.3
2.4
8.1
The value of a is: A 2.4 C 2.7 E 0.3
4 19.2
B 1.2 D 0.9
20 For certain data the values of y were plotted against 1 and the line of best fit was drawn as seen on x the diagram below. The model that relates the variables x and y is: y A y = 20x 20 − 1 (1, 19) B y = 19x + 1 x −1 C y= 20 19 D y = −1 x (0.1, 1) 20 1 −1 E y= x– x
EXTENDED RESPONSE
1 The graph of y = f (x ( ) is shown at right. a Sketch the graph of each of the following functions on the same set of axes with the original graph and give the coordinates of the points A, B, C and D. i y = −f (x) ii y = f (−x) iii y = f (xx − 2) iv y = f (x) + 3 v y = 2f 2 (x) vi y = 1 − f (x + 1) b Maya, a fabric designer, wishes to use the curve of y = f (x) (red) to create a ‘wavy’ pattern as shown in the diagram at right. If she wants the waves to be 2 units apart vertically, suggest the best way she could alter the equation of y = f (x). (Remember a fabric has a fixed width!)
y
D(4, 6)
B C(2, 3) A 2 −3 −2 7 x
2 units apart
2 Consider the function f:R → R, f( f x) = (x − 1)2 (x − 2) + 1. a The coordinates of the turning points of the graph of y = f( f x) are (a, 1) 25 and (b, 27 ). Find the values of a and b. b Find the real values of p for which the equation f( f x) = p has exactly one solution. c For the following, k is a positive real number. x i Describe a sequence of transformations that maps the graph of y = f (x) onto the graph of y = f − 1. k x y = f − 1 ii Find the xaxis intercepts of the graph of in terms of k. k d Find the real values of h for which only one of EXAM TIP the solutions of the equation f (x + h) = 1 is positive. Students must ensure that they show [© VCAA 2004]
3 The graph of the function f: 1 f ( x ) = 2 is shown below. x (−0.5, 4)
y 4
(–2, –0.5)
∪ (0.5, 2) → R,
(0.5, 4)
3
[Assessment reports 1 and 2 2007 © VCAA]
f x) = 12 f(
2 (−2, 0.25)
126
x
1
−2.5 −2 −1.5 −1 −0.5 0
their working — if a question is worth more than one mark, students risk losing all available marks if only the answer is given and it is incorrect. The instructions at the beginning of the paper state that if more than one mark is available for a question, appropriate working must be shown. When students present working and develop their solutions, they should use conventional mathematical expressions, symbols, notation and terminology.
(2, 0.25) 0.5
1 1.5
2 2.5 x
Maths Quest 12 Mathematical Methods CAS for the TINspire
The graph of y = f (x) is to be transformed to become the graph of y = f (2x (2 ) + 1. a Describe these two transformations. b Create matrices to represent these transformations. c Use these matrices to find the images of the points (1, 1) and (2, 0.25) under these transformations, and use these values to deduce the images of the points (−1, 1) and (−2, 0.25). d On the same axes, sketch this transformed function, showing the coordinates of the four points from c above. e Using any method, find a rule for this transformed function. 4 A proposed section of a ride at an amusement park is to be modelled on the curve y = 1 (600 x + 25 x 2 − x 3 ), 500 where y is the height (in metres) of the ride above ground level and x is the horizontal distance (in metres). The xaxis represents ground level. It will travel through a tunnel from A to C; B is the lowest point in the tunnel and D is the highest point on the ride. y
A
D
Ex
C B
a Find the horizontal distance from A to E. b Find the greatest depth below ground level and the maximum height above ground level that the rollercoaster will reach in this section (correct to 2 decimal places). c Describe the impact that a dilation by a factor greater than 1 from the xaxis would have on: i the maximum depth and maximum height from b ii the point at which the rollercoaster would emerge from the tunnel iii the gradient of the slope at this point. 5 Lena and Alex are planning to buy a new house. They’ve been watching the prices of 3bedroom houses in a specific area, where they want to live, for the whole year. During each month they collected the data and then, at the end of the month, they calculated the average price for that month. The results of their calculations are shown in the table below. (The prices given are in thousands of dollars.) Month Price
1
2
3
4
5
6
7
8
9
10
11
12
240 248 255 261 266 271 273 274 275 274 272 270
a Plot the prices against the months. What model does the graph suggest? b If the model of the form y = a(x − b)2 + c is to be used for these data, what is (judging from the graph) the most suitable value for h? c Plot the values of y (the prices) against (x − b)2, where b is the value you’ve selected in part b . Comment on the shape of the graph. d Draw a line of best fit and find its equation. Hence, state the values of a and c in the model. e Write the equation of the model. f According to the Real Estate Institute, the property market is on a steady rise (that is, the prices are going up and are likely to rise further). Do the data collected by Lena and Alex support this theory? g Use the model to predict the average price for the next 2 months.
Chapter 2
Functions and transformations
127
h Lena and Alex were planning to spend no more than 250 000 for their new house. Several months ago the prices were in their range, but they could not find what they wanted. If the prices are going to behave according to our model, how long do they have to wait until the prices fall back into their range? 6 An eagle soars from the top of a cliff that is 48.4 metres above the ground and then descends towards unsuspecting prey below. The eagle’s height, h metres above the ground, at time t seconds can be modelled by a the equation h = 50 + , where 0 ≤ t < 25 and a is a constant. t − 25 a Find the value of a. b Find the eagle’s height above the ground after i 5 seconds ii 20 seconds. c After how many seconds will the eagle reach the ground? d Comment on the changes in speed during the eagle’s descent. e Sketch the graph of the equation. After 24 seconds, the eagle becomes distracted by another bird and reaches the ground exactly 2 seconds later. For this second part of the journey, the relationship between h and t can be modelled by the equation h = a(t − 24)2 + c. f Find the values of a and c. g Fully define the hybrid function that describes the descent of the eagle from the top of the cliff to the ground below. eBoo k plus eBook Digital doc
Test Yourself Chapter 2
128
Maths Quest 12 Mathematical Methods CAS for the TINspire
eBook plus
ACTIVITIES
Chapter opener Digital doc
• 10 Quick Questions: Warm up with ten quick questions on functions and transformations. (page 57) 2A
Transformations of the parabola
Digital docs
• Spreadsheet 132: Investigate transformations. (page 59) • Spreadsheet 108: Investigate the quadratic function in power form. (page 61) 2B
The cubic function in power form
Digital docs
• Spreadsheet 015: Investigate the cubic function in power form. (page 65) • Spreadsheet 236: Investigate graphs of functions. (page 70) 2C
The power function (the hyperbola)
Tutorial
• WE 9 int0522: Watch a worked example on sketching the graph of a hyperbola. (page 74) Digital docs
• Spreadsheet 051: Investigate the hyperbola. (page 72) • Spreadsheet 236: Investigate graphs of functions. (page 71) • WorkSHEET 2.1: Find the domain, range, coordinates of turning points and equations of asymptotes of various graphs. (page 78) • History of Mathematics: Investigate the history of major curves. (page 78) 2D
The power function (the truncus)
Digital doc
• Spreadsheet 236: Investigate graphs of functions. (page 85) 2E
The square root function in power form
Tutorial
• WE 16 int0523: Watch a worked example on implied domain and range. (page 88) Digital doc
• Spreadsheet 236: Investigate graphs of functions. (page 92) 2F
The absolute value function
Tutorials
• WE 20 int0524: Watch a worked example on sketching the graph of an absolute value function. (page 93) • WE 21 int0525: Watch a worked example on expressing an absolute value function as a hybrid function. (page 94)
Digital docs
• WorkSHEET 2.2: Identify transformations, state domain and range, sketch graphs of power functions and absolute value functions. (page 96) • Spreadsheet 002: Investigate graphs of absolute value functions. (page 96) 2G
Transformations with matrices
Interactivity
• Transformations with matrices int0247: Consolidate your understanding of using matrices to transform functions. (page 96) Tutorial
• WE 25 int0526: Watch how to use matrices to determine the resultant equation after transformations. (page 100) 2H
Sum, difference and product functions
Tutorial
• WE 27 int0257: Watch how to use addition of ordinates to sketch the sum of two functions. (page 104) 2J
Modelling
Digital docs
• Spreadsheet 076: Investigate modelling with functions. (page 117) • Investigation: Goal accuracy. (page 119) Chapter review Digital doc
• Test Yourself Chapter 2: Take the endofchapter test to test your progress. (page 128) To access eBookPLUS activities, log on to www.jacplus.com.au
Chapter 2
Functions and transformations
129
EXAM PRACTICE 1 SHORT ANSWER
15 minutes
1 The functions f g are graphed below. On the same axes sketch a graph of f + g. f
g
CHAPTERS 1 TO 2
1 −a x−b 1 C y= +b x+a 1 E y= +a x+b A y=
B y=
−1
+b a−x −1 D y= −a x−b
2 The graph sketched below is best represented by the rule:
0
1 mark
y
2 Write down the maximal domain of f : X → R where f ( x ) = x 2 − 3 .
a
1 mark
3 For y = (2x (2 − 3)2 − 1: a write down the ycoordinate of the turning point b determine the equation of the axis of symmetry. 2 marks
4 a Using the factor theorem, show that x + 2 is a factor of x3 − 7x − 6. b Given x3 − 7x − 6 = (x + 2)Q(x) where Q(x) is a quadratic factor, determine Q(x). 2 marks −2x 2
5 For what values of c does the graph of y = +c intersect the graph of y = −x2 + x − 2 at two distinct points? 2 marks
6 Sketch the graph of the function f :[−2,5) − R, f (x ( ) =  3(x 3( − 3)2 − 10 . 2 marks
MULTIPLE CHOICE
10 minutus
Each question is worth 1 mark. 1 Given a, b ∈ R, this graph could have the rule:
A B C D E
b
c
x
y = (xx – a)(xx – b)2(xx – c)2 y = (xx – a)(xx – c)2 y = (xx – a)3(xx – c)2 y = (xx – a)(xx – c) y = (xx – a)(xx – b)(xx – c)
3 The coefficient of the term in x4 in the expansion of 3 (2 x 2 − )5 is: x A −240 B −72 C −60 D 72 E 720 4 What is the equation of the horizontal asymptote of the graph of y = A y = −4 3 D y= 4
3x − 2 ? 4−x B y = −3
C y = −1
E y=3
5 The quadratic function f : D → R, f (x ( ) = 2(x 2( + 1)2 − 5 has a domain, D,, of [0, 3]. The range of f is: A [−1, 3] C [−5, 27] B R D [−3, 27] E [−5, ∞)
y
y=b x=a 0
130
x
6 The simultaneous linear equations 3x + ay = 12 and ax + 3y = 4a have infinitely many solutions for: A a ∈ [−3,3] B a=3 C a = 3 or a = −3 D a=9 E a=0
Maths Quest 12 Mathematical Methods CAS for the TINspire
EXTENDED RESPONSE
40 minutes
1 Severe tropical cyclone Vance devastated the town of Exmouth in northern WA in 1999. It produced a measured wind gust of 267 km/h, which is the highest wind speed ever recorded on mainland Australia. In order for it to be first categorised as a cyclone, its wind speed needed to exceed 119 km/h. Several years later, it is a peaceful day, with no wind detectable at 1 pm. By 3 pm, however, the wind speed is gusting to 200 km/h, and the residents know they are in trouble. a Let t be the time in hours after noon and v be the wind speed. Establish a linear model of the form v = at + b to represent the relationship between the wind speed and time. 2 marks b Using this model, determine to the nearest minute when the wind speed will be high enough for classification as a cyclone. 1 mark c Predict to the nearest minute when the cyclone will break the record for the highest wind speed ever recorded. 1 mark
d Explain why the linear model is unsatisfactory as a model for the cyclone’s behaviour. e
1 mark
i The wind speed actually peaks at 256 km/h at 5 pm. Use this data and the wind speed at 1 pm to help to create a quadratic model relating wind speed (v) to hours after noon (t). How well does this model match all the data provided?
2 marks
ii The wind speed actually returns to zero at 1 am the following morning, 12 hours later. Discuss how well this result matches the quadratic model. f
1 mark
i Establish a second quadratic model that exactly matches the following data: Time
Wind speed
1 pm
0 km/h
5 pm
256 km/h
1 am
0 km/h 1 marks
ii Evaluate how well this model represents the relationship between time and wind speed. g It has been suggested that a cubic function would be a better model for the data. Use all the data provided in order to establish a cubic model.
1 mark 2 marks
1020 1010 1000 990 980 970 960 950 0 km
Eye
Pressure Wind speed
175 150 125 100 75 50 25
Wind speed (km/h)
Pressure (hPa)
h The graph below shows how air pressure in hectopascals (P hPa) and wind speed (v km/h) relate to 1 distance (x) across a cyclone. It has been suggested that a truncus (general form v ( x ) 2 ) could represent x the shape of the relationship between speed and distance across the cyclone.
500 km
i What is the equation to the vertical asymptote for this relationship? ii Determine a rule for the relationship between speed (v) and distance (x). iii Determine a rule for the relationship between air pressure (P) and distance (x).
1 mark 2 marks 2 marks
eBoo k plus eBook Digital doc
Solutions Exam practice1
Exam practice 1
131
3
3A 3B 3C 3D 3E 3F
The index laws Logarithm laws Exponential equations Logarithmic equations using any base Exponential equations (base e) Equations with natural (base e) logarithms 3G Inverses 3H Literal equations 3I Exponential and logarithmic modelling
Exponential and logarithmic equations arEas oF sTudy
• Logarithm and exponent laws • Identification of appropriate solution processes for solving logarithmic and exponential equations • Numerical solutions of exponential and logarithmic equations • Solution of literal equations such as emx + n = k
• Application of algebraic and logarithmic properties to the simplification of expressions and the solution of equations • Finding the rule of an inverse function involving logarithmic and exponential equations • Application of logarithmic and exponential functions in modelling practical situations eBook plus
3a
Digital doc
The index laws
10 Quick Questions
A number in index form has two parts, the base and the index, power, exponent or logarithm. A number in index form is represented like this: Index, power, exponent or logarithm
ax Base
The index laws are summarised below. ax = x a
ax ÷ ay = ax  y
a y = ax =
(ax)y = axy
(ab)x = axbx
a0 = 1 (a ≠ 0)
ax a b = b x , b ≠ 0
a
132
1
ax × ay = ax + y
−x
=
1 (a ≠ 0) ax
maths Quest 12 mathematical methods Cas for the Tinspire
x
y
x
( a) y
x
Worked Example 1
Simplify
( 2 x 2 y3 ) 3 × 3( xy4 )2 . 6 x 4 × 2 xy4
Think
Write
1
Remove the brackets by multiplying the indices.
2
Add the indices of x and add the indices of y. Simplify 23 to 8 and multiply the whole numbers.
3
Subtract the indices of x and y. Divide 24 by 12.
(2 x 2 y 3 )3 × 3( xy 4 )2 23 x 6 y 9 × 3 x 2 y8 = 6 x 4 × 2 xy 4 12 x 5 y 4 =
24 x8 y17 12 x 5 y 4
= 2x3y13
For negative indices and fractional or decimal indices, the same rules apply.
Worked Example 2
Write in simplest form: 2
− 0.4
a 64 3 b 32 −2
c 125
3
using a CAS calculator.
THINK
WRITE/display x
a
b
2 y
1
Rewrite using the index law a y = a x .
2
Rewrite using a x = ( y a ) x .
= ( 3 64 )2
3
Simplify by taking the cube root of 64.
= 42
4
Square 4.
a 64 3 =
y
3
64 2
= 16 −0.4
1
Write as a fraction with a positive index.
2
Change 0.4 to 104 .
3
Simplify the fractional index.
4
Rewrite using the index law a y = a x .
5
Simplify by taking the 5th root of 32.
6
Square 2.
b 32
= = =
1 320.4 1 4
3210 1 2
32 5 x
y
=
1 5
( 32 )2 1 = 2 2 1 = 4
Chapter 3 Exponential and logarithmic equations
133
−2
c
1
c
To simplify 125 3 with a CAS calculator, open a Calculator page. −2
Complete the entry line as: 125 3 . Then press ENTER ·. Note: If the calculator is set to Approximate, the answer will be displayed as a decimal.
2
−2 3
Write the answer.
125
1
= 25
Worked Example 3
Simplify, leaving your answer with positive indices: a
−
−
a 2 b4 × ( a 3 b 4 )
1 −1 a2 b b − 1 2 3 b
−1
−1
using a CAS calculator.
THINK a
1
Remove the brackets by multiplying the indices.
2
Add the indices of a and of b. Place a5 in the denominator with a positive index.
3
b
WRITE/display
1
On a Calculator page, complete the entry line using the fraction template 12 1 a b as: 3 1 b 2

 1
a a 2b4 × (a3b 4)


= a 2b4 × a 3b4 
= a 5b8 =
b8 a5
b
−1
−
−
Then press ENTER ·.
2
Write the answer. The answer is expressed with positive powers.
12 −1 a b −1 2 3 b
−1
=
b
3
3 a
If the expression contains different numbers that do not have the same base, write each number as a product of prime factors.
134
Maths Quest 12 Mathematical Methods CAS for the TINspire
WorkEd ExamplE 4
Simplify
3 n × 6 n + 1 × 12 n − 1 : 32 n × 8 n
a showing working, and
Think a
WriTE/display
1
Write each number as the product of prime factors.
2
Remove the brackets.
3
In the numerator, add the indices of numbers with base 3 and add indices of numbers with base 2. Subtract the indices.
a
3n × 6 n + 1 × 12 n − 1 32 n × 8 n 3n × (3 × 2) n + 1 × (22 × 3) n − 1 = 32 n × ( 2 3 ) n 3n × 3n + 1 × 2 n + 1 × 2 2 n − 2 × 3n − 1 32 n × 2 3 n 3 n 3 n − 1 3 ×2 = 2n 3 × 23 n =
1
5
Write the term with a negative index in the denominator with a positive index.
= 3n × 2 1 = 3n × 2
6
Simplify.
=
1
On a Calculator page, complete the entry line using the fraction template as:
4
b
b with a CAS calculator.
3n 2
b
3n × 6 n + 1 × 12 n − 1 32 n × 8 n Then press ENTER ·.
2
Write the answer.
3n × 6 n + 1 × 12 n − 1 3n = 2 32 n × 8 n
WorkEd ExamplE 5
eBook plus
Simplify, leaving your answer with positive indices. 3 3 1 − a x 2− − b − + −1 2 1 x x +2 x −2 Think a
1
Tutorial
int0528 Worked example 5
WriTE
Rewrite the expression with positive indices.
a x
3 − x 2 1 = 2 − 3x 2 x
−2
−
Chapter 3
Exponential and logarithmic equations
135
2
Find the lowest common denominator.
b
=
1 3x 4 − 2 x2 x
1 − 3x 4 x2 1 3 b + = 1 1 x + 2 x − 2
Simplify.
1
Rewrite the question using positive indices.
2
Find the common denominator for the terms in the brackets and simplify.
=
Follow the process for division of fractions (change the division sign to a multiplication sign and invert the second fraction).
4
Find the common denominator.
5
Expand the brackets on the numerator.
6
Simplify the numerator by adding like terms.
Factorise the numerator and write the answer.
7
1 x2 − 3x 2 × 2 2 x x
=
3
3
=
3 x + 1 + 2x 1 − 2x x x
=
3x x + 1 + 2x 1 − 2x
=
3 x (1 − 2 x ) + x (1 + 2 x ) (1 + 2 x ) (1 − 2 x )
=
3x − 6 x 2 + x + 2 x 2 (1 + 2 x ) (1 − 2 x )
=
4x − 4x2 (1 + 2 x ) (1 − 2 x )
∴
4 x (1 − x ) (1 + 2 x ) (1 − 2 x )
REMEMBER
1. axay = ax + y ax 2. ax ÷ ay = ax − y or y = a x − y a 3. (ax)y = axy 4. a0 = 1, a ≠ 0 1 1 5. a x = x and − x = a x a a −
1
x
y
y
y
6. a y = a and a y = a x = ( a ) x Exercise
3A
The index laws 1 WE1
136
Simplify: ×
a
x3
x4
e
( x 2 )3 × x 5 ( x 5 )2
b x7 ÷ x2 f
5x 2 y 4 × 4 x 5 y 22 x 3 y 2
Maths Quest 12 Mathematical Methods CAS for the TINspire
c (x2)5 2 3 4 2 g (2 xy ) × 5( x y ) 4 x 5 y 3 × 3x 2 y 3

d (x 3)2
2 WE2 a
Simplify: −3
2 27 3
b 25 2
27 3 d 64
c 810.25
2
243 e 32
−3 5
3 Simplify: a
x4 × x5 x3
3
c 361.5
b 16 4
1
e 256 81
d 9 2 49 4 WE3
−3 4
Simplify, leaving your answer with positive indices. 1
a
3x 3y2
4
× (x2y)
−3 − 1 2 x 2y2
c
b 5
3 2 × 9x 2 y2
1 2 2x 2 y 3
3 2 a2b c d − 1 3a 2 bc 2
−1 −1 2 × 9x 5 y 2
−2
−
2 a 3 b3 ÷ 3 1 2 a c −
−
5 Simplify: 4y1
a x
c
1 16 2
×
(x 2y3) 1
−1 2 −1 2 x5y 4
b
6 WE4 Simplify: a 2n × 4n + 1 × 8n  1 d
32 × 2
−3
3 92
1 2 −1 × 83 x 3 y 2
−3 3 a 2b4 d ab 2
1 2 12 × 4x 5 y 2
−1 3 5x 3 y 4
b 3n × 9n  1 × 27n + 1
3
2
−2
1
9a − 3 b 2 ÷ 2 3 4a b
c 2n × 3n + 1 × 9n
× 16
7 Simplify: a 2n  1 × 3n × 6n + 1
b
52 × 3
−1
125 × 9
−2
÷
27 5
8 WE5 Simplify, writing your answer as a single fraction with positive indices. −
a x 1 +
1


b (x 1 + x 2) 2
−1
x 1 1 + − c −1 x +1 x 1 −1

1
d 2x(x2  y2) 1  (x  y)
−
9 MC 3 x + 3x is equal to: A 1
B
1 + 32 x 3x
C 3
− x2
D 6
x E 1 + 3 3x
Chapter 3 Exponential and logarithmic equations
137
3B
Logarithm laws If a > 0, then N = ax ⇔ loga (N) = x. For example, an expression in index form can also be rewritten in logarithmic form. 8 = 23 ⇔ log 2 (8) = 3 • Since ≠ 0 then loga (0) is undefined. • a0 = 1 ∴ loga (1) = 0 • a1 = a ∴ loga (a) = 1 Let m = ax ⇔ loga (m) = x and n = ay ⇔ loga (n) = y. • mn = ax × ay ∴ loga (mn) = x + y = ax + y = loga (m) + loga (n) x m m a • = ∴ log a = x − y n n ay = ax  y = loga (m) − loga (n) • mp = (ax)p ∴ loga (mp) = px = axp = p loga (m) ax
Changeofbase rule Suppose b = ax, then loga (b) = x. Consider N = by, then logb (N) = y. But N = by = (ax)y = axy. Therefore, loga (N) = xy = loga (b) logb (N). Thus, log b ( N ) =
log a ( N ) . log a (b)
This is called the changeofbase rule.
Worked Example 6
Evaluate: a log2 (1) b log5 (5). Think
Write
a
Log of 1 to any base is equal to zero: loga (1) = 0.
a log2 (1) = 0
b
If the number and base are equal the answer is 1: loga (a) = 1.
b log5 (5) = 1
Worked Example 7
Write in index form: a log2 (8) = 3 b logx (81) = 4. Think
138
Write
a Use ax = y ⇒ loga (y) = x.
a log2 (8) = 3 ⇔ 23 = 8
b Use ax = y ⇒ loga (y) = x.
b logx (81) = 4 ⇔ x4 = 81
Maths Quest 12 Mathematical Methods CAS for the TINspire
WorkEd ExamplE 8
Simplify: a log10 (5) + log10 (2)
b log4 (20)  log4 (5)
c log2 (16)
Think a
b
c
d
d log 5
( x ). 5
WriTE a log10 (5) + log10 (2) = log10 (5 × 2)
1
Rewrite using loga (mn) = loga (m) + loga n.
2
Simplify.
= log10 (10)
3
Simplify using loga (a) = 1. m Rewrite using log a = log a (m) − log a ( n). n
=1
1
20 5
b log 4 (20) − log 4 (5) = log 4
2
Simplify.
= log 4 (4)
3
Simplify using loga (a) = 1.
=1
1
Rewrite 16 as a number with base 2.
2
Rewrite using loga (mp) = p loga (m).
= 4 log2 (2)
3
Simplify using loga (a) = 1.
=4
1
Rewrite using y a = a y .
2
Rewrite using loga (mp) = p loga (m).
1
c log2 (16) = log2 (24)
d log 5
( x ) = log x 5
5
1 5
1
= 5 log 5 ( x )
WorkEd ExamplE 9
Simplify log3 (27) + log3 (9)  log3 (81). Think
WriTE/display
1
On a Calculator page, complete the entry line as: log3 (27) + log3 (9) + log3 (81) Then press ENTER ·.
2
Write the answer.
log3 (27) + log3 (9) +log3 (81) = 1
WorkEd ExamplE 10
eBook plus
Simplify: a 2 + log10 (3)
b 3 log3 (6)  3 log3 (18)
log 3 ( 9 ) c . log 3 ( 27 )
Chapter 3
Tutorial
int0529 Worked example 10
Exponential and logarithmic equations
139
Think a
b
1
c
Write
Write 2 as 2 log10 (10) because log10 (10) = 1. (mp)
a 2 + log10 (3) = 2 log10 (10) + log10 (3)
= log10 (102) + log10 (3)
= p loga (m).
2
Rewrite using loga
3
Rewrite using loga (mn) = loga (m) + loga (n).
= log10 (102 × 3)
4
Write 102 as 100.
= log10 (100 × 3)
5
Multiply the numbers in the brackets. (mp)
= log10 (300) b 3 log3 (6) − 3 log3 (18) = log3 (63) − log3 (183)
= p loga (m).
1
Rewrite using loga
2
Rewrite using m log a = log a (m) − log a ( n). n
3
Write 63 as 6 × 6 × 6 and 183 as 18 × 18 × 18.
4
Simplify.
1 = log3 3 3
5
Write the numbers with the base 3.
= log3 (3 3)
6
Rewrite using loga (mp) = p loga (m).
= −3 log3 (3)
7
Simplify using loga a = 1.
= −3 × 1 = −3
1
Write the numbers with the same base. It is not possible to cancel the 9 and the 27 because they cannot be separated from the log.
2
Rewrite using loga (mp) = p loga (m).
=
2 log3 (3) 3 log3 (3)
3
Cancel the logs because they are the same.
=
2 3
63 = log3 3 18 6×6×6 = log3 18 × 18 × 18
−
c
log3 (9) log3 (32 ) = log3 (27) log3 (33 )
Worked Example 11
Calculate the value of log2 (18), correct to 2 decimal places. Think
140
1
On a Calculator page, complete the entry line as: log2 (18) Then press ENTER ·. Pressing Ctrl / ENTER · will give a decimal approximation.
2
Write the answer.
Write/display
log2 (18) = 4.17, correct to 2 decimal places.
Maths Quest 12 Mathematical Methods CAS for the TINspire
REMEMBER
For a > 0: 1. loga (1) = 0 2. loga (a) = 1 3. loga (0) is undefined. 4. loga (x) does not exist when x ≤ 0. 5. loga (mn) = loga (m) + loga (n) Exercise
3B
m 6. log a = log a (m) − log a ( n) n 7. loga (mp) = p loga (m) log a ( N ) 8. log b ( N ) = log a (b)
Logarithm laws 1 WE6 Evaluate the following. a log3 (1) b log5 (1)
c log2 (2)
2 WE7 Write the following in index form. a log2 (16) = 4 b logx (25) = 2 d log3 (x) = 5 3
c log5 (125) = x
1 e log 5 5 = −1
Write the following in logarithmic form. a 23 = 8 b 34 = 81 − d 5x = 125 e 2 1 = 12
4 WE8
c 43 = x f x3 = 27
Simplify:
a log6 (3) + log6 (2) d log3 (81) 5
d log6 (6)
b log2 (10) − log2 (5)
c log2 (32)
log 5 1 5
f log3
b 3 log3 ( 3 x )
c log 2
e
1 27
Simplify: a log 2 ( x )
6 WE9 Simplify: a log4 (10) + log4 (2) − log4 (5) c 1 log10 (16) + log10 (52 )
x4 y 2
b log5 (25) + log5 (125) − log5 (625) d log3 (2) − log3 (10) + log3 (15)
2
e log2 (16) + log2 (8) + log2 (4) 7 WE10 Simplify: a 4 log2 (12) − 4 log2 (6) d
log 2 (64) log 2 (8)
b 2 + log5 (10) − log5 (2) e
log a ( x ) log a ( x )
8 WE11 Evaluate correct to 3 decimal places. a log10 (3) b log5 (4) d log2 (0.8) 9
c 1 + log2 (5)
e log4 (20)
c log10 (0.5) f log3 (60)
Simplify: a 5 log3 (x) + log3 (x2) − log3 (x7)
b 4 log2 (x) + log2 (x3) − log2 (x6)
c 3 log4 (x) − 5 log4 (x) + 2 log4 (x) e log10 (x2) + 3 log10 (x) − 2 log10 (x)
d 4 log6 (x) − 5 log6 (x) + log6 (x) f 4 log10 (x) − log10 (x) + log10 (x2)
g log5 (x + 1) + log5 (x + 1)2
h log4 (x − 2)3 − 2 log4 (x − 2) Chapter 3 Exponential and logarithmic equations
141
10 MC 2 log10 (5) − log10 (20) + log10 (8) is equal to: B −log10 (2) C 1 A log10 (2)
D −1
E log10 (4)
11 MC If loga (b) = 2, then b is equal to: A 0 B 1 C 2 D a E a2 12 If y = a log10 (x), find x when a = 2 and y = 3. Give your answer correct to 3 decimal places.
3C
Exponential equations The equation ax = b is an example of a general exponential (or indicial) equation and 2x = 32 is an example of a more specific exponential equation. To solve one of these equations it is necessary to write both sides of the equation with the same base if the unknown is an index or with the same index if the unknown is the base.
Worked Example 12
Solve for x in each of the following. 1 a 2x = 32 b 3 x = c 2 × 3x = 162 d 2(1 − x) = 16 27 Think a
1
b
The indices are equal because the base is 2 on each side of the equation.
1
Write 27 with base 3.
3 1
d
Write 32 with base 2, the same as the lefthand side.
2
2
c
Write a 2x = 32
2x = 25 x=5
1 27 1 = 3 3 − 3x = 3 3
b 3x =
Write 1 as a number with base 3. 33 Equate the indices. Divide both sides by 2 to leave 3x on the lefthand side.
2
Write 81 as a number with base 3.
3
Equate the indices.
1
Write 16 with base 2.
2
Equate the indices.
3
Solve for x.
x = −3 c 2 × 3x = 162
3x = 81
3x = 34 x=4 d
= 16 21 − x = 24
21 − x
1−x=4 x = −3
Worked Example 13
Solve 5x × 252x − 3 = 625 for x: a using index laws b with a CAS calculator. Think a
1
Write/display
Write all numbers with the same base.
5x × 252x − 3 = 625 5x × (52)2x − 3 = 54
2
142
Simplify.
5x × 52(2x − 3) = 54
Maths Quest 12 Mathematical Methods CAS for the TINspire
5x × 54x  6 = 54
3
Remove the brackets in the index.
4
Add the indices on the lefthand side.
55x  6 = 54
5
Equate the indices.
5x  6 = 4
6
Solve the equation.
5x = 10 x=2
b
1
On a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve(5x × 25(2x  3) = 625,x) Then press ENTER ·.
2
Write the solution.
Solving 5x × 25(2x  3) = 625 for x gives x = 2.
Sometimes it is possible to use the methods for solving quadratic equations to help solve indicial equations. Remember that 22x = (2x)2. WorkEd ExamplE 14
eBook plus
Solve for x in the following. b 32x  12 × 3x + 27 = 0
a (2x  16)(2x + 4) = 0 Think a
b
1
c 4x  2x + 3 + 16 = 0 WriTE
Use the Null Factor Law to solve by making each bracket equal to zero.
(2x  16) = 0 or (2x + 4) = 0
Solve each equation.
2x = 16 or 2x = 4
3
Write 16 as a number with base 2 but 4 can not be written with base 2.
2x = 24 or no real solution
4
Solve by equating the indices.
1
Write
as
(3x)2.
int0530 Worked example 14
a (2x  16)(2x + 4) = 0
2
32x
Tutorial
x=4 b 32x  12 × 3x + 27 = 0
(3x)2  12 × 3x + 27 = 0
2
Let 3x = a to make a simpler quadratic equation to solve.
a2  12a + 27 = 0, where a = 3x
3
Factorise.
(a  3)(a  9) = 0
4
Use the Null Factor Law by making each bracket equal to zero.
a  3 = 0, a  9 = 0
5
Solve for a.
6
Substitute back a = 3x.
3x = 3, 3x = 9
7
Write numbers with base 3.
3x = 31, 3x = 32
a = 3, a = 9
Chapter 3
Exponential and logarithmic equations
143
c
8
Equate the indices.
1
4x
Rewrite
as
(2x)2
x = 1, x = 2 and
2x + 3
as
2x
×
23.
4x – 2x + 3 + 16 = 0 − 2x × 23 + 16 = 0
c
(2x)2
(2x)2 − 2x × 8 + 16 = 0
2
Rewrite 23 as 8.
3
Let 2x = a to make a simpler quadratic equation to solve.
4
Replace a × 8 with 8a because the coefficient precedes the pronumeral.
5
Factorise.
(a − 4)(a − 4) = 0
6
Use the Null Factor Law.
a − 4 = 0, a − 4 = 0
7
Solve for a.
a = 4 and a = 4
8
The two factors are equal because a2 − 8x + 16 = 0 is a perfect square.
9
Substitute back a = 2x.
2x = 4
10
Write 4 as a number with base 2.
2x = 22
11
Solve by equating the indices.
a2 − a × 8 + 16 = 0 where a = 2x a2 − 8a + 16 = 0
x=2
Remember to always make the righthand side equal to zero when solving quadratic equations. It is a good idea to substitute your answer back into the original equation to check the accuracy of your work. If the base is not the same and the numbers cannot be written with the same base, then logarithms can be used. It is possible to take the logarithm of both sides of an equation provided the same base is used.
Worked Example 15
Solve for x in the following. Give your answers in exact form using base 10 and correct to 3 decimal places. a 5x = 10 b 2(x + 1) = 12 Think a
b
144
Write a
5x = 10 log10 (5x) = log10 (10)
1
Take the logarithm of both sides to base 10.
2
Use loga (mp) = p loga (m) and loga (a) = 1.
x log10 (5) = 1
3
Divide both sides by log10 (5).
x=
4
Use a calculator to simplify.
x ≈ 1.431, correct to 3 decimal places.
1
Take the logarithm of both sides to base 10.
2
Use loga (mp) = p loga (m) to simplify.
3
Divide both sides by log10 (2).
1 (exact form) log10 (5)
2(x + 1) = 12 log10 (2(x + 1)) = log10 (12)
b
(x + 1) log10 (2) = log10 (12)
Maths Quest 12 Mathematical Methods CAS for the TINspire
( x + 1) =
log10 (12) log10 (2)
4
Use a calculator to simplify the righthand side.
5
Solve for x.
∴ x=
log10 (12) − 1 (exact form) log10 (2)
x ≈ 2.585, correct to 3 decimal places.
Note: Logarithms in bases other than 10 may be used. Inequalities are worked in exactly the same way except that there is a change of sign when dividing or multiplying both sides of the inequality by a negative number. Worked Example 16
Solve the following equations for x, giving your answers both in exact form and correct to 3 decimal places. a 2x > 5 b 0.5x ≤ 1.4 Think
Write/display
1
On a Calculator page, press: • Menu b • 3:Algebra 3 • 1:Solve 1 Complete the entry lines as: solve(2x>5,x) solve(0.5x≤1.4,x) Press ENTER · after each entry.
2
Write the answers in exact form. The calculator defaults to base e when solving exponential equations in exact form. Note: ln (A) ⇔ loge (A). This will be discussed later in this chapter. Note: In part b , the inequality ≤ has been 1 changed to ≥ because log e 2 < 0.
a Solving 2x > 5 for x gives
Press Ctrl / ENTER · after solving each equation to obtain a decimal approximation. Write the answers correct to 3 decimal places.
a Solving 2x > 5 for x gives x > 2.322,
3
x>
log e (5) . log e (2)
b Solving 0.5x ≤ 1.4 for x gives −
x≥
7 log e 5 log e (2)
.
correct to 3 decimal places.
b Solving 0.5x ≤ 1.4 for x gives x ≥ −0.485, correct
to 3 decimal places.
REMEMBER
1. The equations ax = y and 2x = 32 are exponential equations. 2. Write numbers with the same base to help simplify problems. The most commonly used bases are 2, 3 and 5. 3. If the base is the same, equate the indices. 4. If the indices are the same, equate the bases.
Chapter 3 Exponential and logarithmic equations
145
5. Use the Null Factor Law to solve quadratic equations. 6. A negative number cannot be expressed in index form, for example, 4 cannot be expressed with base 2. 7. a2x = (ax)2 8. Take the logarithm of both sides of an equation or inequation using the same base. 9. Change the sign of an inequality when multiplying or dividing by a negative number. 10. loga (x) > 0 if x > 1 11. loga (x) < 0 if 0 < x < 1 ExErCisE
3C
Exponential equations 1
a
eBook plus
Index form
3x
b 10x = 1000
= 81
d 7x = 1 49
Digital doc
SkillSHEET 3.1
Solve for x in each of the following.
WE12
2
Digital doc
c 2x × 4x  1 = 16 4
×
Digital doc
b 5x × 52x + 1 = 625
243
d
6
7
Solve for x in the following. a  1) = 0 b 22x  6 × 2x + 8 = 0 2x x c 6 7×6 +6=0 d 4x  6 × 2x  16 = 0 x x e 9 =2×3 +3 b 42x  20 × 4x = 64
WE15 Solve for x correct to 3 decimal places.
a 2x = 5
b (0.3)x  1 = 10
c (1.4)2  x = 6
d 3 × 5x = 27
e 5 × 7x = 1
f 2x × 3x + 1 = 10
WE16 Solve for x correct to 3 decimal places. a 3x > 5 b 22x ≤ 7
d 7x ≥ 0.5 8
33 x + 1 = 81 9x − 2
9)(3x
5 Solve for x in the following. a 25x + 4 × 5x  5 = 0
SkillSHEET 3.3 Solving indicial equations by equating the bases
3x  1 =
WE14
(3x
eBook plus
mC The value of x for which 5 × 2x = 1255, correct to 3 decimal places, is:
A 7.971
B 897.750
Digital doc
D 7.972
E 2.059
Solving liner inequations
146
9
c (0.2)x > 3
e (0.4)x > 0.2
eBook plus SkillSHEET 3.4
d 22x  6 = 1
Solve for x in each of the following.
WE13
3x
c 52 x − 1 = 1 125
b 6x  2 = 216
a
Solving equations
= 32
Solve for x in each of the following.
eBook plus SkillSHEET 3.2
1 2x
e 243x = 3
a 3 × 2x = 48 3
c
C 897.749
mC The solution to the equation 102x = 3 × 10x + 4 is:
A log10 (1), log10 (4)
B
1,
D 0, 0.602
E log10 (4)
4
maths Quest 12 mathematical methods Cas for the Tinspire
C 10x + 1, 10x  1
3d
logarithmic equations using any base The equation loga (y) = x is an example of a general logarithmic equation. Laws of logarithms and indices are used to solve these equations.
WorkEd ExamplE 17
Solve for x in the following equations. a log2 (x) = 3 b log3 (x4) = 16
c log5 (x  1) = 2
Think a
1
b
c
WriTE
Rewrite using ax = y ⇔ loga (y) = x.
23 = x
x=8
2
Rearrange and simplify.
1
Rewrite using loga
(mp)
2
Divide both sides by 4.
3
Rewrite using ax = y ⇔ loga (y) = x.
4
Rearrange and simplify.
1
Rewrite using ax = y ⇔ loga (y) = x.
2
a log2 (x) = 3
= p loga (m).
b
log3 (x4) = 16 4 log3 (x) = 16 log3 (x) = 4 
3 4 = x 1 34 1 = 81
x=
c log5 (x  1) = 2
52 = x  1
x  1 = 25 x = 26
Solve for x.
The base of a logarithmic function and the base of an exponential function must be a positive real number other than 1. In the expression loga (x), a ∈ R+\{1}.
WorkEd ExamplE 18
eBook plus
Solve for x in each of the following: a logx (4) = 2
without technology
Think a
1 125
b log x
= − 3 with a CAS calculator.
Tutorial
int0531
Worked example 18
WriTE/display
1
Rewrite using ax = y ⇔ loga (y) = x.
2
Solve the quadratic equation.
logx (4) = 2 x2 = 4
a
x2  4 = 0 (x  2)(x + 2) = 0 x  2 = 0 or x + 2 = 0 x = ± 2
Chapter 3
Exponential and logarithmic equations
147
3
Check to see if solutions are valid.
x = 2 is the only solution.
This is the only solution. The solution x = −2 is not valid because the base of a logarithmic function must be a positive real number other than 1. b
1
On a Calculator page, press: • Menu b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve(log x
1 125
= − 3, x )
Then press ENTER ·.
2
Solving log x
Write the solution.
1 125
= − 3 for x gives x = 5.
Worked Example 19
Solve for x in the following. a log2 (16) = x b log 3 1 = x c log9 (3) = x 3
Think a
b
1
Write
Rewrite using
ax
= y ⇔ loga (y) = x.
2
Write 16 with base 2.
3
Equate the indices.
1
Rewrite using ax = y ⇔ loga (y) = x.
a log2 (16) = x
2x = 16 2x = 24 x=4
b log 3 1 = x 3
1 3 1 = 1 3
3x =
c
148
−1
3x = 3
1
2
Write 3 with base 3.
3
Equate the indices.
1
Rewrite using ax = y ⇔ loga (y) = x.
x = −1 c log9 (3) = x
9x = 3
(32)x = 3
2
Write 9 with base 3.
3
Remove the brackets.
32x = 31
4
Equate the indices.
2x = 1
5
Solve.
Maths Quest 12 Mathematical Methods CAS for the TINspire
x=
1 2
WorkEd ExamplE 20
Solve for x in the following. a log2 (4) + log2 (x)  log2 (8) = 3
b log10 (x) + log10 (x  3) = log10 (4)
Think a
b
1
WriTE
Simplify the lefthand side. Use loga (mn) = loga (m) + loga (n) and m log a = log a (m) − log a ( n) . n
2
Rewrite using ax = y ⇔ loga (y) = x.
3
Solve.
1
Simplify the lefthand side by using loga (mn) = loga (m) + loga (n).
a log2 (4) + log2 (x)  log2 (8) = 3
4 × x log 2 =3 8 x log 2 = 3 2 23 =
x 2
x = 2 × 23 =2×8 = 16 b log10 (x) + log10 (x  3) = log10 (4)
log10x (x  3) = log10 (4) x(x  3) = 4
2
Equate the logs.
3
Expand.
4
Solve the quadratic equation.
x2  3x  4 = 0 (x  4)(x + 1) = 0 x = 4 or x = 1
5
x > 0, x  3 > 0 because it is not possible to take the logarithm of a negative number, x > 3.
x = 4 is the only solution.
x2  3x = 4
rEmEmBEr
1. In a logarithmic equation, the unknown can be: (a) the number, log2 (x) = 5 (b) the base, logx (8) = 3 (c) the logarithm, log2 (4) = x. 2. In the expression loga (x), a is a positive real number other than 1. 3. Use laws of logarithms and indices to solve the equations. 4. Check that all solutions are valid. ExErCisE
3d eBook plus Digital doc
WorkSHEET 3.1
logarithmic equations using any base 1
WE17 Solve for x in the following.
a
i log5 (x) = 2
i i i log10 (x2) = 4 v log4 (2x  3) = 0
i i log2 (x) = 3 i v log3 (x + 1) = 3 vi log2 (x) = 5
v i i log5 (1  x) = 4
Chapter 3
Exponential and logarithmic equations
149
i i log4 (x) = −2
i log3 (x) = 4
b
i i i log2 (x3) = 12
i v log5 (x − 2) = 3 vi log3 (−x) = −2
v log10 (2x + 1) = 0 v i i log10 (5 − 2x) = 1
2 WE18 Solve for x in the following. a i logx (9) = 2
2 3
i i i log x 1 = − 3
i v logx (62) = 2
i logx (16) = 4
i i log x (125) =
8
b
i i log x (25) =
i i i log x
1 64
3 4
i v logx (43) = 3
= −2
3 WE19 Solve for x in the following. a
i log2 (8) = x
i i log 5 1 = x
i i i log4 (2) = x
i v log6 (1) = x
5
v log 1 (2) = x 2
b
1 16
i log3 (9) = x
i i log 4
i i i log8 (2) = x
=x
i v log8 (1) = x
v log 1 (9) = x 3
4 WE20 Solve for x in the following. a
i log2 (x) + log2 (4) = log2 (20) i i i log3 (x) − log3 (2) = log3 (5)
i i log5 (3) + log5 (x) = log5 (18) i v log10 (x) − log10 (4) = log10 (2)
v log4 (8) − log4 (x) = log4 (2) b
i log3 (10) − log3 (x) = log3 (5) i i i log2 (x) + log2 (5) = 1
i i log6 (4) + log6 (x) = 2 i v 3 − log10 (x) = log10 (2)
v 5 − log4 (8) = log4 (x)
5 Solve for x in the following. i log2 (x) + log2 (6) − log2 (3) = log2 (10) i i log2 (x) + log2 (5) − log2 (10) = log2 (3) a i i i log3 (5) − log3 (x) + log3 (2) = log3 (10) i v log5 (4) − log5 (x)+ log5 (3) = log5 (6) v log5 (x) + log5 (x − 2) = log5 (3) b
i log3 (x) + log3 (x + 2) = log3 (8) i i i log5 (x) + log5 (x + 20) = 3
i i log4 (x) + log4 (x − 6) = 2 i v log5 (x + 1) + log5 (x − 3) = 1
v log6 (x − 2) + log6 (x + 3) = 1
6 MC If loga (x) = 0.7, then loga (x2) is equal to: A 0.49
B 1.4
D 0.837
E 0
C 0.35
7 MC If log10 (x) = (a), then (log10 x)2 + log10 (x) − 6 becomes:
150
A (log10 (a))2 + log10 (a) − 6
B a2 + a + 6
D (a − 2)(a + 3)
E log10 (106x3)
Maths Quest 12 Mathematical Methods CAS for the TINspire
C log10 (x3) − 6
8 Solve for x in the following. a (log10 (x))2 + log10 (x) − 2 = 0 (Hint: Let a = log10 (x)) c (log2 (x))2 − 2 log2 (x) = 8 e (log3 (x))2 − log3 (x4) + 3 = 0 g log2 (x4) = (log2 (x))2 i log10 (x2 + 2x − 5) = 1 log10 ( x ) = 4, find x. 9 If log10 (2)
3E
b d f h j
(log10 x)2 − 2 log10 (x) − 3 = 0 (log2 (x))2 + 3 log2 (x) = 4 (log5 (x))2 − log5 (x3) + 2 = 0 log3 (x3) = (log3 (x))2 log3 (x2 − 3x − 7) = 1
Exponential equations (base e)
Euler’s number e, named after an 18th century Swiss mathematician, is a very important number used in problems involving natural growth and natural decay. Like π, it is irrational and has to be approximated: e = 2.718 281 828 459 . . . The number e can be used to find the value of an investment after a period of time, or the temperature of a liquid after it has been cooling. n 1 To find the value of e, take the expression 1 + and evaluate it for increasing values of n. n n 1 1 1 n=1 1 + n = 1 + 1 = 2 n
2
n
3
n
5
n = 2
1 1 1 + n = 1 + 2 = 2.25
n = 3
1 1 1 + n = 1 + 3 = 2.370 37
n = 5
1 1 1 + n = 1 + 5 = 2.48832
n = 10
1 1 1 + n = 1 + 10
n = 100
1 1 1 + n = 1 + 100
n
n
n
10
= 2.59374 100
1 1 n = 1000 1 + = 1 + 1000 n n
= 2.70481 1000
1 1 n = 10 000 1 + = 1 + n 10 000
= 2.716 92 10 000
= 2.71815
n
1 As n increases, 1 + becomes closer and closer to 2.718 281 or e, or n n 1 e = lim 1 + n→ ∞ n An answer given in terms of e is an exact answer. The laws of indices apply in the same way if e is the base, that is: • ex × ey = ex + y • ex ÷ ey = ex − y • (ex)y = exy • e0 = 1 1 −x • e = x e •
x ey
y
= ex
Chapter 3 Exponential and logarithmic equations
151
Worked Example 21
Solve for x in e3x = e. Think
Write
1
Write the equation.
e3x = e
2
Write e with a power of 1.
e3x = e1
3
Equate the indices.
4
Solve for x.
3x = 1 1 x= 3
CAS calculators have an ex function, which is treated in the same way as any other number.
Worked Example 22
Solve for x, showing working and with a CAS calculator. Express your answers in exact form and correct to 3 decimal places. a ex = 3 b ex − 3e−x = 2 Think
Write/display
Method 1: Technologyfree a
b
1
Write the equation.
2
As loge (e) = 1, take loge of both sides of the equation.
3
Rewrite using loga (xp) = p loga (x).
4
Solve for x.
1
Write the equation. 1 − Write e x as . ex
2
152
ex = 3
a
loge (ex) = loge (3) x loge (e) = loge (3) x = loge (3) ≈ 1.099, correct to 3 decimal places. −x
b ex − 3e
ex
=2 3 − x =2 e
3
Multiply every term by ex.
(ex)2 − 3 = 2ex
4
Make the righthand side equal to zero.
(ex)2 − 2ex − 3 = 0
5
Let ex = a.
a2 − 2a − 3 = 0 where a = ex
6
Factorise and solve for a.
(a − 3)(a + 1) = 0 a − 3 = 0 or a + 1 = 0 a = 3 or a = −1
7
Substitute ex for a.
ex = 3 or ex = −1
8
Solve for x by taking the log of both sides to base e.
loge (ex) = loge (3)
9
ex = 3 is the only solution because ex = −1 has no real solution.
x ≈ 1.099, correct to 3 decimal places.
Maths Quest 12 Mathematical Methods CAS for the TINspire
Method 2 : Technologyenabled 1
On a Calculator page, press: • Menu b • 3:Algebra 3 • 1:Solve 1 Complete the entry lines as: solve(ex = 3,x) solve(ex − 3e−x = 2,x) Press ENTER · after each entry.
2
Write the answers in exact form.
a Solving ex = 3 for x gives x = loge (3). b Solving ex − 3e−x = 2 for x gives
x = loge (3).
3
Write the answers in approximate form, correct to 3 decimal places.
a Solving ex = 3 for x gives x = 1.099, correct to
3 decimal places.
b Solving ex − 3e−x = 2 for x gives x = 1.099,
correct to 3 decimal places.
REMEMBER
1. Euler’s number e is an irrational number, which is approximated to 2.718 (correct to 3 decimal places). 2. Evaluate e by using the ‘ex’ button on the calculator. 3. The number e is an exact answer. Use the calculator to give an approximation if required. 4. The laws of indices apply in the same way if e is the base. 5. On a calculator, use the LN button to take the log of a number to base e. The LOG button defaults to base 10 if not specified. 6. loge (x) = ln (x). 7. ex > 0, that is, ex = −1 has no real solution. Exercise
3E
Exponential equations (base e) 1 Evaluate the following, giving your answer correct to 3 decimal places. 1
1
4
a e2
b e4
c e 2
d e 3
e
f
g ln (4)
h ln (5)
i loge (1.5)
j loge (3.6)
5
e
2 WE21 a ex = e
Solve for x in each of the following. b ex = e2 c ex − 2 = e4 1 − 1 g e3 x + 6 = e e e x + 1 = f e x − 2 = 2 e e
e
−1
d e2x = e
h e 2 x − 1 = e3
Chapter 3 Exponential and logarithmic equations
153
3 WE22 Solve for x in each of the following, giving your answer correct to 3 decimal places. a ex = 2
b ex = 5
c e x =
e ex = 1.3
f ex = 2.6
g 2ex = 6
1 2
d e x =
1 4
h 3ex = 12
4 Solve for x in each of the following, giving your answer correct to 3 decimal places. a (ex − 1)(ex + 2) = 0 b (e−x − 2)(e2x − 3) = 0 c (3e−x − 2)(2ex − 1) = 0 d (ex)2 − ex = 0
e (ex)2 − e × ex = 0
g 6 − 11ex + 3e2x = 0
h 18 − 23ex + 7e2x = 0
f (ex)2 − 7ex + 10 = 0
5 Solve for x in each of the following, giving your answer correct to 3 decimal places. a ex − 4e−x = 0
b ex − 15e−x − 2 = 0
c 5ex − 12e−x − 11 = 0
d 3ex + 6e−x − 11 = 0
e 4ex + 6e−x − 11 = 0
f ex + 2e−x = 3
6 Solve for x in each of the following, giving your answer correct to 3 decimal places. a ex > 1
b ex < e
c ex < 2
d e2x ≥ 4
e ex + 1 ≤ 6
f e1 − x ≤ 10
−x
g e > 0.75 7 If y = Ae−kt, and y = 19.6 when t = 2, and y = 19.02 when t = 5, find the value of the constants A and k. Give your answers correct to 2 decimal places. 8 For a body that has a higher temperature than its surroundings, Newton’s Law of Cooling is given by the formula θ = θ0e−kt, where θ is the difference between the temperature of the body and its surroundings after t minutes and θ 0 is the difference between the original temperature of the body and its surroundings. If the temperature of a freshly poured cup of coffee is 90 °C in a room with a constant temperature of 18 °C, and it cools to 65 °C after 10 minutes, find the value of k. Give your answer correct to 2 decimal places.
3F
Equations with natural (base e) logarithms CAS calculators have an LN function that can be used to find the log to base e in the same way that the LOG key is used to find the log of a number to base 10. CAS calculators default to base 10 when using the log key. In general, aloga(x) = x, where a ∈ R+\{1}. The laws of logarithms apply in the same way for base e as they do for base 10. ex = y ⇔ loge (y) = x.
Worked Example 23
Solve for x, giving your answer both in exact form and correct to 3 decimal places, given that loge (x) = 3. Think 1
154
Rewrite using ex = y ⇔ loge (y) = x.
Write/display
loge (x) = 3 e3 = x
Maths Quest 12 Mathematical Methods CAS for the TINspire
2
Write the answer in exact form.
3
On a Calculator page, press: • Menu b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve(ln (x) = 3,x) Then press ENTER ·.
4
Write the solution in approximate form, correct to 3 decimal places.
∴ x = e3
Solving loge (x) = 3 for x gives x = 20.086, correct to 3 decimal places.
Worked Example 24
Solve for x giving your answer correct to 3 decimal places where appropriate. a loge (3) = loge (x) b loge (x) + loge (3) = loge (6) Think a
b
Write a loge (3) = loge (x)
Since the base is the same, equate the numbers. 1 2 3
x=3
Rewrite using loge (mn) = loge (m) + loge (n).
b loge (x) + loge (3) = loge (6)
loge (3x) = loge (6) 3x = 6
Equate the number parts.
x=2
Solve for x.
REMEMBER
Exercise
3F
1. ex = y ⇔ loge (y) = x
2. e log
4. loge (1) = ln (1) = 0
5. loge (e) = ln (e) = 1
e
( x)
=x
3. 10 log
10
( x)
=x
6. loge (0) = ln (0) is undefined.
7. loge (mn) = loge (m) + loge (n) or
ln (mn) = ln (m) + ln (n)
m 8. log e = log e (m) − log e ( n) n
or
m ln = ln (m) − ln ( n) n
9. loge (mp) = p loge (m)
or
ln (mp) = p ln (m)
Equations with natural (base e) logarithms 1 WE23 Solve for x in each of the following giving, exact answers. a loge (x) = 1 b loge (x) = 2 c loge (x) = −2
d loge (x) = −1
Chapter 3 Exponential and logarithmic equations
155
2 Solve for x, giving exact answers when appropriate, otherwise, correct to 3 decimal places. a ln (2x) = 2 b ln (3x) = 1 c ln (x3) = 3 d ln (x2) = 2
e ln (x2) = 0.4
f ln (x3) = 0.9
g ln (x  1) = 1
h ln (2x + 1) = 2
3 WE24 Solve for x, giving exact answers when appropriate, otherwise, correct to 3 decimal places. a loge (x) = loge (2) b loge (x) = loge (5) c loge (x) + loge (3) = loge (9)
d loge (x) + loge (2) = loge (8)
e loge (x)  loge (5) = loge (2)
f 1 + loge (x) = loge (6)
4 Solve for x, giving exact answers. a loge (x) + loge (5)  loge (10) = loge (3)
b 2 loge (3) + loge (x)  loge (2) = loge (3)
c 3 loge (2) + loge (x)  loge (4) = loge (5)
d loge (4) + loge (3)  loge (x) = loge (2)
e loge (x) + loge (x + 1) = loge (2)
f ln (x + 1) + ln (2x  1) = ln (5)
5 mC If ln (y) = ln (x) + ln (a), then an equation relating x and y that does not involve logarithms is: x a A y=x+a B y = ax C y=xa D y= E y= a x 6 mC In the equation 2 loge (x)  loge (3x) = a, x = A 3ea B a C 3a
D loge (6a)
E no solution
7 Write the following equation without logarithms and with y as the subject. 2 loge (x) + 1 = loge (y) 8 If loge (x) = a and y = ea, express y in terms of x. 9 Solve for x the equation eln (x) = 2. eBook plus Digital doc
WorkSHEET 3.2
3G eBook plus Interactivity
int0248 Inverses
156
10 Five grams of a radioactive substance is decaying so that the amount, A grams, that is left after t days, is given by the formula A = 5ekt. a Find the value of A when the number of grams of the radioactive substance has been halved. b Rewrite the equation with the new value of A. c Rearrange the equation so that t is the subject of the equation. d If k = 0.005, find how long it will take for the number of grams of the radioactive substance to be halved. Give your answer correct to the nearest day.
inverses Inverse operations are opposite operations. Addition and subtraction are inverse operations and multiplication and division are inverse operations. Squaring and taking the square root are also inverse operations. The equation of the inverse of the function y = ex can be found by interchanging the x and y so that y = ex becomes x = ey. Using ax = y ⇔ loga (y) = x, x = ey becomes loge (x) = y or y = loge (x). Therefore y = ex and y = loge (x) are the equations of inverse functions. Thus two important properties follow: aloga(x) = x, x ∈ R+ and loga a(x) = x, x ∈ R.
maths Quest 12 mathematical methods Cas for the Tinspire
Worked Example 25
Calculate the inverse of y = 3ex + 1: a without the use of technology b using a CAS calculator. Think a
b
Write/display
1
Interchange x and y to write the inverse equation.
2
Divide both sides by 3.
3
In order to make y the subject, begin by rewriting the equation using ax = y ⇔ loga (y) = x.
4
Make y the subject.
1
Interchange x and y to write the inverse equation.
2
On a Calculator page, press: • Menu b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve(x = 3ey + 1,y) Then press ENTER ·.
3
Write the equation of the inverse.
y = 3ex + 1 Inverse is x = 3ey + 1 ey + 1 =
x 3
x log e = y + 1 3 x y = log e − 1 3 x = 3e y + 1
The inverse of y = 3ex + 1 is x y = log e − 1 where x > 0 3
Worked Example 26
Calculate the inverse of f(x) = 2 log10 (x − 1) + 1: a without the use of technology b using a CAS calculator. Think a
1 2
Write
Interchange x and y to write the inverse equation. In order to make y the subject, begin by subtracting 1 from both sides.
3
Divide both sides by 2.
4
Rewrite using ax = y ⇔ loga (y) = x.
Let y = 2 log10 (x − 1) + 1 Inverse is x = 2 log10 (y − 1) + 1 2 log10 (y − 1) = x − 1 log10 ( y − 1) = 10
x −1 2
x −1 2
= y −1
Chapter 3 Exponential and logarithmic equations
157
b
5
Add 1 to both sides.
6
Write the answer.
f −1 ( x ) = 10
1
Interchange x and y to write the inverse equation.
x = 2 log10 (y  1) + 1
2
On a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve(x = 2 log10 (y  1) + 1,y) Then press ENTER ·.
y = 10
x −1 2
x
Write the equation of the inverse.
4
The solution from the CAS can be simplified to obtain the solution found in a . Rewrite the answer with a negative power.
y = 10
5
Add the powers when multiplying indices of the same base.
y = 10
6
Simplify the powers and write the answer.
1 2
+
10 2 ∴y = 10
3
∴y
−
x −1 2
× 10 2
−
1+
+1
+1 x
1
x −1 = 10 2
f −1 ( x )
+1
x 1 + 2 2
+
1 2
+1
+1
+1
x −1 = 10 2
+1
rEmEmBEr
1. The equation y = loge (x) is an inverse function of y = ex. 2. To find the inverse of a function, interchange x and y and then make y the subject. ExErCisE
3G
inverses 1
eBook plus Digital doc
SkillSHEET 3.5 Inverses
158
WE25 Find the inverse of the following. a y = 2ex b y = ex + 1 c y = ex  1
d y = e2x  1
2 Find the equation of the inverse of the following. a y = 2 + ex b y = 2  ex d y = 2 + ex + 1 e y = 3  2ex  2 3
WE26 Find the inverse of the following. a f (x) = 2 loge (x) b f (x) = loge (x + 1) d f (x) = loge (2x  1) e f (x) = loge (2  x)
maths Quest 12 mathematical methods Cas for the Tinspire
e y = e2  x
f y = e2  3x
c y = 1  2ex f y = 2  3ex + 1 c f (x) = loge (x  1) f f (x) = loge (2  3x)
4 Find the equation of the inverse of the following. b y = 2  loge (x) a y = 2 + loge (x) d f (x) = 2  ln (x  1) 5
D
7
e y = 3 + 2 ln (x  1)
x −1 5
3
x −1 e 5
B e
+2 −2
E
3
x −1 5
3
x+2 e 3
5
C e
−2
x −1 5
+2
3
−1
mC If y = 5e 2x + 1  1, the equation of the inverse is:
A
1 x + 1 +1 ln 2 5
B
1 x + 1 −1 ln 2 5
D
1 x + 1 1 + ln 2 5 2
E
1 x + 1 1 − ln 2 5 2
(2x)
mC If eloge
A x2
C 5 ln ( x − 1) + 1 2
= y , then y equals: B loge (2x)
C e2x
E 2ex
D 2x
3h
f f (x) = 1  3 ln (x + 2)
mC If y = 5 loge (3x  2) + 1, the equation of the inverse is:
A e
6
c y = 2 + 3 loge (x)
literal equations An equation such as ekx = a, where k ∈ R and a ∈ R+, is called a literal equation. It does not have a numerical solution. The solution will be expressed in terms of the other pronumerals, in this case a and k, often called parameters. For this equation, the solution is: kx = ln (a) ∴ x =
1 ln (a), k ≠ 0, a ∈ R + k
WorkEd ExamplE 27 kx
Solve ekx = 5 + 2e
eBook plus
for x, where k ∈ R\{0}.
Think
Tutorial
WriTE/display
2 e kx
1
Rewrite the equation with positive powers.
e kx = 5 +
2
Multiply both sides by ekx.
(ekx)2 = 5ekx + 2
3
Let y = ekx and make the righthand side zero to obtain a quadratic equation in terms of y.
y2  5y  2 = 0
4
Solve for y using the quadratic formula.
y= y=
int0532 Worked example 27
− ( − 5) ±
( − 5)2 − 4 × 1 × − 2 2 ×1
5 ± 33 2
∴y =
Chapter 3
5 + 33 or 5 − 33 y= 2 2
Exponential and logarithmic equations
159
5
Substitute ekx for y.
e kx =
5 + 33 or kx 5 − 33 e = 2 2
6
Only the first solution is valid as ekx > 0.
e kx =
5 + 33 2
7
Rewrite the exponential equation in logarithmic form using ax = y ⇔ loga (y) = x or take logs of both sides. Divide both sides by k.
8
Write the solution and state the restriction for k.
9
On a Calculator page, press: • Menu b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: − solve(e k × x = 5 + 2e k × x,x) Note: You must put in a multiplication sign between the k and the x. Then press ENTER ·.
10
Write the answer.
5 + 33 kx = log e 2 ∴x =
5 + 33 1 log e , k ∈ R \ {0}. k 2
∴x =
33 + 5 1 log e k 2
Worked Example 28
1 Solve for x, given that log 2 ( x ) − 5 log 2 ( p) = log 2 ( 6 ) where p > 0, 2 a manually b with a CAS calculator. Think a
160
1
Write/display
Simplify the lefthand side using p loga (m) = loga (mp) and m log a (m) − log a ( n) = log a n
1 log 2 x 2 − log 2 ( p5 ) = log 2 (6) x log 2 5 = log 2 (6) p x =6 p5
2
Equate both sides.
3
Multiply both sides by p5.
x = 6 p5
4
Square both sides to obtain x.
x = (6p5)2
5
Write the solution and state the restriction for p.
∴ x = 36p10 where p > 0
Maths Quest 12 Mathematical Methods CAS for the TINspire
b
1
On a Calculator page, press: • Menu b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: 1 solve log 2 ( x ) − 5 log 2 ( p) = log 2 (6),x 2 Then press ENTER ·.
2
Write the answer. Note that the CAS gives the incorrect restriction p ≥ 0. The correct restriction is p > 0 for log2 (p) to be defined.
∴ x = 36p10
Some equations can only be solved using a CAS calculator, that is, no algebraic method exists. The following example demonstrates this. Worked Example 29
Solve the following equations using a CAS calculator. Give your answers correct to 3 decimal places. a ex = x3 b loge (x) = x − 2 Think
Write/display
1
On a Calculator page, press: • Menu b • 3:Algebra 3 • 1:Solve 1 Complete the entry lines as: solve(ex = x3,x) solve(ln (x) = x − 2,x) Press ENTER · after each entry.
2
Write the solutions correct to 3 decimal places.
a Solving ex = x3 for x gives x = 1.857 or x = 4.536. b Solving loge (x) = x − 2 for x gives x = 0.159 or
x = 3.146.
REMEMBER
1. An equation such as ekx = a, where k ∈ R and a ∈ R+, is called a literal equation. 2. Literal equations do not have numerical solutions. 3. The solution of a literal equation is expressed in terms of the other pronumerals, in this case a and k, often called parameters.
Chapter 3 Exponential and logarithmic equations
161
ExErCisE
3h
literal equations 1 Solve 5e2x = a for x, where a ∈ R+. 2 If log3 (D) = cy + log3 (Z), solve for D. 3 Solve emx + n = 3k for x, where m ∈ R\{0} and k ∈ R+. 4 Solve for q given that 2 log3 (p + 5q) = 4. x2 z2 x5 y4 y 5 Prove that log10 3 2 + log10 3 − log10 3 4 − log10 4 = 0. z x y z z z x 6 Consider the exponential equation 9xb × 273a = 81. a Find x in terms of a and b, where a ∈ R, b ∈ R\{0}. b Hence find the value of x if a = 2 and b = 3. 7 Solve 42x  b = 20 for x, where b ∈ R. 8 Solve 2x  1 = 3x + a for x, where a ∈ R. 9 If y = m + Rebx, solve for x. 10 Solve for a given that (log2 (5a))2 = 16b2. a 11 WE27 Solve e kx = 3 + kx for x, where a ≥ 0, k ∈ R\0. e 1
12 WE28 Solve for x given that 2 log 4 ( x ) − 3 log 4 ( y ) = log 4 (3). 13 Solve for b given that 2 loge (a)  5 loge (b)  2 = 0, where a, b ∈ R+. 14 WE29 Solve the following equations using a CAS calculator. Give your answers correct to 3 decimal places. a ex = 3x b x+2=e x c x2  1 = e2x 15 WE29 Solve the following equations using a CAS calculator. Give your answers correct to 3 decimal places. a ln (x) = 2  x b ln (x  2) = x  4 c x2  1 = ln (2x) 16 Solve for x given that log7 (x) = log4 (p).
3i eBook plus eLesson
eles0091 Exponential and logarithmic modelling
162
Exponential and logarithmic modelling Exponential and logarithmic functions can be used to model many real situations involving natural growth and decay. Continuous growth and decay can be modelled by the equation A = A0ekt, where A0 represents the initial value, t represents the time taken and k represents a constant. For continuous growth, k is positive, but for continuous decay, k is negative. Logarithms to base 10, often called common logarithms, are used in scientific formulas for measuring the intensity of earthquakes, the acidity of solutions and the intensity of sound.
maths Quest 12 mathematical methods Cas for the Tinspire
WorkEd ExamplE 30
eBook plus
In the town of Ill Ness, the number of cases of a particular disease, Tutorial D, can be modelled by the equation D = D0ekt, where t is the time in years. int0533 Using available medication the number of cases is being reduced Worked example 30 by 20% each year. There are 10 000 people with the disease today. a How many people will have the disease after one year? b Find the value of k correct to 3 decimal places. c Write the equation substituting values for k and D0. d Find how long it would take for the number of people with the disease to be halved. Give your answer correct to the nearest year. e How long would it take for the number of people with the disease to be reduced to 100? Give your answer correct to the nearest year. Think a
b
d
a (100  20)% = 80%
1
Find the percentage of people with the disease after one year.
2
Find 80% of the original number.
80% of 10 000 = 8000
3
Write a sentence.
Therefore, 8000 people will have the disease after one year.
1
Substitute t = 0 and D = 10 000 into the given equation.
2
c
WriTE
Substitute t = 1 and D = 8000 into [1], and solve for k.
Use the given equation D = D0ekt. 1
Substitute 5000 for D.
2
Simplify by dividing both sides by 10 000.
3
Take loge of both sides.
4
Solve for t.
b
D = D0 ekt When t = 0 and D = 10 000, 10 000 = D0 ek × 0 10 000 = D0 × 1 10 000 = D0 So D = 10 000ekt
[1]
When t = 1 and D = 8000, 8000 = 10 000ek × 1 8000 = ek 10 000 0.8 = ek loge 0.8 = loge (ek) 0.223 = k log (e) e 0.223 = k × 1 k = 0.223 0.223t
c D = 10 000e
0.223t
d D = 10 000e
When D = 5000, 5000 = 10 000e 0.223t 0.223t
0.5 = e
loge (0.5) = 0.223t t=
log e (0.5) − 0.223
≈ 3.108 (3 decimal places) 5
Write a sentence.
It would take about 3 years.
Chapter 3
Exponential and logarithmic equations
163
e
−0.223t
D = 10 000e
e
1
Write the equation.
2
Substitute 100 for D.
3
Simplify by dividing by 10 000.
4
Take loge of both sides.
5
Solve for t.
6
Write the answer in a sentence.
When D = 100, − 100 = 10 000e 0.223t −0.223t
0.01 = e
loge (0.01) = −0.223t t ≈ 20.651 (3 decimal places) It would take about 21 years.
REMEMBER
1. Make a note of all the information that has been given. 2. Give your answers to the correct number of decimal places. 3. Continuous growth and decay is modelled by the equation A = A0ekt, A0 represents the initial value (that is, when t = 0) and k represents a constant.
Exercise
3I
Exponential and logarithmic modelling 1 WE30 Changing δgluconolactone into gluconic acid can be modelled by the equation − y = y0e 0.6t, where y is the number of grams of δgluconolactone present t hours after the process has begun. Suppose 200 grams of δgluconolactone is to be changed into gluconic acid. a Find the value of y0. b Write the equation replacing y0 with your answer. c How many grams of δgluconolactone will be present after 1 hour? Give your answer correct to the nearest gram. d How long will it take to reduce the amount of δgluconolactone to 50 grams? Give your answer correct to the nearest quarter of an hour. −
2 The decay of radon222 gas is given by the equation y = y0e 0.18t, where y is the amount of radon remaining after t days. When t = 0, y = 10 g. Give all answers to the nearest whole number. a Find the value of y0. b Write the equation substituting your value of y0. c What will be the mass after 1 day? d How many days will it take for the mass to reach 1 g? 3 The equation y = A + B loge (x) relates two variables x and y. The table below shows values of x and y. x y a b c d
164
1 3
2 4.386
3 m
Find the value of A and B correct to the nearest whole number. Write the equation relating x and y substituting values for A and B. Using your new equation, find the value of m correct to 3 decimal places. If y = 7.6, find x correct to the nearest whole number.
Maths Quest 12 Mathematical Methods CAS for the TINspire
4 An amount of $1000 is invested in a building society where the 5% p.a. interest paid is compounded continuously. The amount in the account after t years can be modelled by the equation A = A0ert, where r is the continuous interest rate. a Find the value of A0 and r. b Write the equation substituting values of A0 and r. c Find the amount in the bank after i 1 year ii 10 years. Give your answer correct to the nearest dollar. d How long will it take for the investment to double in value? Give your answer to the nearest year. 5 The number of people living in Boomerville at any time, t years, after the first settlers arrived can be modelled by the equation P = P0ekt. Suppose 500 people arrived on 1 January 1850, and by 1 January 1860 there were 675 people. a What is the value of P0? b Find the value of k correct to 2 decimal places. c Write the equation substituting values for P0 and k. d What will be the population on 1 January 1900? Give your answer to the nearest 10 people. e When will the population be 2000? 6 A cup of soup cools to the temperature of the surrounding air. Newton’s Law of Cooling can be − written as T − TS = (T0 − TS)e kt, where T is the temperature of the object after t minutes, and TS is the temperature of the surrounding air. The soup cooled from 90 °C to 70 °C after 6 minutes in a room with an air temperature of 15 °C. a Find the values of TS, T0 and k correct to 2 decimal places. b Write the equation substituting the values for TS, T0 and k. c Find the temperature of the soup after 10 minutes. Give your answer to the nearest degree. d How long would it take for the soup to be 40 °C? Give your answer to the nearest minute. e If the soup is placed in a refrigerator in which the temperature is 2 °C, how long will it take for the soup to reach 40 °C? Use the same value of k and give your answer to the nearest minute. 7 The diameter of a tree for a period of its growth can be modelled by the equation D = D0ekt, where t is the number of years after the beginning of the period. The diameter of the tree grew from 50 cm to 60 cm in the first 2 years that measurements were taken. a Find the values of D0 and k. b Write the equation using these values. c How much will it have grown in the first 5 years? Round to the nearest centimetre. d How long will it take for the tree’s diameter to double? Round to the nearest year. −
8 The decay of a radioactive substance can be modelled by the equation M = M0e kt, where M grams is the mass of the substance after t years. After 10 years the mass of the substance is 98 grams and after 20 years the mass is 96 grams. a What was the mass of the substance initially? Give your answer to the nearest gram. b Find the value of k. Give your answer to 3 decimal places. c Write the equation using these values. d Find the mass of the substance after 50 years. e How long would it take for the mass to be halved? 9 The number of bacteria present in a culture at any time, t hours, can be modelled by the equation N = N0ekt. a If the original number is doubled in 3 hours, find k correct to 2 decimal places. b Write the equation substituting the value of k. c Find the original number of bacteria if there were 2500 bacteria after 4 hours. Give the answer correct to the nearest thousand.
Chapter 3 Exponential and logarithmic equations
165
d Write the equation substituting your value for the original population. e Find the number of bacteria present after 8 hours. Give your answer correct to the nearest thousand. 10 The intensity of light d metres below the surface of the sea can be modelled by the equation − I = I0e kd. Divers in the Sea of Loga have found that the intensity of light is halved when a diver is 5 metres below the surface of the water. a Find the value of k correct to 4 decimal places. b Write the equation substituting the value of k. c Find the percentage of light available at a depth of 10 metres. d If artificial light is necessary when the intensity of light is less than 0.1 of the intensity at the surface (I < 0.1I0), find how deep a diver can go before artificial light is necessary.
166
Maths Quest 12 Mathematical Methods CAS for the TINspire
Summary Index laws
• ax × ay = ax + y • ax ÷ ay = ax − y x y xy • (a ) = a • a0 = 1 1 1 − • a x = x and − x = a x a a x
1
• a y = y a and a y = y a x • ax = y ⇔ loga (y) = x Logarithm laws
loga (1) = 0 loga (a) = 1 loga (0) is undefined. loga (mn) = loga (m) + loga (n) m • log a = log a (m) − log a ( n) n • loga (mp) = p loga (m) log a ( N ) • log b ( N ) = log a (b) • • • •
Exponential equations
• To solve exponential equations: 1. write all terms with the same base, write terms with the smallest possible base or take the logarithm of both sides of the equation 2. then solve the equation. • A negative number cannot be expressed in index form. • If 0 < x < 1, then loga (x) < 0 and if x > 1 then loga (x) > 0. • It is not possible to take the logarithm of a negative number. Exponential equations (base e) n
1 • Euler’s number e = lim 1 + = 2.718 281 828459 ... h →∞ n • The laws of indices and logarithms apply in the same way when using e. Equations with natural (base e) logarithms
• To solve logarithmic equations use the laws of logarithms and indices. Inverses
• • • •
The equation y = loge (x) is an inverse function of y = ex. To find an inverse, interchange x and y. a log ( x ) = x, x ∈ R + log a (a x ) = x, x ∈ R a
Literal equations
• An equation such as ekx = a, where k ∈ R and a ∈ R+, is called a literal equation. • Literal equations do not have numerical solutions. • The solution of a literal equation is expressed in terms of the other pronumerals, in this case a and k, often called parameters.
Chapter 3 Exponential and logarithmic equations
167
chapter review 2 2 log3 (x) + 4 log3 (x) − log3 (x6) is equal to:
Short answer −2
3 2 − 4 1 Simplify 4 x 5 y 3 × 2 x 3 y 5 , leaving your answers with positive indices.
A 0
6x B log3 x 6
2 If log2 (5) = 2.321 and log2 (9) = 3.17, find
C log3 (6x  x6)
8x D log3 x 6
E 6log3 (x  x6)
log 2 95 . 3 Solve 3 × 2x − 7 = 17 for x. log 2 (32) . log 2 (8) 1 5 Solve log x (2) = for x. 3 6 If 4e(2 − x) = 128, find x, giving your answer in exact form. 4 Evaluate
7 Solve for x in loge (5) + loge (x) − loge (2) = loge (10). 8 Find the rule of the inverse function to y = 3e2x − a. 9 Find the rule of the inverse function of y = loge (1 − x) + 3. 10 Solve the equation loge (3x + 5) − loge (2) = 2 for x.
Exam tip Common mistakes include cancelling out the logarithms, adding to obtain 3x + 7, or multiplying to get 6x + 5. [Assessment report 1 2007]
[© VCAA 2007]
11 Solve 6e3x = k for x, where k ∈ R+. 12 Solve 3eax + b − 6k = 0 for x and state the necessary restrictions for the parameters a, b and k. 13 Solve for x given that 4 log2 (ax + b) = 12. 14 Solve for x given that log2 (x) = y + log2 (z). State the restrictions for the parameters. Multiple choice
1 If a > 1, the solution of x for the equation x = a2 is: A 1 B a negative number less than 1 C a positive number less than 1 D a negative number greater than 1 E a positive number greater than 1
168
3 The solution of the equation 3e2x = 4 is closest to: B −0.405 A −0.406 C 0.143 D 0.144 E 0.575 [© VCAA 2005] 4 Evaluated to 3 decimal places, log3 (24) is: A 2.892 B 2.893 C 0.345 D 0.346 E 1.380 5 The solution set of the equation (2x − 1)(22x − 4) = 0 over R is: A {0, 1} B {0, 2} C {1, 2} D {1, 4} E {2, 4} −
6 The solution set of the equation ex − 12e x = −4 over R is: A loge (2), loge (6) B ex − 2, ex + 6 − C 2, 6 D loge 2 E loge 6 7 If loge (x) = a, then e2a + 3ea − 2e−a is equal to: 2 a 2 B x 2 + 3 x − x
A a 2 + 3a −
C 2 log e (a) + 3 log e (a) −
2 log e (a)
d log e ( x 2 ) + 3 log e ( x ) −
2 log e ( x )
E (ea + 2)(e + 1) 8 If loge (2x) = a, then x is equal to: A 2ea ae D 2
Maths Quest 12 Mathematical Methods CAS for the TINspire
B 2ae E e2a
C
ea 2
9 If ex + 4 = e2x − 1, then x is equal to: − 5 B 5 A 3 D 5
C e5 −
14 If y = loga (7x − b) + 3, then x is equal to: 1 A 1 a y − 3 + b B (a y − 3) + b 7 7 b 1 y−3 + b) D a y − 3 − C (a 7 7 y−3 E log a (7 − b) [© VCAA 2007]
5
E e 3 10 The solution(s) to the equation 2 ln (x) = ln (x + 4) + ln 2 is/are: B 2, −4 A 2, 4 C 1 D 2 E 4 11 If 2a =
x2 , then y is equal to: y
x 2a x2 C a 2 A
x B 2a x D 22 a
2
x2 4a 12 The equation which is the inverse of y = ex − 1 is: A y = loge (x) − 1 B y = loge (x − 1) C y = loge (x) + 1 D y = loge (x + 1) − E y = e x − 1 E
13 The air pressure P cm of mercury at h km above sea level can be modelled by the equation − P = 76e 0.13h. One kilometre above sea level the pressure has: A increased by approximately 9 cm B decreased by approximately 9 cm C increased by approximately 41 cm D decreased by approximately 41 cm E neither increased nor decreased significantly
15 If 2 loge (x) − loge (x + 2) = 1 + loge (y), then y is equal to: x2 A 10( x + 2) 2x −1 B x+2 x2 C x+2 2x D x+2 x2 E e( x + 2) [© VCAA 2003] 16 If 3log ( x + 4 ) = y , then y equals: A log (x + 4) B x + 4 C (x + 4)3 D 3 log (x + 4) E −4 3
17 The solution set of the equation e4x − 5e2x + 4 = 0 over R is: A {1, 4} B {−4, −1} C {−2, −1, 1, 2} D {−loge (2), 0, loge (2)} E {0, loge (2)} [© VCAA 2007]
Extended response
a 1 log e = x . If loge (a) = 0.6932, find the value of x, giving your answer correct to 2 decimal places. 3 x x 2 x3 + + ... 2 e x = 1 + + 1! 2! 3! a Write the next 3 terms. b Substitute x = 1 in the equation using the first 7 terms. c Show that e ≈ 2.7182. 3 The apparent brightness of a star can be found using the formula B = 6  2.5 log10 A, where A is the actual brightness of that star. Find the apparent brightness of a star with actual brightness of 3.16. a 4 Earthquake magnitude is often reported on the Richter scale. The magnitude, M, is given by M = log10 + B, T where a is the amplitude of the ground motion in microns at the receiving station, T is the period of the seismic wave in seconds, and B is an empirical factor that allows for the weakening of the seismic wave with the increasing distance from the epicentre of the earthquake.
Chapter 3 Exponential and logarithmic equations
169
Find the magnitude of the earthquake if the amplitude of the ground motion is 10 microns, the period is 1 second and the empirical factor is 6.8. 5 Five grams of a radioactive substance is decaying so that the amount, A grams, that is left after t days, is given by the formula A = 5e kt. a Find the value of A when the number of grams of the radioactive substance has been halved. b Rewrite the equation with the new value of A. c Rearrange the equation so that t is the subject of the equation. d If k = 0.005, find how long it will take for the number of grams of the radioactive substance to be halved. Give your answer correct to the nearest day. 6 A school in the suburb of Bienvenue opened with 30 students in February 1995. It has been found for the first years after opening that the number of students enrolled in the school t years after opening can be modelled by the equation N = N0ekt. There were 45 students enrolled in February 1996. a Find the values of N0 and k. b Write the equation substituting the values for N0 and k. c How many students will there be 5 years after the opening? d How many years will it take for the school to have 1000 pupils? Another school in the suburb of Enbaisse has a declining student population. The number of students enrolled at any one time can be modelled by the equation E = E0e rt. There are 1000 students enrolled in February 1995 and 900 in February 1996. e Find the values of E0 and r. f Write the equation substituting the values for E0 and r. g How many students will be enrolled after 5 years? h How many years will it take for the two schools to have approximately the same number of pupils? i What will the population be then? Use the calculator value in the working and do not round off until the final answer. kx
7 Solve ekx = 4 + ke
for x, where k ∈ R+.
eBook plus Digital doc
Test Yourself Chapter 3
170
maths Quest 12 mathematical methods Cas for the Tinspire
eBook plus
aCTiviTiEs
Chapter opener Digital doc
• 10 Quick Questions: Warm up with ten quick questions on exponential and logarithmic equations. (page 132) 3A
The index laws
Tutorial
• WE 5 int0528: Watch a worked example on writing expressions with positive indices. (page 135) 3B
Logarithm laws
Tutorial
• WE 10 int0529: Watch a worked example on simplifying logarithmic expressions. (page 139) 3C
Exponential equations
Tutorial
• WE 14 int0530: Watch a worked example on solving exponential equations. (page 143) Digital docs
• SkillSHEET 3.1: Practise writing expressions using index form. (page 146) • SkillSHEET 3.2: Practise solving equations. (page 146) • SkillSHEET 3.3: Practise solving indicial equations by equating the bases. (page 146) • SkillSHEET 3.4: Practise solving linear inequations. (page 146) 3D
Logarithmic equations using any base
Tutorial
• WE 18 int0531: Watch a worked example on solving logarithmic equations using a CAS calculator. (page 147) Digital doc
• WorkSHEET 3.1: Simplify exponential and logarithmic expressions, and solve logarithmic and exponential equations. (page 149) 3F
Equations with natural (base e) logarithms
3G
Inverses
Interactivity
• Inverses int0248: Consolidate your understanding of inverses using the interactivity. (page 156) Digital docs
• SkillSHEET 3.5: Practise finding inverses. (page 158) 3H
Literal equations
Tutorial
• WE 27 int0532: Watch a worked example on solving literal equations. (page 159) 3I
Exponential and logarithmic modelling
eLesson
• Exponential and logarithmic modelling eles0091: Learn about how exponential and logarithmic modelling is used. (page 162) Tutorial
• WE 30 int0533: Watch a worked example on exponential modelling. (page 163) Chapter review Digital doc
• Test Yourself Chapter 3: Take the endofchapter test to test your progress. (page 170) To access eBookPLUS activities, log on to www.jacplus.com.au
Digital doc
• WorkSHEET 3.2: Solve logarithmic and exponential equations and application questions. (page 156)
Chapter 3
Exponential and logarithmic equations
171
4
4A Graphs of exponential functions with any base 4B Logarithmic graphs to any base 4C Graphs of exponential functions with base e 4D Logarithmic graphs to base e 4E Finding equations for graphs of exponential and logarithmic functions 4F Addition of ordinates 4G Exponential and logarithmic functions with absolute values 4H Exponential and logarithmic modelling using graphs
Exponential and logarithmic graphs AREAS OF STUDY
• Graphs and identification of key features of graphs of the following functions: – exponential functions: exponential equations, y = ax – logarithmic functions, y = loge (x) and y = log10 (x), the relationship a = ek, where k = loge (a) • Key features of functions, including: – axis intercepts – domain (including maximal domain) and range – asymptotic behaviour – symmetry • Transformations for y = f (x) to y = Af (n(x + b)) + c, where A, n and c ∈ R and f is an exponential or logarithmic function and relation between the graph of the original
• •
• •
•
function and the graph of the transformed function Families of transformed functions for a single transformation parameter Graphs of sum, difference, product and composite functions of f and g, where f and g are exponential or logarithmic equations such as y = x2ekx, k ∈ R Graphical and numerical solutions of equations Recognition of the general form of possible models for data presented in graphical or tabular form using exponential and logarithmic functions Applications of simple combinations of exponential or logarithmic functions and interpretation of features of the graphs of these functions in modelling practical situations eBook plus
4A
Graphs of exponential functions with any base
Digital doc
10 Quick Questions
The function f (x) = ax is an exponential function where a is a positive, real number which is not 1 (that is, a ∈ R+\{1}). In this chapter, a will take the values of 2, 10 and Euler’s number e. An exponential function f (x) can be written f: R → R where f (x) = ax and a ∈R+\{1}. In the previous chapter it was seen that if x has a positive coefficient, f (x) is an increasing function and may be used to describe physical growth. Examples of this include population and bacterial growth, and increases in investment values, light intensity and temperature. If x has a negative coefficient, f (x) is a decreasing function and may be used to describe physical decay. Examples of this include population and bacterial decline, radioactive decay, temperature cooling and decreases in light intensity and vehicle values. Graphs with a positive coefficient of x are considered first.
172
Maths Quest 12 Mathematical Methods CAS for the TINspire
Graphs of f (x) = 2x and f (x) = 10x The graphs of f (x) = 2x and f (x) = 10 x are shown below. y 12 10 8 6 (0, 1) 4 Asymptote 2 (1, 2) y= 0 x 0 −3 −2 −1 1 2 3
y 12 (1, 10) 10 8 6 4 Asymptote 2 (0, 1) y= 0 x −3 −2 −1 0 1 2 3
f (x) = 2x f (x) = 10x These graphs have a number of common features: 1. They both cross the yaxis at the point (0, 1) because a0 = 1. 2. The graph does not cross the xaxis; therefore, there are no xintercepts. 3. There is a horizontal asymptote along the xaxis (y = 0). 4. For f (x) = a x, another point on the graphs is (1, a). 5. The maximal domain is R. 6. The range is R+. 7. They are both increasing functions. That is, as x → ∞, y → ∞. 8. It can be seen that the greater the value of a, the steeper the graph.
Dilation A dilation changes the shape of the graph, making it wider or narrower.
Dilation from the xaxis If the coefficient of ax is changed to a positive real number greater than 1, the graph is stretched vertically and is said to be dilated from the xaxis. This could be written f (x) = Aax where A is the dilation factor. The graph is stretched vertically (along the yaxis) away from the xaxis because each yvalue is being multiplied by the constant A. In mapping notation, a dilation factor of A from the xaxis, where A > 0, is given by (x, y) → (x, Ay). If A = 3 and a = 2 the function becomes f (x) = 3 × 2x. A comparison can be made more easily if graphs are drawn on the same axes. The graphs of f (x) = 2x, g(x) = 3 × 2x and h(x) = 4 × 2x are shown. The dilation factor, A, is 1, 3 and 4 respectively. f (x ( ) = 2x
g(x ( ) = 3 × 2x (x
h(x ( ) = 4 × 2x (x
xintercept
—
—
—
yintercept
(0, 1)
(0, 3)
(0, 4)
R
R
R
R+ = (0, ∞)
R+ = (0, ∞)
R+ = (0, ∞)
y=0
y=0
y=0
1
3
4
(1, 2)
(1, 6)
(1, 8)
Domain Range Horizontal asymptote Dilation factor from the xaxis A point on the graph Mapping
y (0, 4)
h(x) = 4 × 2x g(x) = 3 × 2x
4 3 (0, 3) f( f x) = 2x 2 1 (0, 1) Asymptote x y= 0 −3 −2 −1 0 1 2 3
(x, y) → (x, 3y) (x, y) → (x, 4y)
Chapter 4
Exponential and logarithmic graphs
173
If A is a real number between 0 and 1, the yvalues are multiplied by a constant which is less than 1 and the graph becomes less steep. A comparison can be made more easily if the graphs are drawn on the same axes. The graphs of f (x) = 2x, g(x) = 12 × 2x and h(x) = 14 × 2x are drawn below. y (1, 2)
2
The dilation factor is 1,
1 2
and
1 4
(0, 1–2) −1
1– 2
× 2x
h(x) = 1–4 × 2x
(1, 1)
(0, 1) Asymptote y= 0
f x) = 2x f( g(x) =
–1
(0, 1–4) (1, 2 ) x 0 1
respectively.
f (x ( ) = 2x
g( x ) = 12 × 2 x
h( x ) = 14 × 2 x
xintercept
—
—
—
yintercept
(0, 1)
1 0, 2
1 0, 4
Domain Range
R R+
Horizontal asymptote
= (0, ∞)
R R+
R R+
= (0, ∞)
= (0, ∞)
y=0
y=0
y=0
Dilation factor from the xaxis
1
1 2
1 4
A point on the graph
(1, 2)
(1, 1)
Mapping
(
(x, y) → x, 1 y 2
(1, ) 1 2
)
(
(x, y) → x, 1 y 4
)
For any positive real value of A, as A increases, the graph of f (x (x) = A × 2x becomes steeper and closer to the yaxis. Similarly, as A decreases the graph becomes less steep and further from the yaxis. The domain, range and horizontal asymptote stay the same as for f (x (x) = 2x. The mapping is ((x, y) → (x ( , Ay).
Dilation from the yaxis If the coefficient of x changes, the graph is stretched horizontally and is said to be dilated from the yaxis. This can be written f (x) = akx where 1 is the dilation factor. The dilation factor is the k reciprocal of the coefficient of x. In mapping notation, a dilation factor of 1 from the yaxis, where k > 0, is given by (x, y) → ( 1 x, y). k k To show the effect of a dilation from the yaxis, the three graphs x 2x and h(x) = 2 are drawn at right. The dilation f (x) = 2x, g(x) = 22x 2 1 factors are 1, 2 and 2, respectively.
174
Maths Quest 12 Mathematical Methods CAS for the TINspire
y 4 2 (0, 1) −3 −2 −1
2 g(x) = 22x f x) = 2x –x f( h(x) = 22
Asymptote y=0 01 2 3 x
x
f (x ( ) = 2x
2 g(x ( ) = 22x (x
h(x ( ) = 22 (x
xintercept
—
—
—
yintercept
(0, 1)
(0, 1)
(0, 1)
R
R
R
R+ = (0, ∞)
R+ = (0, ∞)
R+ = (0, ∞)
y=0
y=0
y=0
1
1 2
2
(2, 4)
(1, 4)
(4, 4)
Domain Range Horizontal asymptote Dilation factor from the yaxis A point on the graph
(x, y) →
Mapping
(
1 2
x, y
)
(x, y) → (2x (2 , y)
For all the graphs discussed so far (that is, of the form f (x (x) = A × akx, a ∈ R+\{1}, k > 0), the + maximal domain is R, the range is R , the xaxis is the horizontal asymptote and they are all increasing functions. The dilations have affected the steepness of the graphs. The mapping is ( , y) → ( 1 x, Ay). (x k WORKED EXAMPLE 1 x
Sketch the graph of f (x (x) = 2 × 2 2 , showing the intercepts and the asymptote, and stating the domain and the range. THINK
WRITE/DRAW x
1
Write the rule.
2
State the basic shape and transformations.
An exponential curve with basic shape f (x) = 2x. Dilation of 2 units from the xaxis and 2 units from the yaxis.
3
Find the yintercept. Either let x = 0 or use the fact that the yintercept is A in the function
If x = 0, then
x
f (x) = A × a k .
f (x) = 2 × 2 2
0
y = 2 × 22 = 2 × 20 =2×1 =2 so the yintercept is 2.
4
Find the horizontal asymptote.
The horizontal asymptote is the xaxis. There are no xintercepts.
5
Find another point on the graph.
If x = 2, y = 2 × 2 2 = 4.
6
Sketch the graph.
2
y 4
(2, 4)
(0, 2)
x
x −3 −2 −1 0 1 2 3 7
State the domain and the range.
f (x) = 2 × 2 2 Asymptote y = 0
The domain is R and the range is R+.
Chapter 4
Exponential and logarithmic graphs
175
Translation Vertical translation If a constant is added to the function, the graph is moved up or down and is said to be translated vertically. In mapping notation, a vertical translation of B units is given by (x, y) → (x, y + B). The graph of g(x) = 2x + 1 is shown below with the graph of f (x) = 2x. The graph of f (x) = 2x has been moved up 1 unit. Every yvalue has been increased by 1. The line y = 1 is the horizontal asymptote. f (x ( ) = 2x
g(x ( ) = 2x + 1 (x
xintercept
—
—
yintercept
(0, 1)
(0, 2)
R
R
Range
(0, ∞)
(1, ∞)
Horizontal asymptote
y=0
y=1
A point on the graph
(1, 2)
(1, 3)
Domain
y
g(x) = 2x + 1 f x) = 2x f(
4 3 (0, 2)
(1, 3)
(0, 1) 0 1
−1
(1, 2) Asymptote y=1 2
x
(x, y) → (x, y + 1)
Mapping
The graph of g(x) = 2x − 1 is the graph of f (x) = 2x translated 1 unit down. This means that the horizontal asymptote is now the line y = −1 and the graph crosses the xaxis. Therefore, there is an xintercept. Every yvalue has been decreased by 1. The graph of y = 2x − 1 is shown with the graph of y = 2x. y
f (x ( ) = 2x
g(x ( ) = 2x − 1 (x
xintercept
—
(0, 0)
4
yintercept
(0, 1)
(0, 0)
R
R
2 (0, 1)
Range
(0, ∞)
(−1, ∞)
Horizontal asymptote
y=0
y = −1
A point on the graph
(1, 2)
(1, 1)
Domain
−33 −2 −11
(1, 2)
g(x) = 2x − 1
(1, 1) x 0 1 2 3 Asymptote y = −1
(x, y) → (x, y − 1)
Mapping
Horizontal translation In mapping notation, a horizontal translation of b units is given by (x, y) → (x + b, y). The graph of g(x) = 2x − 1 is the graph of f (x) = 2x translated 1 unit to the right. The graph of h(x) = 2x + 1 is the graph of f (x) = 2x translated 1 unit to the left. The graph of g(x) = 2x − 1 is shown at right with the graph of f (x) = 2x. 176
f x) = 2x f(
Maths Quest 12 Mathematical Methods CAS for the TINspire
y
h(x) = 2x + 1 f(x) = 2x f(
4 2 (0, 1) −3 −2 −1
(1, 1) (0, 1–2 )
01 2 3
g(x) = 2x − 1 Asymptote y=0 x
f (x ( ) = 2x
g(x ( ) = 2x − 1 (x
h(x ( ) = 2x + 1 (x
xintercept
—
—
—
yintercept
(0, 1)
(0, )
(0, 2)
R
R
R
R+ = (0, ∞)
R+ = (0, ∞)
R+ = (0, ∞)
Horizontal asymptote
y=0
y=0
y=0
A point on the graph
(1, 2)
(2, 2)
(0, 2)
(x, y) → (x + 1, y)
(x, y) → (x − 1, y)
Domain Range
Mapping
1 2
Horizontal and vertical translations have the same effect regardless of the base. g(x ( ) = 2 + 10x + 1 is f (x (x (x) = 10 x translated 1 unit to the left and 2 units up. Horizontal translations keep the same asymptote, domain and range as the original function; vertical translations keep the same domain, but the range changes. It is interesting to note that f (x) = 0.5 × 2x can be written as f (x) = 2x − 1 because f (x) = 0.5 × 2x 1
= 2 × 2x −
= 2 1 × 2x = 2x − 1 A dilation of 0.5 from the xaxis has the same effect on f (x) = 2x as a translation of 1 unit to the right. WORKED EXAMPLE 2
eBoo k plus eBook
Sketch the graph of f (x (x) = 2x − 1 − 2, showing intercepts and asymptotes, and stating the domain and range. THINK
WRITE/DRAW
Tutorial
int0534 Worked example 2
f (x) = 2x − 1 − 2
2
Write the rule. State the basic shape.
3
State the translations.
Horizontal translation of 1 unit to the right Vertical translation of 2 units down
4
Find the horizontal asymptote by translating y = 0 down 2 units.
The horizontal asymptote is y = −2.
5
Find the yintercept. Make x = 0.
yintercept: If x = 0, then
1
An exponential curve with the same shape as f (x) = 2x
−
y=2 1−2 = 12 − 2 = −1 12
6
Find the xintercept. Make y = 0.
7
Equate the indices.
8
Solve for x.
xintercept: If y = 0, then
Chapter 4
2x − 1 − 2 = 0 2x − 1 = 21 x−1=1 x=2
Exponential and logarithmic graphs
177
9
Sketch the graph.
y 1 −3 −2 −1 −1 −2
10
(2, 0) 0 1 2 3 x
f (x) = 2x − 1 − 2
1 (0, −11–2 ) Asymptote y = −2
The domain is R and the range is (−2, ∞).
State the domain and the range.
For all the graphs of the form f (x (x) = ax + b + B, where b, B ∈ R and a ∈ R+\{1}, the maximal domain is R, the range is (B, ∞), the horizontal asymptote is y = B and they are all increasing functions. The graph remains exactly the same shape as f (x (x) = ax. The mapping is ((x, y) → (x ( − b, y + B).
Reflections If a negative sign is in front of the a the graph is reflected in the xaxis. (Remember a > 0.) The mapping is (x, y) → (x, −y). The graph of g(x) = −2x is shown with the graph of f (x) = 2x. f (x ( ) = 2x
g(x ( ) = −2x (x
y
xintercept
—
—
4
yintercept
(0, 1)
(0, −1)
R
R
R+ = (0, ∞)
R− = (−∞, 0)
Horizontal asymptote
y=0
y=0
A point on the graph
(1, 2)
(1, −2)
Domain Range
f x) = 2x f(
2
Asymptote (1, 2) y = 0 −3 −2 –1 0 1 2 3 x (1, −2) −2 (0, 1)
(0, −1) g(x) = −2x
(x, y) → (x, −y)
Mapping
If there is a negative sign before the x term, the graph is reflected in the yaxis. The graph of g(x) = 2−x is shown with the graph of f (x) = 2x. f (x ( ) = 2x
g(x ( ) = 2 −x (x
xintercept
—
—
yintercept
(0, 1)
(0, 1)
R
R
R+ = (0, ∞)
R+ = (0, ∞)
Horizontal asymptote
y=0
y=0
A point on the graph
(1, 2)
(−1, 2)
Domain Range
Mapping
f x) = 2x f(
– y g(x) = 2–x
3 (−1, 2) 2 1 −3 −2 −1
(1, 2) (0, 1) 0
Asymptote y= 0 x 1 2 3
(x, y) → (−x, y) For all the graphs of the form g(x) = −ax, where a ∈ R+\{1}, the maximal domain is R and the range is R−. The horizontal asymptote is y = 0 and the functions are all decreasing. It is the reflection of f (x) = ax in the xaxis. The mapping is (x, y) → (x, −y).
178
Maths Quest 12 Mathematical Methods CAS for the TINspire
For all the graphs of the form g(x) = a−x, where a ∈ R+\{1}, the maximal domain is R and the range is R+. The horizontal asymptote is y = 0 and the functions are all decreasing. It is the reflection of f (x) = ax in the yaxis. The mapping is (x, y) → (−x, y). WORKED EXAMPLE 3
Sketch the graph of f (x (x) = 2 − 2x − 1, showing intercepts and asymptotes, and stating the domain and the range: a showing all working b with a CAS calculator. THINK a
WRITE/DRAW
1
Write the rule.
f (x) = 2 − 2x − 1
2
State the basic shape.
An exponential curve with basic shape f (x) = 2x
3
State the transformations.
Horizontal translation of 1 unit to the right Vertical translation of 2 units up Reflection in the xaxis
4
Find the horizontal asymptote by translating The horizontal asymptote is y = 2. y = 0 up 2 units. − If x = 0, f (x) = 2 − 2 1 Find the yintercept by making x = 0.
5
=2−
1 2
= 1 12 The yintercept is 1 12. 6
Find the xintercept by making y = 0.
7
Sketch the graph.
2 − 2x − 1 = 0 2x − 1 = 21 x−1=1 x=2 The xintercept is 2. If y = 0,
y 2
Asymptote y=2 (0, 3–2 )
1 0
−2 −1
1
(2, 0) x 2
f (x) = 2 − 2x − 1
−1
b
8
State the domain and the range.
1
To graph y = 2 − 2x − 1 on a Graphs page, complete the function entry line as: f1(x) = 2 − 2x − 1 Press ENTER ·. Note: The horizontal asymptote at y = 2 is not displayed.
The domain is R and the range is (−∞, 2).
Chapter 4
Exponential and logarithmic graphs
179
2
To locate the xintercept on a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve(f 1(x) = 0, x) Press ENTER ·.
3
Write the xintercept in coordinate form.
∴ (2, 0) are the coordinates of the xintercept.
4
To calculate the yintercept, complete the entry line as: ff1(0) and press ENTER ·. Write the yintercept in coordinate form.
∴ 0,
( ) are the coordinates of the yintercept. 3 2
REMEMBER
1. Graphs of the form f (x) = ax, where a ∈ R+\{1} (a) The maximal domain is R. (b) The range is R+. (c) The xaxis is the horizontal asymptote. (d) The yintercept is 1. (e) They are all increasing functions. 2. Dilation (a) Graphs of the form f (x) = Aax, where a, A ∈ R+\{1}, have a yintercept of A. A changes the steepness and proximity of the graph to the yaxis when compared with the graph of f (x) = ax. As A increases, the graph becomes steeper. So f (x) = ax is dilated by a factor of A from the xaxis. (b) Graphs of the form f (x) = akx, where a ∈ R+\{1} and k > 0, have a different steepness than graphs of the form f (x) = ax. As k increases, the graph becomes more steep (that is, it gets closer to the yaxis). The yintercept remains the same. So f (x) = akx is dilated by a factor of 1 from the yaxis. k 3. Translation (a) Graphs of the form f (x) = ax + B, where a ∈ R+\{1}, B ∈ R, have a yintercept of 1 + B, a horizontal asymptote of y = B and a range of (B, ∞). If B < 0 there is one xintercept. If B > 0 there are no xintercepts. (b) Graphs of the form f (x) = ax + b, where a ∈ R+\{1}, b ∈ R, have a yintercept of ab. They are the same shape as f (x) = ax but have been translated b units to the left if b > 0 and to the right if b < 0. 4. Reflection (a) Graphs of the form f (x) = −ax, where a ∈ R+\{1}, have a yintercept of −1, a range of R− and they are decreasing functions. They are the same shape as f (x) = ax but have been reflected in the xaxis. (b) Graphs of the form f (x) = a−x, where a ∈ R+\{1}, have the same key features as f (x) = ax but are decreasing functions; they have been reflected in the yaxis. 5. Combinations of transformations For all the graphs of the form f (x) = A × ax + b + B, where b, B, A ∈ R and a ∈ R+\{1}, the maximal domain is R, the range is (B, ∞), the horizontal asymptote is y = B and they are all increasing functions.
180
Maths Quest 12 Mathematical Methods CAS for the TINspire
EXERCISE
4A eBoo k plus eBook Digital doc
Spreadsheet 034 Exponential graphs
Graphs of exponential functions with any base 1 WE1 Sketch the graph of each of the following functions, showing the intercepts and the asymptote, and stating the domain and range of each one. a f (x) = 2x b f (x) = 10x c f (x) = 2 × 10x x x d f (x) = 3 × 2 e f (x) = 0.3 × 2 f f (x) = 0.5 × 10x 3x 2 2x 2 g f (x) = 2 h f (x) = 10 i f (x) = 5 × 32x x x 3x j f (x) = 4 × 2 k f (x) = 2 × 10 2 l f (x) = 2 × 2 3 2 WE2 Sketch the graph of each of the following functions, showing intercepts and asymptotes, and stating the domain and range of each one. a f (x) = 2x + 1 b f (x) = 3x + 2 c f (x) = 3x − 3 x x + 2 d f (x) = 2 − 4 e f (x) = 2 f f (x) = 10x + 1 x − 4 x − 3 g f (x) = 3 h f (x) = 2 i f (x) = 2x + 1 − 8 x − 2 j f (x) = 10 +1 3 WE3 Sketch the graph of each of the following functions, showing intercepts and asymptotes, and stating the domain and range of each one. a f (x) = 2x b f (x) = −10x −x c f (x) = 10 d f (x) = 2−x x e f (x) = 1 − 3 f f (x) = 10 − 10x −x g f (x) = 2 + 10 h f (x) = 1 + 2−x 1 − x i f (x) = 2 − 2 j f (x) = 1 − 32 − x 4 Sketch the graph of f (x (x) = 2 × 31 − x, showing the intercepts and asymptotes and stating the domain and range. 5 Sketch the graph of the function f: f R → R where f (x (x) = −3 × 2x − 1, showing the intercepts and asymptotes and stating the domain and range. 6 Sketch the graph of the function f: f R → R where f (x (x) = 5 − 4 × 31 − x showing the intercepts and asymptotes and stating the domain and range. 7 State the transformation of y = 2x needed to sketch the graphs of the following functions. Give details of each transformation. a y = 23x b y = 24x c y = 2 × 2x x − x d y=3×2 e y= 2 f y = 2−x x x g y=2 +1 h y=2 −3 i y = 2x − 1 x + 5 j y=2 8 Each of the following functions is a translation of f (x (x) = 10x. State how far each graph is translated, and in which direction. a f (x) = 10 x + 4 b f (x) = 10 x − 2 c f (x) = 10 x + 2 x − 3 x d f (x) = 10 e f (x) = 10 − 3 f f (x) = 2 + 10 x x − 1 2 + x g f (x) = 10 h f (x) = 10 i f (x) = 5 + 10 x + 1 x − 4 x − 3 j f (x) = 10 +2 k f (x) = 10 −4 l f (x) = 10 x + 2 − 3 9 Write down the domain and range for each of the following graphs and the equations for the horizontal asymptotes. a
y
b
y
c
y
6 4 2 (0, 2) y=0 x 0 −3 −2 −1 1 2
6 4 4 (0, 4) 2
(0, 2) y=1
−3 −2 −1
0 1
Chapter 4
2
x
y=3
2 –3 −2 −1
0 1
2
x
Exponential and logarithmic graphs
181
y 2
d
y
e y=1
−3 −2 −1 0
1
2
2
6
x
−2
−1 0
2
−3 −2 −1
y=1
(1, 0)
4 (0, 4) y=2
y
f
1 2
3
x
−2 (0, −2) 0
1 2 3
x
10 If the graph of f (x (x) = 2x is translated 1 unit up and reflected in the xaxis, what is the new equation? 11 If the graph of f (x (x) = 1 − 2x + 3 is translated 2 units to the right and dilated by factor 3 from the xaxis, what is the new equation? 2 − 1 + 3 are respectively: 12 MC The domain and range of the graph of f (x (x) = 102x − A R, R B R, [ 1, ∞) C [3, ∞), R D [1, ∞), R E R, (3, ∞)
13 MC The yintercept of the graph of f (x (x) = 10−x + 1 is: A 1 B 2 C −1
D −2
E 11
2x − 1
14 MC When the graph of f (x (x) = − 4 is translated 3 units down and 2 units to the left, it becomes: A f (x) = 2x − 3 − 7 B f (x) = 2x − 3 − 1 x + 1 C f (x) = 2 −7 D f (x) = 2x + 1 − 1 x − 2 E f (x) = 2 −6 15 Under certain conditions a mathobacillus bacterial colony doubles its numbers every minute. The population can be modelled by: P = 50 × 2t where P is the number of bacteria t minutes after counting has begun. a Find the number of bacteria when t = 0. b Find the number of bacteria after 3 minutes. c Sketch the graph of the population as a function of time.
4B
Logarithmic graphs to any base The function f (x) = loga (x) is a logarithmic function where a is a positive, real number which is not 1 (that is, a ∈ R+\{1}). Logarithms to base 10 appear in many scientific formulas. An example of this is the intensity of earthquakes which is measured on the Richter scale. Remember that loga (x) does not exist when x < 0. In this course only graphs of functions where a > 1 are considered. Graphs with a positive coefficient of x will be considered first.
Graphs of f (x) = log2 (x) and f (x) = log10 (x) The graphs of f (x) = log2 (x) and f (x) = log10 (x) are shown below. Asymptote y x= 0 (2, 1) 2 (1, 0) 1 −1 0 −1 −2
1
2 3
f (x) = log2 (x) 182
y
Asymptote x= 0
1
x
0
(10, 1)
(1, 0) 2 4 6 8 10
−1
f (x) = log10 (x)
Maths Quest 12 Mathematical Methods CAS for the TINspire
x
These graphs have a number of common features: 1. They are both increasing functions; that is, as x → ∞, y → ∞. 2. Both graphs cross the xaxis at the point (1, 0), as loga (1) = 0 3. The yaxis is a vertical asymptote, so there is no yintercept. This is because loga (0) is undefined. 4. There are no negative values of x because the log of a negative number does not exist. 5. Another point on the graph is (a, 1), where a is the base because loga (a) = (1). 6. The domain is R+. 7. The range is R. 8. It can be seen that the smaller the value of a, the steeper the graph or the more rapidly the graph rises.
Dilation Dilation from the xaxis If the coefficient of loga (x) is changed to a positive real number greater than 1, the graph is stretched along the yaxis and is said to be dilated from the xaxis. This could be written f (x) = A loga (x) where A is the dilation factor. The graph is stretched along the yaxis because each yvalue is being multiplied by the constant, A. In mapping notation, a dilation factor of A from the xaxis, where A > 0, is given by Asymptote (x, y) → (x, Ay). y x= 0 (10, 2) g(x) = 2 log10 (x) The asymptote, xintercept, domain and range are the same for all graphs of the form f (x (x) = A loga (x ( ). Graphs with different values of A can be more easily compared by drawing them on the same axes. The graphs of g(x) = 2 log10 (x), f (x) = log10 (x) and h(x) = 12 log10 (x) are drawn at right. Notice that, as A increases, the graph becomes steeper.
1 0
f x) = log10 (x) (10, 1) f( h(x) = 1–2 log10 (x) (10, 1–2 ) x 2 4 6 8 10
−1
1
f (x ( ) = log10 (x ( )
g(x ( ) = 2 log10 (x (x ( )
h(x ( ) = 2 log10 (x (x ( )
xintercept
(1, 0)
(1, 0)
(1, 0)
yintercept
—
—
—
R+ = (0, ∞)
R+ = (0, ∞)
R+ = (0, ∞)
R
R
R
x=0
x=0
x=0
1
2
1 2
(10, 1)
(10, 2)
(10, )
Domain Range Vertical asymptote Dilation factor from the xaxis A point on the graph Mapping
(x, y) → (x, 2y)
1 2
(
(x, y) → x, 12 y
)
Note: h(x) = 12 log10 (x) can be written as h(x) = log10 ( x ) and is drawn in the same way. 1 2
Chapter 4
Exponential and logarithmic graphs
183
Dilation from the yaxis If the coefficient of x is changed to a positive real number not equal to 1, the graph is stretched along the xaxis and is said to be dilated from the yaxis. This could be written f (x) = loga (kx), where 1 is the dilation k factor. When k > 1, the graph moves away from the xaxis. To see the effect of a dilation from the xaxis the graphs g(x) = log10 (2x (2 ), f (x) = log10 (x) and h(x) = log10 ( 12 x) are drawn at right.
y Asymptote x= 0 1
0
1
h(x) = log10 (–2 xx) 2 (2, 0) (1, 0)
(1, 0)
( , 0)
= (0, ∞)
h(x ( ) = log10 (x
— R+
= (0, ∞)
= (0, ∞) R
x=0
x=0
x=0
Dilation factor from the yaxis
1
1 2
2
Another point on the graph
(10, 1)
(5, 1)
(20, 1)
(x, y) →
1 2
— R+
R
Mapping
( x)
(2, 0)
R
Horizontal asymptote
x
6
1 2
— R+
4
−1
g(x ( ) = log10 (2x (x (2 )
yintercept Range
f x) = log10 (x) f(
( 1–2 , 0)
f (x ( ) = log10 (x ( ) xintercept
Domain
2 ) 2x g(x) = log10 (2x
( x, y ) 1 2
(x, y) → (2x (2 , y)
For all graphs discussed so far (that is, of the form y = A loga (kx), A ∈ R+\{1}, k > 0), the maximal domain is R+, the range is R, the yaxis is the vertical asymptote and they are all increasing functions. The graphs have varied in their xintercepts. The dilation factor from the xaxis is A and the dilation factor from the yaxis is 1 . In mapping notation, k Asymptote ( , y) → 1 x, A (x Ay . y x=0 k f x) = 2 log2 (3x) f( If the base is 2, the graph is steeper than when the base is 10 but the vertical asymptote, intercept, domain and range 2 stay the same as for f (x) = A log10 (kx). f x) = 2 log10 (3x) f( 1 The graphs of f (x) = 2 log10 (3x) and f (x) = 2 log2 (3x) are ( –3 , 0) x shown at right. 0 1 2 3 1. The vertical asymptote is the yaxis. −2 2. The xintercept is 13 . 3. The domain is R+ and the range is R. 4. The basic graph is dilated by factor 2 from the xaxis and factor 13 from the yaxis.
(
184
)
Maths Quest 12 Mathematical Methods CAS for the TINspire
WORKED EXAMPLE 4
Sketch the graph of f (x (x) = 3 log10 (2x (2 ), showing the intercepts and the asymptote, and state the domain, range and the kind of transformation. THINK
WRITE/DRAW
1
Write the rule.
f (x) = 3 log10 (2x (2 )
2
State the basic shape.
A logarithmic graph with basic shape f (x) = log10 (x)
3
Find the vertical asymptote (log10 (0) is undefined). The vertical asymptote is the yaxis.
4
Find the xintercept. Remember that log10 (1) = 0.
xintercept: Let y = 0, 3 log10 (2x (2 ) = 0 2 =1 2x x = 12 The xintercept is 12.
5
Find the yintercept.
There are no yintercepts.
6
Mark another point on the graph. Choose one which makes 22xx equal the base.
Let x = 5, y = 3 log10 (10) =3×1 =3 Coordinates are (5, 3).
7
Sketch the graph.
Asymptote y x=0 3
f x) = 3 log10 (2x f( (2 )
1
0
(5, 3)
( –2 , 0) 1
2
3
4
5
x
−3 8
State the domain and the range.
The domain is R+ and the range is R.
9
State the transformations.
The dilations are of factor 3 from the xaxis and factor 12 from the yaxis.
Translation Translations may be either vertical or horizontal.
Vertical translation If a constant is added to the function, the graph is moved up or down and is said to be translated vertically. The graphs of f (x (x) = log10 (x ( ), g(x ( ) = log10 (x (x ( ) + 1 and h(x ( ) = log10 (x (x ( ) − 1 are drawn and compared. y Asymptote x=0 2
g(x) = log10 (x) + 1 f x) = log10 (x) f(
1 0 −1
h(x) = log10 (x) − 1 2
4
6
8
10
12 x
−2 −3
Chapter 4
Exponential and logarithmic graphs
185
f (x ( ) = log10 (x ( )
g(x ( ) = log10 (x (x ( )+1
(
1 , 10
0
h(x ( ) = log10 (x (x ( )−1
)
xintercept
(1, 0)
yintercept
—
—
—
R+ = (0, ∞)
R+ = (0, ∞)
R+ = (0, ∞)
R
R
R
Vertical asymptote
x=0
x=0
x=0
Vertical translation
—
1 unit up
1 unit down
(1, 0)
(1, 1)
(1, −1)
(x, y) → (x, y + 1)
(x, y) → (x, y − 1)
Domain Range
A point on the graph Mapping
(10, 0)
Horizontal translation If a constant is added to x, the graph of f (x) = log10 (x) is translated horizontally. The graphs of f (x) = log10 (x), g(x) = log10 (x + 1) and h(x) = log10 (x − 1) are drawn. f (x ( ) = log10 (x ( )
g(x ( ) = log10 (x (x ( + 1)
h(x ( ) = log10 (x (x ( − 1)
(1, 0)
(0, 0)
(2, 0)
—
—
xintercept yintercept Domain
— R+
(−1,
= (0, ∞)
Range
R
Horizontal translation A point on the graph
(1, ∞)
R
x=0
Vertical asymptote
∞)
x=
R
−1
x=1
—
1 unit to left
1 unit to right
(10, 1)
(9, 1)
(11, 1)
(x, y) → (x − 1, y)
(x, y) → (x + 1, y)
Mapping y 2
g(x) = log10 (x + 1)
1 −2 −1
−1
0 1
3
4
5
x
f x) = log10 (x) f( h(x) = log10 (x − 1)
−2 x = −1
2
x=1
For all graphs of the form f (x (x) = loga (x ( + b) + B, where b and B ∈ R and a ∈ R+\{1}, − the maximal domain is ( b, ∞), the range is R, the vertical asymptote is x = −b and they are all increasing functions. The graph has exactly the same shape as f (x (x) = loga (x ( ). The horizontal translation is −b and the vertical translation is B. In mapping notation, ( , y) → (x (x ( − b, y + B).
186
Maths Quest 12 Mathematical Methods CAS for the TINspire
WORKED EXAMPLE 5
eBoo eBook k plus
Sketch the graph of f (x (x) = 1 + log10 (x ( + 2), showing intercepts and asymptotes, and stating the domain, range and transformations. Give intercepts to 1 decimal place. THINK
Tutorial
int0535 Worked example 5
WRITE/DRAW
1
Write the rule.
f (x) = 1 + log10 (x + 2)
2
State the basic shape.
Logarithmic graph, basic shape f (x) = log10 (x)
3
Find the vertical asymptote (log10 (0) is undefined). x + 2 = 0 x = −2 is the vertical asymptote.
4
Find the xintercept.
If y = 0,
5
Use the law ax = y ⇔ loga(y) = x.
x + 2 = 10 1 x = 0.1 − 2 = −1.9 − The xintercept is 1.9.
6
Find the yintercept.
If x = 0, y = 1 + log10 (0 + 2) (exact answer) ≈ 1 + 0.301 03 ≈ 1.3, correct to 1 decimal place The yintercept is 1.3.
7
Sketch the graph.
1 + log10 (x + 2) = 0 log10 (x + 2) = −1 −
Asymptote y x = –2 2
(0, 1.3)
(−1.9, 0) −2
0
2
x 4 6 8 f x) = 1 + log10 (x + 2) f(
8
State the domain and the range.
The domain is (−2, ∞) and the range is R.
9
State the translations.
The horizontal translation is 2 units to the left and the vertical translation is 1 unit up.
Reflections If there is a negative sign in front of the loga (x) term, the graph is reflected in the xaxis. The mapping is (x, y) → (x, −y). The graphs of f (x) = log10 (x) and g(x) = −log10 (x) are shown at right.
Asymptote y x=0 f x) = log10 (x) f( 1 (10, 1) (1, 0) 0 −1
Chapter 4
2
4
g(x) =
6 8 10
x
(10, –1) 10 (x))
–log
Exponential and logarithmic graphs
187
f (x ( ) = log10 (x ( )
g(x ( ) = −log10 (x (x ( )
(1, 0)
(1, 0)
xintercept yintercept
— R+
Domain
= (0, ∞)
Range
= (0, ∞)
R
R
x=0
x=0
(10, 1)
(10, −1)
Vertical asymptote A point on the graph
— R+
(x, y) → (x, −y)
Mapping
If there is a negative sign in front of the x term, the graph is reflected in the yaxis. The mapping is (x, y) → (−x, y). The graphs of f (x) = log10 (x) and g(x) = log10 (−x) are shown below right. f (x ( ) = log10 (x ( )
g(x ( ) = log10 (−x) (x
(1, 0)
(−1, 0)
xintercept yintercept Domain Range Vertical asymptote A point on the graph
— R+
— R−
= (0, ∞)
= (−∞, 0)
R
R
x=0
x=0
(10, 1)
(−10, 1)
Mapping
Asymptote y x=0 f x) = log10 (x) g(x) = log10 (−x) f( 1 (−10, 1) (10, 1) (−1, 0) −10
−2
(1, 0) 2
−1
(x, y) → (−x, y)
WORKED EXAMPLE 6 a Sketch the graph of f (x (x) = 2 log10 (3 − x) − 2, showing intercepts and asymptotes, and stating
the domain, range and transformations. Give exact values or round to 3 decimal places.
b Sketch the same graph using a CAS calculator. THINK a
188
WRITE/DRAW
1
Write the rule.
f (x) = 2 log10 (3 − x) − 2
2
State the basic shape.
Logarithmic graph, f (x) = log10 (x)
3
Find the vertical asymptote (log10 (0) undefined).
For the vertical asymptote, 3−x=0 x = 3 is the vertical asymptote.
4
Find the xintercept.
If y = 0,
2 log10 (3 − x) − 2 = 0 2 log10 (3 − x) = 2 log10 (3 − x) = 1 3 − x = 101 x = −7 − The xintercept is 7.
Maths Quest 12 Mathematical Methods CAS for the TINspire
10
x
5
Find the yintercept.
6
Sketch the graph.
If x = 0, y = 2 log10 (3) − 2 = −1.046 (to 3 decimal places) The yintercept is −1.046. (–7, 0)
y
x −6 −44 −2 0 2 Asymptote (0, 2 log10 (3) − 2) x=3 −22
f (x (x) x) = 2 log10 (3 − x) x −2 7 8
b
State the domain and the range. State the transformations.
1
To graph y = 2 log10 (3 − x) − 2 on a Graphs page, complete the function entry line as: f1(x) = 2 log10 (3 − x) − 2 Press ENTER ·. Note: The vertical asymptote at x = 3 is not displayed.
2
To locate the xintercept on a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve (f 1(x) = 0, x) Press ENTER ·.
3
Write the xintercept in coordinate form. To calculate the yintercept, complete the entry line as: ff1(0) and press ENTER ·. Write the yintercept in coordinate form.
4
(−∞,
The domain is 3) and the range is R. Reflection in the yaxis, dilation 2 units from the xaxis, vertical translation 2 units down, horizontal translation 3 units to the right
The coordinates of the xintercept are (−7, 0). The coordinates of the yintercept are (0, −1.046) (correct to 3 decimal places).
REMEMBER
1. The function f (x) = loga (x) is a logarithmic function where a ∈ R+\{1}. (a) The vertical asymptote is the yaxis, so there are no yintercepts. (b) The graph crosses the xaxis at (1, 0) because loga (1) = 0. (c) The domain is R+; that is, there are no negative values of x. (d) The range is R.
Chapter 4
Exponential and logarithmic graphs
189
2. Dilation (a) The function f (x) = A loga (x) dilates the graph of f (x) = loga (x) by a factor of A units from the xaxis. The vertical asymptote, xintercept, domain and range remain the same. As A increases, the graph becomes steeper. (b) The function f (x) = loga (kx) dilates the graph of f (x) = loga (x) by a factor of 1 k from the yaxis. The vertical asymptote, domain and range stay the same, but the xintercept is 1 . As k increases, the graph becomes steeper and the xintercept k becomes smaller. 3. Translation (a) The function f (x) = loga (x) + B translates the graph of f (x) = loga (x) vertically B units. The vertical axis, domain and range remain the same but the xintercept changes. There is no change in the shape of the graph. (b) The function f (x) = loga (x + b) translates the graph of f (x) = loga (x) horizontally −b units. The shape and the range remain the same but the vertical asymptote, the xintercept and the domain change: (i) The vertical asymptote becomes x = −b. (ii) The graph crosses the xaxis at (1 − b, 0). (iii) The domain is (−b, ∞). 4. Reflection (a) The function f (x) = −loga (x) reflects the graph of f (x) = loga (x) in the xaxis. All key features remain the same but the graph is a decreasing function instead of an increasing function. (b) The function f (x) = loga (−x) reflects the graph of f (x) = loga (x) in the yaxis. The vertical asymptote and the range remain the same but the xintercept and the domain change. (i) The graph crosses the xaxis at (−1, 0). (ii) The domain is (−∞, 0).
EXERCISE
4B
Logarithmic graphs to any base 1 a c e g
Find the vertical asymptote of each of the following functions. f (x) = log2 (x) b f (x) = log10 (x) f (x) = 2 log10 (x) d f (x) = 5 log2 (x) f (x) = log2 (3x) f f (x) = log10 (4x) f (x) 3 log10 (2x (2 ) h f (x) = 2 log2 (3x)
2 WE4 Sketch the graph of each of the functions in question 1, showing the intercepts and asymptotes, and stating the domain and range of each one. 3
State the horizontal and vertical translations required to transform f (x (x) = log2 (x ( ) into the following functions. a f (x) = log2 (x + 2) b f (x) = log2 (x + 1) c f (x) = log2 (x − 3) d f (x) = log2 (x − 4) e f (x) = 1 + log2 (x − 5) f f (x) = log2 (x − 3) + 2 g f (x) = 2 + log2 (x + 1) h f (x) = 3 + log2 (x + 1) i f (x) = log2 (x + 3) − 2 j f (x) = log2 (x + 1) − 2
4 WE5 Sketch the graph of each of the functions in question 3, showing the intercepts and asymptotes, and stating the domain and the range of each one.
190
Maths Quest 12 Mathematical Methods CAS for the TINspire
eBoo k plus eBook Digital doc
Spreadsheet 069 Logarithmic graphs
5 Sketch the graph of each of the following functions showing intercepts and asymptotes, and stating the domain, range and the transformation of f (x (x) = log10 (x). a f (x) = −log10 (x) b f (x) = −2 log10 (x) c f (x) = log10 (−x) d f (x) = 3 log10 (−x) e f (x) = 1 − log10 (x) f f (x) = 1 + log10 (x) g f (x) = 2 + log10 (−x) h f (x) = 2 − log10 (−x) 6 Sketch the graph of f (x (x) = log2 (2x (2 − 1), showing intercepts and asymptotes, and stating the domain, range and transformations of the graph f (x (x) = log2 (x ( ). 7 WE6 Sketch the graph of each of the following functions, showing intercepts and asymptotes, and stating the domain and range of each one. Give exact values or round to 1 decimal place. a f (x) = 2 log2 (x) + 3 b f (x) = 3 log10 (x) + 1 c f (x) = log10 (1 − x) d f (x) = log10 (3 − x) e f (x) = 1 − 2 log2 (x) f f (x) = 2 − 5 log10 (x) g f (x) = 3 log10 (x + 1) − 2 h f (x) = 4 log2 (x − 2) − 1 i f (x) = 1 + 2 log10 (3x) j f (x) = 3 − 4 log10 (2x (2 ) 8 Each graph in question 7 is a transformation of the graph of f (x (x) = loga (x), a = 2, 10. State the base of each graph and the kinds of transformation that have taken place. 9 Sketch the graph of f: f R → R where f (x (x) = 3 log2 (2 − x) + 1, showing all key features and stating the domain and range. 10 Sketch the graph of ff: [0, 10] → R where f (x (x) = 3 log10 (x ( + 1) − 2, showing key features.
eBoo k plus eBook Digital doc
WorkSHEET 4.1
4C
11 MC When the function f (x (x) = log2 (x ( + 2) is translated 2 units up and 3 units to the left, the function becomes: A f (x) = log2 (x + 4) + 3 B f (x) = log2 (x + 5) + 2 C f (x) = log2 (x − 1) + 2 D f (x) = log2 (x + 5) − 3 E f (x) = log2 (x − 1) + 2
Graphs of exponential functions with base e Graphs of exponential functions with base e are drawn in exactly the same way as with any other base. The graph of f (x) = ex can be dilated, translated and reflected in the same way. The graphs of f (x) = ex, g(x) = 2x and g(x) = 10x are shown at right. The graph of f (x) = ex is shown in red. It can be seen that h(x) = 10x is steeper than f (x) = ex and h(x) = 2x is less steep than f (x) = ex.
Features common to all three graphs
y 4
−2
h(x) = 10x f x) = ex f(
g(x) = 2x 3 (1, e) 2 (1, 2) Asymptote (0, 1) y=0 x 0 2 −1 1
1. The graphs all cross the yaxis at the point (0, 1). 2. The horizontal asymptote is the xaxis or the line y = 0. 3. The domain is R. 4. The range is R+. 5. They are all increasing functions.
WORKED EXAMPLE 7 3 − 1. State the transformations of f (x (x) = ex needed to form the graph of f (x (x) = 2 e3x
THINK 1
Write the rule.
WRITE/DRAW
f (x) = 2e3x − 1
Chapter 4
Exponential and logarithmic graphs
191
2
3
State the dilation. The coefficient of ex gives the dilation from the xaxis. The reciprocal of the coefficient of x gives the dilation from the yaxis.
Dilation by a factor of 2 units from the xaxis
State the translation. The vertical translation is given by the constant added to the ex term.
The graph is translated 1 unit down.
Dilation by a factor of
1 2
units from the yaxis
WORKED EXAMPLE 8
Sketch the graph of f (x (x) = ex. On the same set of axes sketch the graph of f (x (x) = ex − 2, marking the asymptote and yintercept, and state the transformation, the domain and the range. Give exact answers. THINK
WRITE/DRAW
1
Write the rule for the first graph.
f (x) = ex
2
State the basic shape.
Exponential curve
3
State the horizontal asymptote.
The horizontal asymptote is the xaxis.
4
Find the yintercept by making x = 0.
If x = 0, y = e0 y = 1 is the yintercept.
5
Draw the graph.
y
f x) = ex f(
2
Asymptote y = 0
(0, 1) −3 −2 −1 0
1
x
6
Write the rule for the second graph.
f (x) = ex − 2
7
State the transformation. The horizontal translation is given by the constant added to the x term.
The horizontal translation is 2 units to the right.
8
State the horizontal asymptote, which is the same as for f (x) = ex.
The horizontal asymptote is the xaxis.
9
Find the yintercept by making x = 0.
If x = 0, y = e0 − 2 − y=e 2 =
10
Sketch the graph on the same set of axes.
y 2 (0, 1)
1 e2 f x) = ex f( f x) = e (x − 2) f( (0, e−2) Asymptote y = 0. (2, 1)
−3 −2 −1 0
192
2 3
1
2 3
x
11
Check the accuracy of your graph by transforming the point (0, 1) and marking it on your graph.
(0, 1) → (2, 1)
12
State the domain and the range which are the same for both graphs.
The domain is R and the range is R+.
Maths Quest 12 Mathematical Methods CAS for the TINspire
For graphs of y = Aekx, k > 0 the coefficient of the ex term shows the dilation of A from the xaxis and the reciprocal of the coefficient of the x term shows the dilation of 1 from the yaxis. k The mapping is (x, y) → ( 1 x, Ay). k WORKED EXAMPLE 9 2 , marking the asymptote and intercept, and state the transformations, Sketch the graph of y = 3e2x domain and range.
THINK
WRITE/DRAW
1
Write the rule.
2 f (x) = 3e2x
2
State the basic shape.
Exponential curve
3
State the transformations.
Dilation by a factor of 3 units from the xaxis and dilation by a factor of 12 unit from the yaxis
4
Find the asymptote, recognising that it is unchanged by dilation.
The horizontal asymptote is the xaxis.
5
Find the yintercept by making x = 0 or by multiplying yvalues of f (x) = ex by 3. (0, 1) → (0, 3)
If x = 0,
6
Sketch the graph.
y = 3e2 × 0 =3×1 =3 so yintercept is 3. y 4 2
(0, 3)
−3 −2 −1 0 1 2 3 7
State the domain and range.
2 f (x) = 3e2x Asymptote y = 0
x
The domain is R and the range is R+.
The vertical translation is given by the constant added to the ex term. The horizontal translation is given by the constant added to the x term. WORKED EXAMPLE 10
Sketch the graph of f (x (x) = ex − 2 + 1, marking the asymptote and intercept, and state the transformations, domain and range. Find the yintercept correct to 2 decimal places. THINK
eBoo k plus eBook Tutorial
int0536 Worked example 10
WRITE/DRAW
1
Write the rule.
f (x) = ex − 2 + 1
2
State the basic shape.
Exponential curve
3
State the transformations.
The horizontal translation is 2 units to the right and the vertical translation is 1 unit up.
4
Find the asymptote by translating the asymptote of f (x) = ex up one unit.
The horizontal asymptote is y = 1.
5
Find the yintercept.
If x = 0, y ≈ 1.14 (to 2 decimal places) so the yintercept is 1.14.
6
Find another point by translating the yintercept of (0, 1) → (2, 2) f (x) = ex to the right by 2 units and up by 1 unit.
Chapter 4
Exponential and logarithmic graphs
193
7
y
Sketch the graph.
(2, 2)
(0, 1.14) 2 −1 0 8
State the domain and the range.
1
Asymptote y=1
2
f (x) = ex − 2 + 1
x
3
The domain is R and the range is (1, ∞).
A reflection in the xaxis is shown by a negative sign before the ex term. A reflection in the yaxis is shown by a negative sign before the x term. WORKED EXAMPLE 11
Sketch the graph of f (x (x) = 2 − e−x, marking the asymptote and intercepts. State the transformations, domain and range. Give exact answers. Check using a CAS calculator. THINK
WRITE/DRAW
1
State the rule.
f (x) = 2 − e−x
2
State the basic shape.
Exponential curve
3
State the transformations.
A reflection in the xaxis and a reflection in the yaxis. The vertical translation is 2 units up.
4
Find the horizontal asymptote by translating the asymptote of f (x) = ex up 2 units.
The horizontal asymptote is y = 2.
5
Find the yintercept by making x = 0 or by reflecting (0, 1) in the xaxis and translating it up 2 units.
If x = 0,
6
Find the xintercept by making y = 0 and solving the equation. 1 e−x = x e loge (eex) = x loge (e) =x×1
If y = 0,
−
y=2−e 0 =2−1 =1 or (0, 1) → (0, −1) → (0, 1) The yintercept is 1. 2 − e−x = 0 − =2 e−x 1 =2 ex ex = 12 loge (eex) = loge 1 2
x=
loge 1 2
so the xintercept is loge 1 . 2
7
Sketch the graph.
y (loge( 1–2 ), 0)
2
Asymptote y=2 (0, 1) 0 1
−1
f (x) = 2 − e−x x
−2 8
194
State the domain and the range.
The domain is R and the range is (−∞, 2).
Maths Quest 12 Mathematical Methods CAS for the TINspire
9
To graph y = 2 − e−x, on a Graphs page, complete the function entry line as: f1(x) = 2 − e−x Then press ENTER ·. Note: The horizontal asymptote at y = 2 is not displayed.
2 , g(x) = e2x 2 − 1 and h(x) = e2x 2 −2 The graphs of f (x) = e2x are drawn at right.
y
2 f x) = e 2x f( 2 −1 g(x) = e 2x
2
2 −2 h(x) = e 2x
(0, 1) 1 (0, e−1) 0
−1 2 f (x ( ) = e2x
2 −1 g(x ( ) = e2x (x = e 2 x −
xintercept
—
—
yintercept
(0, 1)
Domain Range Horizontal asymptote
(
− 1 1 0, e = 0, e
2 −2 h(x ( ) = e2x (x 2( 2(x = e − 1)
—
)
(
− 1 2 0, 2 = 0, e e
R
R
R
R+ = (0, ∞)
R+ = (0, ∞)
R+ = (0, ∞)
y=0
y=0
y=0
1 2
Horizontal translation A point on the graph
1 2
(0, 1)
unit to right
( , 1) 1 2
(1, 1) ( —12 , 1) Asymptote (0, e−2) y = 0 x 1
)
1 unit to right (1, 1)
In summary: Transformations can be represented in the following way: 1. Dilation by factor A from the xaxis can be written as y = Af (x), A > 0. 2. (a) Dilation by factor 1 from the yaxis can be written as y = f (kx), k > 0. k (b) Dilation by factor k from the yaxis can be written as y = f 1 x , k > 0. k 3. (a) Translation of B units up can be written as y = f (x) + B, B > 0. (b) Translation of B units down can be written as y = f (x) − B, B > 0. 4. (a) Translation of b units to the right can be written as y = f (x − b), b > 0. (b) Translation of b units to the left can be written as y = f (x + b), b > 0. 5. (a) Reflection in the xaxis can be written as y = −f (x). (b) Reflection in the yaxis can be written as y = f (−x).
( )
Chapter 4
Exponential and logarithmic graphs
195
REMEMBER
For all the graphs of the form f (x) = Aex + b + B, where A ∈ R+, b, B ∈ R, the maximal domain is R, the range is (B, ∞), the horizontal asymptote is y = B. When A ∈ R− the range changes to (−∞, B). 1. A is the dilation factor from the xaxis. As A increases, the graph becomes steeper. 2. B is the vertical translation. When B > 0 the graph is translated up B units. When B < 0 the graph is translated down B units. 3. b is the horizontal translation. When b > 0 the translation is to the left. When b < 0 the translation is to the right. 4. When A < 0 the graph is reflected in the xaxis. 5. When f (x) = e−x the graph is reflected in the yaxis. 6. When f (x) = ekx the graph is dilated by factor 1 from the yaxis. k 7. Consider transformations in the following order: reflection, dilation, translation. EXERCISE
4C
Graphs of exponential functions with base e 1 WE7 State the transformations of f (x (x) = ex needed to form each of the following functions. 2 a f (x) = e3x b f (x) = e2x c f (x) = 4 e d f (x) = 2e e f (x) = 1 + ex − 2 f f (x) = 2 + ex + 5 2 2x g f (x) = 3 − e h f (x) = 1 − e3x 2 i f (x) −1 x 2
x 3
eBoo k plus eBook Digital doc
Spreadsheet 034 Exponential functions
2 WE8 For each of the following examples sketch the graph of f (x (x) = ex. On the same set of axes, sketch the graph of the given functions, marking the asymptote and yintercept, and state the transformation, the domain and the range. Give exact answers. a f (x) = ex − 1 b f (x) = ex − 3 c f (x) = 2eex d f (x) = 3eex e f (x) = 1 + ex f f (x) = ex − 2 g f (x) = e2 − x h f (x) = e1 − x i f (x) = 3 − ex 3 WE9 Sketch the graph of each of the following functions, marking the asymptote and intercept, and stating the transformations, domain and range. For the intercepts give exact answers or correct to 1 decimal place where appropriate. 2 a f (x) = e3x b f (x) = e2x c f (x) = 4eex 1 1 4x x d f (x) = 2ee e f (x) = 2 e f f (x) = 4 e 2 x g f (x) = 2e3x h f (x) = 3e4x 4 WE10 Sketch the graph of each of the following, marking the asymptote and intercepts, and stating the transformations, domain and range. Mark the yintercepts as exact values. a f (x) = ex + 3 b f (x) = ex − 3 c f (x) = ex + 1 x x + 1 d f (x) = e + 2 e f (x) = e +2 f f (x) = ex + 2 + 1 x − 1 x − 2 g f (x) = e +3 h f (x) = e +5 i f (x) = ex + 2 − 1 5 WE11 Sketch the graph of each of the following functions, marking the asymptote and intercepts, and stating the transformations, domain and range. For intercepts, give exact answers or correct to 1 decimal place where appropriate. a f (x) = e−x b f (x) = −ex c f (x) = 1 − ex −x −x d f (x) = 2 + e e f (x) = 1 + e f f (x) = 1 + ex −x −x g f (x) = 3 − e h f (x) = 5 − e i f (x) = −2e−x 196
Maths Quest 12 Mathematical Methods CAS for the TINspire
2 − 3 − 1, marking all the key features and stating the domain 6 Sketch f: f R → R where f (x (x) = e2x and range. Write answers correct to 2 decimal places where appropriate.
7 Sketch f: f [0, ∞) → R where f (x (x) = ex − 2 − 1, marking all the key features and stating the domain and range, giving answers correct to 2 decimal places where appropriate. 8 If f (x (x) = ex, sketch the following functions. a y = f (x) + 1 b y = f (x) − 2 1 d y = 2 f (x) e y = 1 − f (x) g y = f (x + 1) h y = f (x − 2) j y = 1 + f (−x)
c y = 2f 2f (x) f y = 2 − f (x) i y = f (−x)
2 + 3 is translated so that the new graph has a horizontal 9 MC The graph with equation y = e2x asymptote of y = −1. The new graph has undergone a translation of: A 1 unit down B 2 units down C 3 units down D 4 units down E 5 units down
10 MC An increasing exponential function has a horizontal asymptote of y = 2 and a yintercept of 4. A possible equation for this exponential is: 2 +4 2 −4 A y = e2x B y = e4xx + 2 C y = e2x x x y D y = 2ee + 4 E y = 2ee + 2 −
11 The graph shown is modelled by the equation y = 2eex b + B. a Find the values of B and b. b If the graph was translated down 1 unit, translated 3 units to the left and dilated by factor 12 from the xaxis, write the equation of the new graph.
4D
4 3 2 −2 −1
(3, 3) Asymptote y=1 0 1 2 3
4
x
Logarithmic graphs to base e Graphs of logarithmic functions with base e are drawn in exactly the same way as with any other base. The function f (x) can be dilated, translated and reflected in the same way. The graphs of f (x) = log2 (x), g(x) = log10 (x) y Asymptote f x) = log2 (x) f( x=0 and h(x) = loge (x) are shown at right. The graph of 2 h(x) = loge (x) h(x) = loge (x) is in blue. (1, 0) 1 f (x) = log2 (x) and h(x) = loge (x) are steeper than g(x) = log10 (x) g(x) = log10 (x). x Remember h(x) = loge (x) −1 0 1 2 3 4 5 −1 ⇔ h(x) = ln (x)
Common features 1. The graphs all cross the xaxis at (1, 0) because loga (1) = 0. 2. The vertical asymptote is the yaxis (x = 0) because loga (0) is undefined. 3. The domain is R+. 4. The range is R. 5. They are all increasing functions. WORKED EXAMPLE 12
State the transformations of f (x (x) needed to form the graph of f (x (x) = 2 loge (x ( − 3) + 1. THINK
WRITE
1
State the rule.
f (x) = 2 loge (x − 3) + 1.
2
State the dilation. The coefficient of loge (x) gives the dilation from the xaxis.
Dilation is 2 units from the xaxis.
Chapter 4
Exponential and logarithmic graphs
197
3
State the translations. (a) Horizontal translation is given by the constant added to the x term.
Horizontal translation is 3 units to the right.
(b) Vertical translation is given by the constant added to the log term.
Vertical translation is 1 unit up.
Dilations do not change the vertical asymptote, the domain or the range of a logarithmic graph. The xintercept does change, however. WORKED EXAMPLE 13
Sketch the graph of f (x (x) = 3 loge (2x (2 ), marking all key features and stating the domain and range. THINK
WRITE/DRAW
1
Write the rule.
f (x) = 3 loge (2x (2 )
2
State the basic shape of the curve.
A log graph with basic shape f (x) = loge (x).
3
State the vertical asymptote, which occurs at loge (0). Vertical asymptote is the yaxis.
4
Find the xintercept by letting y equal 0. Solve for x. Divide both sides by 3. Use loge (x) = a ⇔ ea = x. Use e0 = 1. Divide both sides by 2.
5
Sketch the graph of f (x) = 3 loge (2x (2 ).
If y = 0, 3 loge (2x (2 ) = 0 loge (2x (2 ) = 0 e0 = 2x 2 2 =1 2x x = 12 The xintercept is 12. y Asymptote x=0 6 4 2
6
State the domain and the range.
−1 0 −2
f (x) = 3 loge (2x (2 ) ( 1–2 , 0) 1
2 3
x
The domain is R+, and the range is R.
Translations do not change the shape of the basic graph, only the position. The asymptotes, intercepts and domain may change but the range stays the same. WORKED EXAMPLE 14
Sketch the graph of f (x (x) = ln ((x + 1) − 2, marking the vertical asymptote and the intercepts. THINK
198
WRITE/DRAW
1
State the rule.
f (x) = ln (x + 1) − 2
2
State the basic shape.
Log graph with shape f (x) = ln (x).
3
Find the vertical asymptote by making (x + 1) = 0.
x + 1 = 0 for asymptote so x = −1 is the vertical asymptote.
4
Find the yintercept by making x equal to 0. Remember that ln (1) = 0.
If x = 0, y = ln (0 + 1) − 2 = ln (1) − 2 − = 2 − yintercept = 2.
Maths Quest 12 Mathematical Methods CAS for the TINspire
5
Find the xintercept by making y equal to 0. Use ln (x) = a ⇔ ea = x.
6
Sketch the graph.
If y = 0,
ln (x + 1) − 2 = 0 ln (x + 1) = 2 e2 = x + 1 x = e2 − 1 x ≈ 6.39 xintercept ≈ 6.39 (to 2 decimal places) Asymptote x = −1 y −2
0 2 −2 (0, −2)
(e2 − 1, 0) 4 6 x
f (x) = ln (x + 1) − 2
4
A reflection of f (x) = loge (x) in the xaxis does not change the graph’s asymptote, xintercept, domain or range. If there is a yintercept, the sign changes. A reflection in the yaxis does not change the vertical asymptote or the range, but the xintercept and the domain change.
WORKED EXAMPLE 15
eBoo k plus eBook
Sketch the graph of f (x (x) = 2 − 3 loge (1 − x), marking the asymptote and intercepts. State the domain and range and check using a CAS calculator. THINK
Tutorial
int0537
Worked example 15
WRITE/DRAW
1
State the rule.
f (x) = 2 − 3 loge (1 − x)
2
Find the vertical asymptote by translating the line x = 0 one unit to the right or by making 1 − x = 0.
Vertical asymptote is x = 1.
3
Find the yintercept by making x equal to 0 and solving the equation.
If x = 0, y = 2 − 3 loge (1) =2
4
Find the xintercept by making y equal to 0.
If y = 0, 2 − 3 loge (1 − x) = 0 3 loge (1 − x) = 2 loge (1 − x) = 23 2
e3 = 1 − x
2
x = 1 − e3 x ≈ −0.95 (to 2 decimal places) 5
Sketch the graph, remembering that there is a reflection in both the x and the yaxes.
y 4 (0, 2) 2– (1 − e 3 , 0)
f (x) = 2 − 3 loge (1 − x)
x −22 −1 0 1 −2 Asymptote x=1 6
State the domain and the range.
The domain is (−∞, 1) and the range is R.
Chapter 4
Exponential and logarithmic graphs
199
7
To graph f (x) = 2 − 3 loge (1 − x) on a Graphs page, complete the function entry line as: f1(x) = 2 − 3 ln (1 − x) Then press ENTER ·. Note: The vertical asymptote at x = 1 is not displayed.
To sketch a graph by using transformations it is necessary to dilate, reflect and then translate. REMEMBER
1. The function f (x) = loge (x) is an increasing logarithmic Asymptote y x=0 function: y = loge (x) (a) The vertical asymptote is the yaxis so there are no yintercepts. (b) The graph crosses the xaxis at (1, 0) because loga (1) = 0. x 0 (1, 0) (c) The domain is R+; that is, there are no negative values of x. (d) The range is R. (e) After transformations, the function becomes f (x) = A loga k(x + b) + B. 2. Dilation (a) Dilations from the xaxis are given by the coefficient of loge (x). Then, y = f (x) becomes y = Af (x). (b) Dilations from the yaxis are given by the reciprocal of the coefficient of x. Then, y = f (x) becomes y = f (kx). (c) Dilations do not change the vertical asymptote, the domain or the range of a logarithmic graph. Dilations from the yaxis change the xintercept. 3. Translation (a) Vertical translation is given by the term added to loge (x). Then, y = f (x) becomes y = f (x) + B. If B > 0 the graph is moved up, and if B < 0 the graph is moved down. (b) Horizontal translation is given by the term that is added to x. y = f (x) becomes y = f (x + b). If b > 0 the graph is translated to the left, and if b < 0 the graph is translated to the right. (c) Translations do not change the shape of the basic graph, only the position. The asymptotes, intercepts and domain may change but the range stays the same. 4. Reflection (a) Reflection in the xaxis occurs if the coefficient A is negative. Then, y = f (x) becomes y = −f (x). (b) Reflection in the yaxis occurs if the coefficient of x is negative. Then, y = f (x) becomes y = f (−x). (c) When following a sequence of transformations to sketch a graph, it is necessary to dilate, reflect and then translate.
200
Maths Quest 12 Mathematical Methods CAS for the TINspire
EXERCISE
4D
Logarithmic graphs to base e 1 WE12 State the transformations of f (x (x) needed to form the graph of each of the following. a f (x) = 5 loge (2x (2 ) b f (x) = 2 loge (4x) c f (x) = loge (x + 1) − 3 d f (x) = loge (x − 2) + 1 e f (x) = −loge (−x) f f (x) = −loge (2x (2 ) g f (x) = 1 − loge (x − 2) h f (x) = 2 − loge (x + 3) i f (x) = 3 loge (x + 4) − 1 j f (x) = 1 − loge (x − 4) k f (x) = 2 loge (1 − x) + 3 l f (x) = 3 loge (2 − x) − 1
eBoo k plus eBook Digital doc
Spreadsheet 069 Logarithmic graphs
2 WE13 Sketch the graph of the following functions, marking all key features and stating the domain and range. Give exact values. a f (x) = loge (2x (2 ) b f (x) = loge (3x) c f (x) = 3 loge (x) x d f (x) = 2 loge (x) e f (x) = 2 loge f f (x) = 3 loge (2x (2 )
(3)
3 WE14 Sketch the graph of the following functions, marking the vertical asymptote and the intercepts. Give exact answers for a – f, otherwise round to 1 decimal place. a f (x) = 1 + loge (x) b f (x) = 2 + loge (x) c f (x) = loge (x − 1) d f (x) = loge (x − 2) e f (x) = loge (x + 2) f f (x) = loge (x + 3) g f (x) = loge (x − 3) + 2 h f (x) = loge (x − 1) + 1 i f (x) = loge (x + 3) − 2 j f (x) = loge (x + 1) − 2 4 WE15 Sketch the graph of the following functions, marking the asymptote and intercepts, and stating the domain and range. Give intercepts correct to 2 decimal places where appropriate. a f (x) = loge (−x) b f (x) = −loge (x) − c f (x) = 2 loge (x) d f (x) = loge (−2x 2 ) − − − e f (x) = 3 loge ( 2x 2 ) f f (x) = −2 loge ( 4x) g f (x) = loge (1 − x) h f (x) = loge (2 − x) i f (x) = −loge (2 − x) j f (x) = −loge (3 − x) 5 The graph of f (x (x) = loge (x ( ) undergoes the following transformations. Find the equation of the image of f (x (x) in each case. a Translation of 1 unit to the right. b Translation of 2 units down. c Dilation of 5 units from the xaxis. d Dilation of 3 units from the yaxis. e Reflection in the xaxis. f Reflection in the yaxis. g Reflection in the xaxis and translation of 3 units to the left. h Dilation of 4 units from the yaxis and translation 1 unit up. i Reflection in the xaxis, dilation of 2 units from the xaxis and translation of 3 units down. 6 Sketch the graphs of the equations found in question 5, showing the asymptote and intercepts, and stating the domain and the range. Give intercepts correct to 2 decimal places where appropriate. 7 State the transformations of f (x (x) needed to form the graph of f (x (x) = 2 loge (3x + 6) − 1 and sketch the graph, showing the asymptote and intercepts. Give intercepts correct to 2 decimal places. −
8 Sketch the graph of ff: ( ∞, 1] → R where f (x (x) = −3 loge (2 − x), showing key features. 9 MC The graph of y = loge (x) is transformed into the graph of y = 5 loge (2x (2 ) by: A A dilation of factor 5 from the xaxis and a dilation of factor 2 from the yaxis B A dilation of factor 2 from the xaxis and a dilation of factor 5 from the yaxis C A dilation of factor 15 from the xaxis and a dilation of factor 2 from the yaxis D A dilation of factor 5 from the xaxis and a dilation of factor 12 from the yaxis E A dilation of factor 12 from the xaxis and a dilation of factor 5 from the yaxis. Chapter 4
Exponential and logarithmic graphs
201
10 MC For the function f (x) = 3 loge (x − 2), the vertical asymptote and the coordinates of the xintercept are respectively: A x = 3, (2, 0) B x = 2, (3, 0) C y = 3, (2, 0) D y = 2, (3, 0) E x = 3, (3, 0) a 11 MC The function ff: [0.5, 6] → R where f (x) = is undefined when x is equal to: logg e ( x ) A 0 B 0.5 C 1 D 6 E a 12 a When the function f (x (x) = a + b loge (x ( ) is reflected in the xaxis and translated 2 units to the right, find g(x ( ), the equation of the image of f (x (x (x). b State the domain and range of g(x) and write it using correct function notation. c If g(x) is reflected in the yaxis and dilated by factor 4 from the xaxis, write in correct function notation the equation of h(x), the image of g(x).
4E
eBoo k plus eBook Digital doc
WorkSHEET 4.2
Finding equations for graphs of exponential and logarithmic functions As with other functions it is sometimes necessary to be able to find the equation of an exponential or logarithmic function from a graph. If we know points on the curve, we can substitute the values into the most suitable general equation: 1. For an exponential graph the general equation is y = Ae(x + b) + B. 2. For a logarithmic graph the general equation is y = A loge (x + b) + B. Both of these examples are written with base e but they could be written with any base, for example y = A × 2(x + b) + B or y = A log10 (x + b) + B. If there are two unknowns, two pieces of information are necessary. The coordinates of two points, substituted into a general equation, will give two equations and enable two unknowns to be found.
WORKED EXAMPLE 16 y 4
The equation of the graph shown is of the form f (x (x) = Aex + B. Find the values of A and B correct to 2 decimal places and hence find the equation of the function.
(0, 2) (−2.44, 0) −4 −3 −2 −1
THINK 1
2 3
202
Use the point on the yaxis and substitute values into the given equation. Substitute the coordinates of another point into the given equation. Solve simultaneous equations by subtracting [1] from [2].
0 1 x
WRITE
For (0, 2): 2 = Ae0 + B 2=A+B (−2.44, −2.44
For 0 = Ae
[2] − [1]:
Maths Quest 12 Mathematical Methods CAS for the TINspire
[1]
0): +B
[2] −2
−2
−2.44
= Ae −A − = A(e 2.44 − 1)
4
Find A, rounding the answer correctly.
−2
A=
−
2.444
Substitute in [1]: 2 =
−
2.444
e −1 ≈ 2.19 (to 2 decimal places)
5
Find B by substituting the exact value of A in equation [1] or [2] to find B.
−2
e
B=2−
e − ≈ 0.19 x f (x) = 2.19ee − 0.19
6
−
−1 −2 2.444
+B −1
Rewrite the original equation, substituting values for A and B. Note: You can use a CAS calculator to solve equations [1] and [2]. 1 On a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Press /r to obtain the expression template and choose the simultaneous equation template for two unknowns. Complete the entry line as: a + b = 2 solve , a −2.44 +b=0 a × e Press ENTER ·. 2
Write down the solutions correct to 2 decimal places.
∴ A = 2.19 and B = −0.19
The horizontal asymptote of an exponential equation gives the vertical translation and hence the value of B in the general equation f (x) = Aex + B. WORKED EXAMPLE 17
The equation of the graph shown is of the form f (x (x) = aex + b. Find the values of a and b and hence find the equation of the function.
y 10 (0, 5)
THINK
WRITE
−3 −2 −1
Asymptote y=2 0 1
2
x
1
Use the horizontal asymptote to find the value of b.
The graph of ex has been translated up 2 units, so b = 2.
2 3
Substitute the value of b in the equation. Use the yintercept to find a by substituting in the given equation.
4
Write the equation, using the values of a and b.
f (x) = aex + 2 For (0, 5): 5 = ae0 + 2 ae0 = 3 a=3 f (x) = 3eex + 2
If there are three unknowns, three pieces of information are necessary to solve the equation.
Chapter 4
Exponential and logarithmic graphs
203
The vertical asymptote of a logarithmic graph gives the horizontal translation and hence the value of b in the equation y = A loge (x + b) + B. A translation to the left gives a positive value of b and a translation to the right gives a negative value of b. WORKED EXAMPLE 18
The equation of the graph shown is of the form y = A loge (x ( + b) + B. Find the values of A, b and B and hence find the equation.
eBoo k plus eBook
y
x = −1 5 (0, 5) (2, 0) 1
2
3
Use the vertical asymptote to find the value of b.
2 3
Substitute the value of b into the equation. Use the yintercept to find an equation.
4
Simplify using loge (1) = 0.
5
Substitute the value of B in the equation. Use the xintercept to find A.
x
The graph of f (x) = loge (x) is translated 1 unit to the left, so b = 1. f (x) = A loge (x + 1) + B. For (0, 5): A loge (0 + 1) + B = 5 A loge (1) + B = 5 =5 f (x) = A loge (x + 1) + 5 For (2, 0): A loge (2 + 1) + 5 = 0 A loge (3) + 5 = 0 A loge (3) = −5
Substitute values back into the original equation.
−
5 log e (3) A ≈ −4.55 (to 2 decimal places) f (x) = −4.55 loge (x + 1) + 5 A=
7
4
WRITE
1
6
int0538 Worked example 18
0 THINK
Tutorial
REMEMBER
1. The general equation for an exponential graph is f (x) = Aek(x + b) + B, where A is the dilation factor from the xaxis, 1 is the dilation factor from the yaxis, b is the k horizontal translation and B is the vertical translation. 2. The general equation for a logarithmic graph is f (x) = A loge k(x + b) + B, where A is the dilation factor from the xaxis, 1 is the dilation factor from the yaxis, b is the k horizontal translation and B is the vertical translation. 3. The coordinates of points on the curve can be substituted into the general equation to find the values of unknowns in the equation. 4. One piece of information is required for each unknown in the general equation. 5. The horizontal asymptote of an exponential graph gives the vertical translation and hence the value of B. If the asymptote is above the xaxis B is positive and if it is below the xaxis the value of B is negative. 6. The vertical asymptote of a logarithmic graph gives the horizontal translation and hence the value of b. If the asymptote is to the left of the yaxis b is positive and if it is to the right of the yaxis b is negative.
204
Maths Quest 12 Mathematical Methods CAS for the TINspire
EXERCISE
4E
Finding equations for graphs of exponential and logarithmic functions 1 WE16 The equation of the graph shown is of the form f (x (x) = Aex + B. Find the values of A and B correct to 2 decimal places and hence find the equation of the function. State the equation of the horizontal asymptote.
y 4 2
(1, 4) (0, 3) 0 1
2 Find the values of b and B if the general equation of the graph shown is y = ex + b + B. Give your answers correct to 2 decimal places.
2
3
y 6
4
x
(1, 6)
4 2 (0, 2) −1−0.5 0 0.5 1 1.5
3 Find the values of A and k if the general equation of the graph shown is y = Aekx. Give your answers correct to 2 decimal places.
x
y 0 −2 −1 (−1, −4) −5
1
2
x
−10 (−2, −10)
4 Find the values of A and B if the general equation of the graph shown is y = A × 2x + B. Give exact answers.
y 4 (1, 3) 2 x −33 −2 −1 0 1 2 Asymptote (−3, −1) −2 –– y = −19 15
5 WE17 The equation of the graph shown is of the form f (x (x) = a × 2x + b. Find the values of a and b and hence find the equation of the function. Give exact answers.
4 2
(0, 11–2 ) 1 Asymptote y=1 0 1 2 x −3 −2 −1
6 MC If the horizontal asymptote is y = 1 and the yintercept is −2, the equation for the exponential function of the form y = −e x + b + B is: A y = −ex + 1 − 2 B y = −ex + 2 + 1 − x + 1 C y= e −2 D y = −ex − 1 − 2 − x + 1.1 E y= e +1
Chapter 4
Exponential and logarithmic graphs
205
7 Find the values of b and B if the general equation of the graph shown is y = e x + b + B. Give exact answers.
y 6 4
(2, 4) Asymptote y=3
2 0 1
2
8 Find the values of A and B if the general equation of the graph shown is y = Aex + B . Give exact answers.
3
4
x
4 (0, 3) 2 0 1 2 x Asymptote −2 y = −1
−3 −2 −1
9 WE18 The equation of the graph shown is of the form y = A loge (x ( + b) + B. Find the values of A, b and B and hence find the equation. Give answers correct to 1 decimal place.
y Asymptote 6 − x= 2
10 Find the values of A and k, given that the graph of y = A log10 kx passes through the points (1, 1) and (3, 2). Give your answers correct to 4 decimal places where appropriate.
4
11 If the horizontal translation is 2 units to the left, the vertical translation is 3, the graph passes through the point (2, 6), and the equation is of the form y = A log2 (x ( + b) + B, find the values of A, b and B.
4F
Addition of ordinates
(2, 6)
2 −3 −2 −1
0 1
2
eBoo k plus eBook Interactivity
Sometimes we need to sketch the graph of a function that can be thought int0249 of as the sum of two functions. For example, the function y = x2 + ex can Addition of ordinates be thought of as the sum of the functions y = x2 and y = ex. Such a graph can be drawn by sketching the two individual functions on the same set of axes, then adding the yvalues (ordinates) for each xvalue and plotting the resulting points. This method is convenient to use when we know the basic shape of the individual functions but cannot recognise the basic shape of the given function. We can sketch both graphs on the same set of axes, then plot the resulting points by adding the yvalues for each xvalue, and hence sketch the new curve. For example, the graph of y = x2 + ex can be sketched using the addition of ordinates technique, since the basic shape of the function is not known, but the two individual functions are a basic positive parabola and a basic exponential curve. The graph of y = ex + 1 can also be thought of as the sum of two functions, but since we can recognise its shape as the basic exponential curve translated 1 unit up, there is no need to use the addition of ordinates method. Note that the ‘sum function’ can only be defined for the domain over which both of the individual functions are defined. So the domain of the ‘sum’ function is obtained by finding the intersection of the domains of the individual functions. That is, if h(x) = f (x) + g(x), domain h(x) = domain f (x) ∩ domain g(x). Sometimes a function is defined as the difference between two individual functions. We can still use the method of addition of ordinates to graph such a function, because the second function can be expressed as the addition of a negative function. That is, h(x) = f (x) − g(x) can be written as h(x) = f (x) + (−g(x)). 206
Maths Quest 12 Mathematical Methods CAS for the TINspire
x
So if, for example, we need to graph y = x2 − ex, we could think of it as y = x2 + (−ex). We then would sketch the basic parabola and the basic negative exponential curves on the same set of axes and use the addition of ordinates technique to obtain the graph of the original function. WORKED EXAMPLE 19
State the domain of f (x (x) = 3x 3 2 + loge (x ( ). THINK
WRITE
1
Write f (x) as the sum of the two individual functions g(x) and p(x).
Let f (x) = g(x) + p(x), where g(x) = 3x2 and p(x) = loge (x).
2
State the domain of g(x).
Domain g(x) = R
3
State the domain of p(x).
Domain p(x) = R+
4
State the domain of f (x) by finding the intersection of the domains of individual functions.
Domain f (x) = domain g(x) ∩ domain p(x) = R ∩ R+ = R+
When sketching the graph by hand, it is important to select wisely the values of x for which the ordinates should be added. As a guide, good points to select are: 1. the end points of the graph 2. the points of intersection of the graphs 3. the xintercepts of either of the graphs. WORKED EXAMPLE 20
Using the same scale and axes, sketch the graphs of y1 = ex and y2 = e−x over the domain [−2, 2). Hence, sketch the graph of y = ex + e−x, rounding coordinates to 1 decimal place. THINK 1
Sketch y1 = ex by finding the horizontal asymptote, the yintercept, and the end points.
WRITE/DRAW
The horizontal asymptote is the xaxis. x = 0, y = e0, y = 1 x = 2, y = e2, y = 7.4 − x = −2, y = e 2, y = 0.1 The yintercept is 1. y 8
(2, 7.4)
6
y1 = ex
4 (−2, 0.1)
2
(0, 1) 0
−2 −1 2
On the same axes, sketch y2 = e−x by reflecting y1 in the yaxis and finding the horizontal asymptote, the yintercept, and the end points.
x
2
1
The horizontal asymptote is the xaxis. The yintercept is 1. − y2 = e−x (−2, 2, 7.4)
y1 = ex
y 8
(2, 7.4)
6 4 (−2, 0.1) −2 −1
Chapter 4
2
(0, 1) 0
1
2
(2, 0.1) x
Exponential and logarithmic graphs
207
−
3
Add ordinates at the end point with the lowest xvalue (y = y1 + y2 = ex + e−x).
When x = −2, y = e 2 + e2 ≈ 7.5 Point (−2, 7.5)
4
Add ordinates when the graphs intersect.
When x = 0,
y=1+1 =2
Point (0, 2) When x = 2, Point (2, 7.5)
5
Add ordinates at the other end point.
6
Plot the points (which were obtained by adding ordinates), and join them to sketch the graph. It is necessary to label only end points and intercepts.
−x
y2 = e
(−2, 7.5)
y ≈ 7.5 y 8 6 4 2
−2 −1
y1 = ex
(2, 7.5)
y = ex + e − x
(0, 2)
0 1
2
x
WORKED EXAMPLE 21 y
Given the graphs of f (x (x) and g(x ( ), (x sketch the graph of h(x ( ) = f (x (x (x) + g(x ( ). (x
f x) f( g(x) x
0 THINK 1
2
Add the ordinates at the point where g(x) has the xintercept and mark the resulting point on the set of axes. Note that g(x) = 0 at this point; therefore f (x) + 0 = f (x) (that is, if one of the functions cuts the xaxis, the sum is equal to the yvalue of the other function).
3
Add the ordinates at the first point of intersection of the 2 functions. Note that at the point of intersection the value of y is the same for both functions, so the resulting point is double the yvalue. Mark the point on the set of axes.
4
208
Add the ordinates at the LHS end points of the graph: a large positive value plus a smaller negative value should give a smaller positive value. Mark this point on the axes.
WRITE/DRAW 1
0
3
Add the ordinates at the point where f (x) has the xintercept and mark the result on the set of axes.
Maths Quest 12 Mathematical Methods CAS for the TINspire
2
y
y
f x) f(
f x) f(
g(x)
g(x)
x
y
0
4
y
f x) f(
f x) f(
g(x) 0
x
x
g(x) 0
x
5
6
7
Add the ordinates at the second point of intersection of the 2 graphs and mark the resulting point on the set of axes.
y
5
Add the ordinates at the RHS end points of the graph: 2 positive values together give an even larger positive value.
Join the points with a smooth curve to produce the graph of h(x). Note that the turning point of h(x) is between the yaxis and the point with the same yvalue. The graph begins to turn at the left end point because the graph of g(x) turns there.
y
6
f x) f(
f x) f(
g(x)
g(x)
0
x
y
h(x)
0
x
f x) f( g(x) 0
x
WORKED EXAMPLE 22
Sketch the graph of f (x (x) = x + x , using addition of ordinates. THINK 1
State the two individual functions.
2
State the domain of f (x) by finding the intersection of the domains of individual functions.
3
On the set of axes sketch the graph of g(x) (a straight line, passing through the origin and bisecting the first quadrant) and p(x) (a basic square root curve). Note that only the first quadrant is needed, since the domain is R+ ∪ {0}.
WRITE/DRAW
Let f (x) = g(x) + p(x), where g(x) = x and p(x) = x Domain g(x) = R Domain p(x) = R+ ∪ {0} Domain f (x) = domain g(x) ∩ domain p(x) Domain= R ∩ R+ ∪ {0} Domain= R+ ∪ {0} 1
y g(x) p(x)
0 4
5
6
The first point of intersection of g(x) and p(x) is at the origin, f (x) = 0 + 0 = 0 (that is, f (x) will also start at the origin). Mark this point on the set of axes. Add the ordinates at the second point of intersection of the two graphs and mark the resulting point on the set of axes.
2
x
y
3
g(x)
y g(x)
p(x)
0
x
p(x)
0
x
Add the ordinates at the RHS end points of the graph and mark the resulting point on the set of axes.
Chapter 4
Exponential and logarithmic graphs
209
7
Sketch the graph of f (x) by joining the points. Note, as x increases, f (x) approaches g(x) because p(x) is increasing slowly.
f x) f( y g(x) p(x)
0
x
Further graphs The example below shows how to draw the graph of a product function. WORKED EXAMPLE 23
Sketch the graph of y = x2ex using a CAS calculator. Show all axis intercepts and any asymptotes. THINK
WRITE/DISPLAY
1
To graph y = x2ex on a Graphs page, complete the function entry line as: f1(x) = x2ex Then press ENTER ·. Note: y = x2ex is asymptotic to the negative xaxis, as when x → −∞, x2ex → 0.
2
Substitute x = 0 to locate the yintercept. Write the yintercept in coordinate form (which is also the xintercept).
y = 02e0 y=0 ∴ (0, 0) y = 0 is an asymptote.
REMEMBER
1. A graph of the sum of 2 functions can be drawn by sketching the 2 functions on the same set of axes and adding the yvalues for each value of x. 2. The domain of the ‘sum’ function is obtained by finding the intersection of the domains of individual functions: if h(x) = f (x) + g(x), domain h(x) = domain f (x) ∩ g(x). 3. Suitable points at which to add ordinates are: (a) left end points and right end points (b) the points of intersection of the 2 graphs (c) the xintercepts of either of the 2 functions. 4. Exact values of y can be obtained by substituting the required xvalues into the given function. 5. The ‘difference’ function can be rewritten as the ‘sum’ function as follows: h(x) = f (x) − g(x) = f (x) + [−g(x)], so that addition of ordinates can be used.
210
Maths Quest 12 Mathematical Methods CAS for the TINspire
EXERCISE
4F
Addition of ordinates 1 WE19 State the domain of each of the following functions. 1 a y=x+ b y=3 x −x x d y= x+2 −x g y=
e y = x2 − 3x3
2 − x2 ( x + 1)2
h y = 3x − 2 1 − x
c y = x2 + 1 x − 1 2 1 f y = 2x 2 3− x− 3 2 i y = 3(2 − x)2 + 3x
j y = 2 3− x − 3 x +1 2 MC If f (x) = g(x) + p(x), and if f (x) is defined over the domain [−2, 7] and g(x) is defined for x ≤ 7, the domain of p(x) could not be: A [−2, 7] B [−2, ∞) C [−2, 7) D [−2, 7.5) E [−2, 15] 3 WE20 Using the same scale and axes, sketch the graphs of y1 and y2 over the given domain. Hence, sketch the graph of y = y1 + y2, rounding coordinates to 1 decimal place as this is accurate enough for most sketch graphs. Check all graphs using a CAS calculator. a y1 = 2eex, y2 = 2e−x, [−1, 1] b y1 = 3e−x, y2 = 3eex, [−1, 1] −x − c y1 = e , y2 = x, ( 2, 2] d y1 = ex, y2 = x, (−2, 2] x 2 − e y1 = e , y2 = x , ( 2, 2) f y1 = e−x, y2 = x2, (−2, 3) 4 Using the same scale and axes, sketch the graphs of f (x (x) and g(x ( ). Hence, sketch the graph of (x h(x ( ) = f (x (x (x) + g(x ( ), rounding coordinates to 1 decimal place as this is accurate enough for most (x sketch graphs. State the domain and range of h(x ( ). Check all graphs using a CAS calculator. (x a f (x) = 2 loge (x), g(x) = x b f (x) = 3 loge (x), g(x) = x c f (x) = loge (x), g(x) = 2x 2 d f (x) = loge (x), g(x) = 12 x − − e f (x) = loge (x), g(x) = x f f (x) = 2 loge (x), g(x) = x 5 Using the addition of ordinates, sketch the graph of ff: [−2, 2] → R, where f (x (x) = 12 ex + 12 e−x. State the domain and the range, giving answers to 1 decimal place where rounding is necessary. 6 MC If the domain of y1 is (−2, 2] and the domain of y2 is (0, ∞), the domain of y1 + y2 is: A (−2, ∞) B [2, ∞) C (0, ∞) D (−2, 0) E (0, 2] 7 WE21 Given the graphs of f (x (x) and g(x ( ), sketch the graph of h(x (x ( ) = f (x (x (x) + g(x ( ). (x y y y a b c f x) f(
0
d
g(x)
x
0
g(x)
f x) f(
x y
x f x) f( g(x)
g(x)
e
y
0
f
f x) f(
y
f x) f(
0
g(x)
x 0
x
x
0
f x) f( g(x)
8 On the same set of axes sketch the graphs of f (x (x) = x2 and g(x ( ) = 5x + 6. Use the additionof(x 2 ordinates method to sketch the graph of y = x + 5x + 6.
Chapter 4
Exponential and logarithmic graphs
211
9 On the same set of axes sketch the graphs of f (x (x) = x3 and g(x ( ) = x2 − 1 and hence sketch the (x graph of y = x3 + x2 − 1, using the additionofordinates technique. 10 WE22 Sketch the graph of each of the following functions, using addition of ordinates. Check your answers using a CAS calculator. a y= x + 2− x
b y = 2x 2 − x
c y = x − 3 − x2
d y= x+5 + 5− x y
11 MC The graph at right is likely to represent the sum of which 2 functions?
y
A
B
0
C
x
y
0
D 0
0
y
x
x
y
E
y
x 0
x
0
x
12 Sketch the graph of y = loge x + x by sketching the 2 individual functions on the same set of axes and then adding the ordinates. State the domain of the function. Verify your answer with a CAS calculator. 13 If f (x (x) = 3x − 3 and g(x ( ) = loge x, sketch the graph of h( x ) = f ( x ) + g( x ), using addition of (x ordinates. Check your answer using a CAS calculator. 14 WE23 For each of the following, sketch the graph using a CAS calculator. Show all axis intercepts and any asymptotes. State the domain and range. 2 a y = x2e2x − b y = x2e−x c y = ex 2
15 For each of the following, sketch the graph using a CAS calculator. Show all axis intercepts and any asymptotes. State the domain and range. a y = x2 loge (x) b y = −x2 loge (3x) c y = x loge (x)
4G
Exponential and logarithmic functions with absolute values x for x ≥ 0 The modulus, or absolute value, function is defined as f ( x ) = x = − . x for x < 0 To obtain the graph of y = f x , the graph of y = f (x), where x ≥ 0, is reflected in the yaxis.
212
Maths Quest 12 Mathematical Methods CAS for the TINspire
y
The rule of the composite function y = f x , where f (x) = loge (x), x > 0, can be written as: logg e ( x ) y = log e x = − logg e ( x )
2 1
for x > 0 for x < 0
−2
−1
0 −1
2 x
1
−2 −3
To obtain the graph of y = f ( x ) , negative yvalues of y = f (x) are reflected in the xaxis. The rule of the composite function y = f ( x ) , where f (x) = loge (x), x > 0, can be written as: − logg e ( x ) y = log e ( x) x) = logg e ( x )
y 4 3 2 1
for 0 < x < 1
0
for x ≥ 1
1
2 y
The rule of the composite function y = f x , where f (x) = ex, can be written as: e x for x ≥ 0 y = ex = − x for x < 0 e
(0, 1) −2
y=
−1
0
1
2 x
y
The rule of the composite function y = f ( x ) , where f ( x ) = e x − k, k ∈ R+, can be written as: ex
4 x
3
y=k
− (e x − k ) for x < log e k −k = for x ≥ log e k e x − k
(0, 1 − k)
−2
−1
0
1
2 x
(loge(k), 0) WORKED EXAMPLE 24
For the function y = 2 log e x + 2 − 3: a sketch the graph of y = 2 log e x + 2 − 3, showing any asymptotes b calculate all axis intercepts both in exact form and correct to 2 decimal places c state the domain and range. THINK a
1
WRITE/DRAW y
3 Sketch the graph of y = 2 log e x + 2 − 3, x > −2. The graph is not defined when x = −2, so there is a vertical asymptote at x = −2.
2 1 −7 −6 −5 −4 −3 −22 −1−10
1 2 3 4 5 x
−2 −3 Asymptote x = −2
Chapter 4
Exponential and logarithmic graphs
213
2
y
Reflect y = 2 log e x + 2 − 3 in the vertical asymptote at x = −2.
2 1 −7 −6 −5 −4 −3 −22 −1−10
1 2 3 4 5 x
−2 −3 Asymptote x = −2
3
To graph y = 2 log e ( x + 2) − 3 with a CAS calculator, open a Graphs page and complete the function entry line as:
(
)
f1(x) = 2 log e x + 2 − 3 Then press ENTER ·. Press /r and select the absolute value template.
b
1
To locate the xintercepts, express y = 2 log e x + 2 − 3 as a hybrid function.
2
Substitute y = 0 and solve each equation.
3
Write the equation in exponential form and solve for x.
2 log e ( x + 2) − 3 y= 2 log e ( − ( x + 2)))) − 3 2 loge (x + 2) − 3 = 0 3 logg e ( x + 2) = 2
for x > − 2 for x < − 2
3
4
Repeat for the other xintercept.
x + 2 = e2 3
x = e2 − 2 ∴ x = 2.481 69 2 loge (−(xx + 2)) − 3 = 0 3 logg e ( − ( x + 2)) 2)) = 2 − ( x + 2)
3
= e2
3
x = −e2 − 2 ∴ x = −6.481 69 5
Write down the coordinates of the xintercepts in exact form and correct to 2 decimal places.
3 3 − e 2 − 2, 0 and e 2 − 2, 0 − (2.48, 0) and ( 6.48, 0)
6
Substitute x = 0 to obtain the yintercept and simplify.
y = 2 log e 0 + 2 − 3
y = 2 log e 0 + 2 − 3
y = 2 log e 2 − 3
y = 2 log e 2 − 3
∴y = −1.613 71 7
214
Write down the coordinates of the yintercept (0, 2 loge (2) − 3) in exact form and correct to 2 decimal places. (0, −1.61)
Maths Quest 12 Mathematical Methods CAS for the TINspire
c
1
State the domain.
x ∈ R\{−2}
2
State the range.
y∈R
WORKED EXAMPLE 25
eBoo k plus eBook
a Sketch the graph of y = e − 1 showing all axis intercepts and asymptotes. b State the domain and range. x
THINK a
1
2
b
3
Note that the graph of y = − 1 now has a horizontal asymptote at y = 1.
1
Substitute x = 0 to obtain the yintercept and simplify.
int0539 Worked example 25
WRITE/DRAW
Sketch the graph of y = e x − 1 . The graph has a horizontal asymptote at y = −1. Reflect the negative part (below the xaxis) of y = e x − 1 and the asymptote about the xaxis.
Tutorial
y 2 1 −2
−1
0 −11 − −2
1
Asymptote y=1 x 2 Asymptote y = −1
ex
y = e0 − 1 = 1−1 =0
2
Write the coordinates of the yintercept (which is also the xintercept).
The coordinates of the yintercept are (0, 0).
3
State the domain.
x∈R
4
State the range.
y ∈ R+∪{0}
REMEMBER
x for x ≥ 0 1. The modulus, or absolute value, function is defined as f ( x ) = x = − . x for x < 0 2. For y = f x , the graph of y = f (x), where x ≥ 0, is reflected in the yaxis. 3. For y = f ( x ) , negative yvalues of y = f (x) are reflected in the xaxis. EXERCISE
4G
Exponential and logarithmic functions with absolute values 1
WE24 Sketch the graphs of the following functions, showing all axis intercepts and asymptotes in exact form. For each graph, state the maximal domain and the range of the function.
b y = − log e ( x + 1) − 2 c y = log10 3 − x + 1 y = 3 log e x − 2 + 3 2 Sketch the graphs of the following showing all axis intercepts and asymptotes in exact form. For each graph, state the maximal domain and the range of the function. y = log e ( x − 2)
b y=
− log
e (1 −
x ) − 2) x)
Chapter 4
c y = log10 ( x + 3) + 1
Exponential and logarithmic graphs
215
3 Sketch the graphs of the following showing all axis intercepts in exact form. For each graph, state the maximal domain and the range of the function. a
x 1 y = e x−1 +4
c y = −e 4
− x −1
+2
WE25 Sketch the graphs of the following showing all axis intercepts and asymptotes in exact form. For each graph, state the maximal domain and the range of the function.
y = ex − 3 c y= e
4H
b y = e 4− x − 3
−x
b y = − ex − 2 + 4
−1 + 2
Exponential and logarithmic modelling using graphs As seen in chapter 3, exponential and logarithmic functions can be used to model real situations. Graphs of these functions can be used to illustrate the model and make predictions for future changes. In most cases when modelling real life situations, the domain is restricted to [0, ∞) because t = 0 when the model begins.
WORKED EXAMPLE 26
The population of wombats in Snubnose Gully is increasing according to the equation: W = 100e0.03t where W is the number of wombats t years after 1 January 1998. a Find the initial size of the population. b Find the population 2 years and 10 years after the number of wombats was first recorded. Give answers to the nearest whole wombat. c Graph W against t for 0 ≤ t ≤ 30. d Find the expected size of the population in the year 2020. e Find the year in which the wombat population reaches 250. WRITE/DRAW
THINK a
b
216
a W = 100e0.03t
1
State the rule.
2
Find W when t = 0.
When t = 0, W = 100
3
Write the answer in a sentence.
The initial size of the population is 100 wombats.
1
Find W when t = 2.
b When t = 2, W = 100e0.03
×2
= 100 × 1.0618 ≈ 106 (nearest whole number)
2
Write the answer in a sentence.
After 2 years there are 106 wombats.
3
Find W when t = 10.
When t = 10, W = 100e0.03 10 = 100 × 1.3499 ≈ 135 (nearest whole number)
4
Write the answer in a sentence.
After 10 years there are 135 wombats.
×
Maths Quest 12 Mathematical Methods CAS for the TINspire
c
1
To graph f (x) = 100e0.03x on a Graphs page, complete the function entry line as: f1(x) = 100e0.03x Then press ENTER ·. Note the viewing window settings: Xmin:0, Xmax:30, Ymin:0 and Ymax:250 When drawing your graph, label the axes with the given variables.
c
d
1
Convert the year 2020 to the correct number of years. Find W by substituting t = 22 into the equation.
d t = 2020 − 1998
2
e
= 22 years × When t = 22, W = 100e0.03 22 = 193.479 = 193 (nearest whole number) In the year 2020, there are approximately 193 wombats.
3
Write the answer as a sentence.
1
Let W = 250.
2
Divide both sides by 100.
2.5 = e0.03
3
Take natural logs of both sides.
loge (2.5) = loge (e0.03 t) 0.03t = loge (2.5)
4
Divide both sides by 0.03.
t=
5
Evaluate the answer using a CAS calculator.
t = 30.543 ∴ t = 31 (nearest year)
6
Express t = 31 as a year and write the answer as a sentence.
There will be 250 wombats in the year 2029.
e 250 = 100e0.03
×t
×t ×
1 logg ( 2.5) 0.003 e
REMEMBER
1. In general the number of unknowns indicates the number of points required to substitute into the equation. 2. The equation for exponential growth is A = A0ekt, where A0 represents the initial value, t represents the time taken and k represents the rate constant. This is a rapidly increasing function. 3. The equation for exponential decay is A = A0e−kt, where A0 represents the initial value, t represents the time taken and k represents the rate constant. This is a rapidly decreasing function. EXERCISE
4H
Exponential and logarithmic modelling using graphs 1
WE26 The population of a species of koala found on Eucalyptus Island is increasing according to the equation K = 50e0.04t, where K is the number of koalas t years after 1 January 1998.
Chapter 4
Exponential and logarithmic graphs
217
eBoo k plus eBook Digital doc
Spreadsheet 041 Function grapher
a Find the initial size of the population. b Find the population 2 years and 10 years after the number of koalas was first recorded. Give answers to the nearest whole number. c Plot a graph of K against t. d Use the graph to find the size of the population after 15 years. 2 The population of a species of wallaby found on a reserve is increasing according to the equation W = 150 × 1.08t, where W is the number of wallabies t years after records were first kept. a Find the initial size of the population. b Find the population 1 year and 5 years after records were first kept. Give answers to the nearest whole number. c Plot a graph of W against t. d Use the graph to find the size of the population after 15 years. e Use the graph to find how long it would take for the population to double. 3 A student invests $500 with a company that pays interest of 6% compounded continuously. (Interest paid according to the formula A = A0ert is said to be compounded continuously and r is called the continuous interest rate.) a How much money made up the initial investment? b How much did the student have with the company after 1 year? Give your answer correct to the nearest 5 cents. c How much interest did the student have after 5 years? Give your answer correct to the nearest 5 cents. d Plot a graph of A against t. e Use your graph to find how much the student would have in the account after 8 years. 4 The decay of a radioactive element, E, is given by the equation E = E0 e−kt, where E is the number of radioactive nuclei present t days after the experiment begins. a If 200 radioactive nuclei are present in the element at the beginning of the experiment and there are 33 radioactive nuclei present after 10 days, find the value of k correct to 2 decimal places. Use this rounded value for all working in the other parts of this question. b Find the number of radioactive nuclei, E, present after 1 day and after 5 days. c Plot a graph of E against t. d Use the graph to find how long it would take before 50 radioactive nuclei are left. e Would there ever be no radioactive nuclei left? Give reasons for your answer. f The halflife of a radioactive element is the time taken for half of the radioactive nuclei to decay. Use your graph to find the halflife of this element. 5 A hardboiled egg is placed in water to cool. The cooling process can be modelled by the equation T − T1 = (T T0 − T1)e−kt, where T is the temperature of the egg t minutes after it was placed in the water and T1 is the temperature of the water. Assume that the temperature of the egg is 98 °C when it is first placed in the 18 °C water and it takes 5 minutes for it to cool to 38 °C. a Substitute the values of T0 and T1 into the equation and simplify, making T the subject. b Find the value of k to 3 decimal places. c Substitute it into the equation. d Find how long it would take to reach a temperature of 25 °C. Give your answer to the nearest minute. e Draw a graph of T against t, marking in asymptotes. f Use the graph to find the temperature after 15 minutes. g Assuming that the water does not become significantly warmer, use your graph to determine whether the egg will ever reach the temperature of the water.
218
Maths Quest 12 Mathematical Methods CAS for the TINspire
6 The population of a species of fish in Lake Underwater is declining due to an unknown cause. The number of fish t weeks after the first dead fish was found on the surface of the water can be modelled by the equation P = A loge (t) + B. One week after the first dead fish was discovered, the population was 10 000. The population was 8000 after 5 weeks. a Find the values of A and B. Give answers correct to 2 decimal places where appropriate. b Find how many fish there are after 10 weeks. c Find how long it would take for the population to be 3000. Give your answer to the nearest week. d Sketch a graph of P against t, clearly indicating the asymptote. e Use the graph to find how many fish there would be after 20 weeks. 7
MC The relationship between m and n is modelled by the equation m = log10 (an + b). The graph is shown at right. The values of a and b are: A 3 and 2 B 2 and 3 C 3 and −2 − D 1 and 2 E 1 and 3
m 2 1 −1
(4, 1)
(1, 0) 0 1
2
3
4
n
8 It costs a clothing company $20 to produce a jacket. Production costs are proportional to the number of jackets produced. a If the company produces n jackets, write an equation for the company’s production costs for the jackets, $C. As the company produces more jackets they find that they have to sell them at a lower price. The company’s revenue, $R, is modelled by the equation − R(n) = 2000(1 − e 0.1n) b Show that R(0) = 0. c According to this model, revenues plateau as costs increase. What is the value that the revenue approaches? d On the same axes, sketch the graphs of the cost equation and the revenue equation for 0 ≤ n ≤ 55. e Use addition of ordinates to sketch the graph of the profit the company can expect to make from selling the jackets (profit = revenue − cost). f Write an equation for this profit. g Use a CAS calculator to find the number of jackets that must be sold in order to maximise the profits and what that profit would be. h Use a CAS calculator to find how many jackets the company could make before they began to make a loss. i Is this a reasonable model for a company to use? Explain your reasons.
Chapter 4
Exponential and logarithmic graphs
219
SUMMARY Exponential functions y • For graphs of the form f (x) = ax, where a ∈ R+\{1}: 1. The maximal domain is R. 4 2. The range is R+. 3. The xaxis is the horizontal asymptote. 2 4. The yintercept is 1. (0, 1) 5. They are all increasing functions. x 0 1 2 −2 −1 • Reflection: 1. If f (x) = ax is reflected in the xaxis the result is the graph of f (x) = −ax. The graph is a decreasing function instead of an increasing function. The yintercept changes to (0, −1) and the range becomes R−. 2. If f (x) = ax is reflected in the yaxis the result is the graph of f (x) = a−x. All key features stay the same but the graph is a decreasing function instead of an increasing function. • Translation: For all of the graphs of the form f (x) = ax + b + B, where b, B ∈ R, and a ∈ R+\{1}, the maximal domain is R, the range is (B, ∞), the horizontal asymptote is y = B and they are all increasing functions. b translates the graph horizontally, B translates the graph vertically. • Dilation: If f (x) = Aakx, where A, k ∈ R+, the graph of f (x) = ax is dilated by factor A from the xaxis and by factor 1 k from the yaxis. • Combinations of transformations: For all the graphs of the form f (x) = A × ak(x + b) + B, where b, B ∈ R, A, k ∈ R+ and a ∈ R+\{1}, the maximal domain is R, the range is (B, ∞), the horizontal asymptote is y = B and they are all increasing functions. b translates the graph horizontally, B translates the graph vertically and A dilates the graph by factor A from the xaxis, k dilates the graph by factor 1 from the yaxis. If A, k < 0 the graphs are reflected in the x and k yaxes, respectively. Logarithmic functions
• For graphs of the form f (x) = loga (x), where a ∈ R+\{1}: y 1. The maximal domain is R+; that is, there are no negative values of x. 2. The range is R. 3. The vertical asymptote is the yaxis so there are no x 0 (1, 0) yintercepts. 4. The graph crosses the xaxis at (1, 0) because loga (1) = 0. 5. They are all increasing functions. • Reflection: 1. If f (x) = loga (x) is reflected in the xaxis the result is the graph of f (x) = −loga (x). All key features remain the same but the graph is a decreasing function instead of an increasing function. 2. If f (x) = loga (x) is reflected in the yaxis the result is the graph of f (x) = loga (−x). The vertical asymptote and the range remain the same but the xintercept and the domain change. (a) The graph crosses the xaxis at (−1, 0). (b) The domain is (−∞, 0). • Dilation: 1. The function f (x) = A loga (x) dilates the graph of f (x) = loga (x) by a factor of A from the xaxis. The vertical asymptote, xintercept, domain and range remain the same. As A increases, the graph becomes steeper.
220
Maths Quest 12 Mathematical Methods CAS for the TINspire
2. The function f (x) = loga (kx) dilates the graph of f (x) = loga (x) by a factor 1 from the yaxis. The vertical k asymptote, domain and range stay the same, but the xintercept is 1 . As k increases, the graph becomes k steeper and the xintercept becomes smaller. • Translation: 1. The function f (x) = loga(x) + B translates the graph of f (x) = loga (x) vertically, B units. The vertical axis, domain and range remain the same but the xintercept changes. There is no change in the shape of the graph. 2. The function f (x) = loga (x + b) translates the graph of f (x) = loga (x) horizontally b units. The shape and the range remain the same but the vertical asymptote, the xintercept and the domain change: (a) The vertical asymptote becomes x = −b. (b) The graph crosses the xaxis at (1 − b, 0). (c) The domain is (−b, ∞). • Combinations of transformations: The function f (x) = A loga (x + b) + B has dilation factor A from the xaxis and is translated b units horizontally and B units vertically. If A < 0 there is a reflection in the xaxis and if −x is used there is a reflection in the yaxis. Addition of ordinates
• A graph of the sum of two functions can be drawn by sketching the two functions on the same set of axes and then adding the yvalues for each value of x. • If h(x) = f (x) + g(x), domain h(x) = domain f (x) ∩ domain g(x). • Suitable points at which to add ordinates are: 1. the end points of the graph 2. the points of intersection of the two graphs 3. the xintercepts of the two graphs. • The technique can be used for the difference of the two functions, if it is rewritten as a sum: h(x) = f (x) − g(x) = f (x) + [−g(x)]. Further graphs
• Use a CAS calculator to graph the product of two functions or composite functions. • Obtain the equation of any asymptote of the function by considering asymptotic behaviour of the individual functions. • On your graph, clearly label: 1. asymptotes 2. axis intercepts. Absolute value graphs
x for x ≥ 0 • The modulus, or absolute value, function is defined as f ( x ) = x = − . x for x < 0 • For y = f x , the graph of y = f (x), where x ≥ 0, is reflected in the yaxis. • For y = f ( x ) , negative yvalues of y = f (x) are reflected in the xaxis.
Chapter 4
Exponential and logarithmic graphs
221
CHAPTER REVIEW −
SHORT ANSWER ( + 1) − 4, showing 1 Sketch the graph of x) = 2(x intercepts and asymptotes, and stating the domain and range.
2 Sketch the graph of f (x (x) = 3 log10 (2x (2 ), showing intercepts and asymptotes, and stating the domain and range. 3 Find the equation of the graph below, given that it is of the form f (x (x) = Aex + B. y 4 2 x −4 −3 −2 −1 0 1 2 (0, −1) −2 y = −4 −44 −5
4 The graphs of f (x (x) and g(x ( ) are shown below. (x Sketch the graph of ((f + g)(x )( ), using the addition)(x ofordinates technique. y a g(x) f x) f( x
b
g(x)
y
b determine the rule for g 1(x), and hence state − the domain and range of g 1(x). 8 If h(x ( ) = f (x (x (x) + g(x ( ) where f (x (x (x) = x2 + 1 and g(x ( ) = loge (x (x ( ), sketch the graph of h(x ( )= (x f (x (x) + g(x ( ). State the domain and range (x of h(x ( ). (x 9 Describe a sequence of transformations that maps the graph of y = f (x (x) on to the graph of y = 2f 2f (3 – x) + 5. 10 After a protection program for tigers was introduced in a province of India on 1 January 2006, the population of the tigers is modelled by T t) = 50 × 20.4t, where t is the number of months T( after the start of the program. At the same time, the number of elephants in the province is modelled by − E(t) = 400 × 4 0.1t. Find: a the number of tigers and elephants in the province on 1 January 2006 b which of the two animals has the highest numbers in the province on 1 April 2007 c the date when the population of the tigers will equal the population of elephants in the province. If the number of elephants falls below 25, they are at risk of extinction in this province. d According to the model, will this happen? If so, when will it happen?
f x) f( x
5 a Use the law loga (m mp) = p loga (m) to simplify 2 f (x (x) = log10 (x ( ) and hence sketch the graph of the function ff: (0, ∞) → R where f (x (x) = log10 (x ( 2). b Sketch the graph over the domain, R, noting that it is now possible to take negative values of x. 6 P = P0e−t. If P = 120 when t = 0, find the exact value of P when t = 20. 7 For the function g(x ( ) = 2eex + 1 – 4: (x a sketch the graph of g(x) by finding the equation of any asymptotes and the coordinates of all intercepts
222
11 Sketch the graph of y = − log e ( x + 3) , showing all intercepts and asymptotes in exact form. State the maximal domain and the range.
Maths Quest 12 Mathematical Methods CAS for the TINspire
)
(
12 Let f : D → R, f ( x ) = 2 log log e x + 3 + 1, where D is the maximal domain of f. f a State D. b Find the exact coordinates of the points where the graph of y = f (x) intersects the x and yaxes. c Sketch the graph of y = f (x). Indicate any asymptote with its equation.
6 Which of the following graphs best represents the function f (x (x) = log10 (x ( − 2) + 1? A y x=2
2 −6 −4 −2 0 −2
2 4
6
2 4
6
2 4
6
x
−4
EXAM TIP
R/{−3}
R\{−3}.
a Some students wrote instead of Some also wrote R−3, or just −3, which are insufficient when specifying the domain. b Some students did not give exact values and some missed the second xintercept, because they were not aware of the absolute value function inside the natural logarithm. Instead of writing loge (3), some students appeared to write ‘3’. c Some students only drew the righthand branch of the function. Some did not show asymptotic behaviour correctly, particularly by having branches suddenly stop at a circle somewhere on the asymptote. The function should not curve away from the asymptote, nor touch it.
B
x = −2 y 2 −6 −4 −2 0 −2 − −4 −
C
x = −2 y 2
[Assessment report 1 2003]
[© VCAA 2003]
−6 −4 −2 0 −2 −
MULTIPLE CHOICE
1 The horizontal asymptote for the graph of f (x (x) = 2x − 1 is: A x=2 B x=1 C y=0 D y = −1 E y = −2
4 When the function f (x (x) = log10 (x ( − 1) is translated 2 units up and 1 unit to the left the function becomes: A log10 (x − 2) + 2 B log10 (x − 3) + 3 C log10 (x) + 2 D log10 (x + 1) + 2 E log10 (x + 2) + 1 5 The function f (x (x) = log2 (x ( + 1) has as its domain: A (1, ∞) B [1, ∞) C (−1, ∞) − D [ 1, ∞) E R
x
−4 −
D
2 The domain and range of the graph of f (x (x) = 3 × 10x + 1 are respectively: A R, [1, ∞) B R, (1, ∞) C R, [3, ∞) D R, (3, ∞) E R, R 3 The yintercept for the graph of f (x (x) = 1 − 2x is: A 0 B 1 C 2 D 3 E 4
x
y 2
0
x=2
1 2 3 4 5 6 7
x
−2
E
x = −2 y
2
−3 −2 −1
0 1 2 3 x
7 The graph of f (x (x) = 2eex + 1 is obtained from the graph of f (x (x) = ex by: A a dilation of 2 units from the yaxis and a translation of 1 unit to the left
Chapter 4
Exponential and logarithmic graphs
223
B a dilation of 2 units from the yaxis and a translation of 1 unit to the right C a dilation of 2 units from the xaxis and a translation of 1 unit to the left D a dilation of 2 units from the xaxis and a translation of 1 unit to the right E a dilation of 1 unit from the xaxis and a translation of 2 units to the right 8 If f (x (x) = ex, the function in the graph below is: y
14 If f (x (x) = loge (x ( + 2) + 1, then f (1) and f (0) are respectively (correct to 2 decimal places): A 2.10, 1.69 B 2.09, 0.69 C 2.098, 1.693 D 1.10, 1.69 E 2.10, 0.69 15 The domain of {{ff (x (x) + g(x ( )} is: (x A dom f ∪ dom g B dom f ∩ dom g C dom f D dom g E R −
16 If f (x (x) = g(x ( ) + p(x (x ( ), and dom g(x (x ( ) = ( ∞, 5] and (x − dom p(x ( ) = ( 2, ∞), then the domain of f (x (x (x) is: − − A ( 2, 5] B [ 2, 5) − − C [ 2, 5] D ( 2, 5) E none of the above
2 (0, 1) x −3 −2 −1 0 1 2 3 Asymptote –2 y = −2
17 Which of the following shows the graph of y=
A 3f 3f (x) + 2 C 2f 2f (x) + 3 E 2 + 3f 3f (x)
B 3f 3f (x) − 2 D 2 − 3f 3f (x)
x+4 + y
A
10 The vertical asymptote for the graph of f (x (x) = 3 loge (x ( − 2) + 1 is: − A x= 1 B x=0 C x=1 D x=2 E x=3
4
−4
4 x
4
13 The general equation of the graph shown y is y = Aex + B. The values of 4 A and B are: − − (0, 3) A 3, 1 B 2, 1 − − 2 C 4, 1 D 1, 3 E 4, 1
4 x
−4
4 x
y 4
−
12 If f (x (x) = loge (x ( + 1), the yintercept intercept of 22ff (x (x) + 3 is: A 0 B 1 C 2 D 3 E 4
y
2
E
4 x
–4
D
y
C −4
11 For the function ff: [ 1, ∞] → R where f (x (x) = 3 loge (x ( + 2), the domain and range are respectively: − A ( 2, ∞), (0, ∞) B (2, ∞), R − − D [ 1, ∞), [0, ∞) C ( 1, ∞), R+ E R+, R
−2
2
x
18 The decay of uranium235 is modelled by the equation U = U0ekt, where U is the number of grams of uranium235 after t million years. If a 1000 g mass of uranium235 decays to 907 g in 100 million years, the values of U0 and k are respectively: − A 100, 0.000 976 B 1000, 0.000 976 − − C 10 000, 0.000 907 D 1000, 0.000 907 E 100, 907
19 The function g has the rule g(x ( ) = logex − b, (x where b is a real constant. The maximal domain of g is: A R+ B R\{b} 0 2 x C R D (−b, b) Asymptote [© VCAA 2006] E (b, ∞) −
−2 y = 1
224
y
B
4
9 If the horizontal asymptote is y = 2 and the yintercept is 1, a possible equation for the graph is: A y = ex + 2 B y = 2eex + 1 C y = 2eex − 1 D y = ex − 2 x E y=2−e
−6 −4 −2
4−x ?
Maths Quest 12 Mathematical Methods CAS for the TINspire
20 The maximal domain D of the function f: f D→R with rule f (x ( ) = loge (x − 3) + 6 is: B (3, ∞) A R\{3} D (−3, ∞) C R − [© VCAA 2007] E ( ∞, 3) 21 The function f satisfies the functional equation x + y f ( x) + f ( y) f = where x and y are any 2 2 nonzero real numbers. A possible rule for the function is: A f ( x ) = log e x B f ( x) = 1 x C f ( x) = 2 x D f ( x) = 2 x E f ( x ) = sinn (2 x )
22 The function f has the property that f (x (x) + f (y (y) = (x (x + y) f (xy (xy) for all nonzero real numbers x and y. A possible rule for f is: A f (x (x) = 2x 2 B f (x (x) = 5 − 2x 2 1 C f ( x) = x D f (x (x) = ex E f ( x ) = log e x [© VCAA 2005]
23 Let f (x) = ex. For all positive real numbers x and y, f (x + y) is equal to: A f (x (x) + f (y (y) B f (x (x) f (y (y) C f (xy (xy) D (f (f (x (x)) y y [© VCAA 2003] E f (x (x )
[© VCAA 2006]
EXTENDED RESPONSE
1 Sketch the graph of f (x (x) = 2eex − 1 + 1, showing all key features. State the domain and the range. 2 Sketch the graph of f (x (x) = 1 − loge (2 − x), clearly showing intercepts and asymptotes, and state the domain and range. 3 By adding ordinates, sketch the graph of f (x (x) = log10 (2x (2 ) + log10 (x ( ). State the domain and the range. 4 N is the number of bacteria in a culture where N = 10 000e0.04tt is the formula for the number of bacteria in the culture after t hours. Find: a the initial number of bacteria b the number of bacteria after 10 hours. Give your answer to the nearest thousand. 5 Sketch the graph of y = 4 − e x − 1 showing all axis intercepts in exact form and any asymptotes. State the maximal domain and the range of the function. 6 Part of the graph of the function f : R → R, f (x (x) = a + be−x is shown at right. The line with equation y =1 is an asymptote and the graph passes through the origin. a Explain why a = 1 and b = −1. Let h be the function h:[0, 2] → R, h(x) = f (x). b State the range of h using exact values. − c Find the inverse function h 1.
y y=1
0
1
x
EXAM TIP
a Most students were able to explain why a = 1 by discussing the asymptote; however, many of them were unable to explain why b = −1. The most common error occurred when students tried to discuss reflections and translations but failed to indicate which graph they were transforming. − b This question was generally well done. Some students were unable to obtain full marks because they rounded up 1 − e 2 to 0.8647. Others used round brackets instead of square brackets. c When the inverse function is asked for, the domain must be given. Students will be penalised if the domain is left out. If only the rule for the inverse function is asked for, the domain does not have to be given. Some students gave the incorrect −
answer h 1 ( x ) = − log e 1 − x .
[Assessment report 2 2007]
[© VCAA 2007]
Chapter 4
Exponential and logarithmic graphs
225
log e (5 (5 − x) x ) + 1. 7 A function f is defined by the rule f ( x ) = log a Sketch the graph of f over its maximal domain. Clearly label any intersections with the axes with their exact coordinates and any asymptotes with their equations. − b Find the rule for the inverse function f 1. EXAM TIP
The graph on a calculator screen is of limited use. Many students did not give exact intercepts. Some tried to ‘cheat’ by finding numerical approximations and then writing these as fractions (which were incorrect). Many students failed to identify the vertical asymptote at all, while some identified a horizontal asymptote. If they failed to identify the vertical asymptote, the righthand end of the graph frequently ended abruptly on or close to the xaxis. Many students were also unable to represent asymptotic behaviour correctly. A few students still wrote a circle where their graph ended on the asymptote — this should not be encouraged, as students will lose marks for doing it. [Assessment report 1 2005]
[© VCAA 2005]
8 Kerri invested $5000 with a company that pays interest of 5% compounded continuously. To answer the questions below, use the formula A = A0ert, where A is the amount of the investment, A0 is the original investment, r is the continuous interest rate and t is the number of years since the money was originally invested. Round the answer to the nearest 5c. a How much money did Kerri first invest with the company? b How much money did Kerri have with the company after 1 year? c How much did she have after 5 years? If the company had been paying interest compounded quarterly, the formula used would have been
( )
4t
A = A0 1 + r . 4 d Using this system, how much would Kerri have after 5 years of the investment? e How long would it take to double the investment? Give your answer in years. f Which is the better investment? Give reasons. g What is the difference in the amount of interest after 5 years? h What would be the difference in 5 years if Kerri had invested $10 000? 9 A local council decided to build a new road along the coast. To make it safer it was decided to design the road so that it followed the curve represented by the equation y= 2 log10 (2x (2 − a) + 3 where a > 0. y The grid at right shows the road. In each direction, 1 unit represents Bridge 1 kilometre. Ship a If the new section of road goes 4 over the bridge marked on the graph, find the value of a. b Find the xcoordinate of the point where the road begins. Give the answer correct to 2 2 decimal places. How far would this be from the vertical axis to the nearest metre? c What is the shortest distance x 0 2 4 from the beginning of the road 6 8 1 3 5 7 9 to the coastline if the coastline is on the vertical asymptote at that point? Give the answer correct to the nearest metre. d The main highway is along the xaxis. How far is the road from the main highway when it is in line with the ship marked on the grid? Give the answer in kilometres, correct to 1 decimal place.
226
Maths Quest 12 Mathematical Methods CAS for the TINspire
10 The air pressure P in kilopascals (kPa) at a height of x kilometres (km) above sea level may be modelled by the equation P = aebx. A mountain climber uses an altimeter to record air pressures at known heights on a climb of Mount Kosciuszko (height 2.228 km). These pressures are shown in the table below. x
0
0.5
1.0
1.5
2.0
P
101.3
95.2
89.4
84.0
78.9
−
−
a If b is between 0.1 and 0.2, find the value for b (to 3 decimal places) which produces the best fitting model of the form P = aebx for the above data. b Use your model to predict the air pressure at the top of Mount Kosciuszko. 2 − bex + c is shown below. 11 The graph of the function f (x (x) = e2x
y y=6
(0, 2) 0
a b c d e f g h
x
Find the values of b and c. Show that the exact values of the xintercepts are x = loge (2) and x = loge (3). Use a CAS calculator to find the coordinates of the turning point. Round answers to 2 decimal places. Find the exact values of the coordinates of the point of intersection of the function and the horizontal asymptote. − eBoo k plus eBook If the function is reflected in the xaxis, fully define the new function g(x) = f (x). If the function is reflected in the yaxis, fully define the new function, h(x). Digital doc If the function is reflected in both the x and the yaxis, sketch the graph of Test Yourself the new function, k(x), write its equation and state the domain and the range. Chapter 4 Find the equation of f (2 − x) + 1. State the domain and range, rounding to 2 decimal places where appropriate.
Chapter 4
Exponential and logarithmic graphs
227
eBook plus
ACTIVITIES
Chapter opener Digital doc
• 10 Quick Questions: Warm up with ten quick questions on exponential and logarithmic graphs. (page 172) 4A
Graphs of exponential functions with any base
Tutorial
• WE 2 int0534: Watch how to sketch a graph of an exponential function. (page 177) Digital doc
• Spreadsheet 034: Investigate exponential graphs. (page 181) 4B
Logarithmic graphs to any base
Tutorial
• WE 5 int0535: Watch how to sketch the graph of a logarithmic function. (page 187) Digital docs
• Spreadsheet 069: Investigate logarithmic graphs. (page 191) • WorkSHEET 4.1: Sketch graphs of exponentials and logarithms, identify transformations and determine rules for graphs. (page 191) 4C
Graphs of exponential functions with base e
Tutorial
• WE 10 int0536: Watch how to sketch the graph of an exponential function and state the transformations. (page 193) Digital doc
• Spreadsheet 034: Investigate the exponential function. (page 196) 4D
Logarithmic graphs to base e
Tutorial
• WE 15 int0537: Sketch the graph of a logarithmic function stating the domain and range using a CAS calculator to check. (page 199) Digital docs
• Spreadsheet 069: Investigate logarithmic graphs. (page 201) • WorkSHEET 4.2: Sketch graphs of exponentials and logarithms and determine points of intersection between graphs. (page 202)
228
4E
Finding equations for graphs of exponential and logarithmic functions
Tutorial
• WE 18 int0538: Watch how to find the equation of an exponential given the graph. (page 204) 4F
Addition of ordinates
Interactivity int0249:
• Addition of ordinates: Consolidate your understanding of addition of ordinates using the interactivity. (page 206) 4G
Exponential and logarithmic functions with absolute values
Tutorial
• WE 25 int0539: Watch a worked example on how to sketch the graph of an absolute function. (page 215) 4H
Exponential and logarithmic modelling using graphs
Digital doc
• Spreadsheet 041: Invesigate graphs using a function grapher. (page 218) Chapter review Digital doc
• Test Yourself: Take the endofchapter test to test your progress. (page 227) To access eBookPLUS activities, log on to www.jacplus.com.au
Maths Quest 12 Mathematical Methods CAS for the TINspire
5
5A 5B 5C 5D
Relations and their inverses Functions and their inverses Inverse functions Restricting functions
inverse functions AREAS oF STuDy
• Graphs and identification of key features of graphs of functions studied • Graphs of inverse functions • Functions and their inverses, including conditions for the existence of an inverse function, and use of inverse functions to solve equations involving exponential, logarithmic and power functions
• The concept of an inverse function, connection between the domain and range of the original function and its inverse and the conditions for the existence of an inverse function, including the form of the graph of the inverse function for specified functions • Finding the rule of an inverse function and giving its domain and range eBook plus
5A eBook plus Interactivity
int0248 Relations and their inverses
Relations and their inverses
Digital doc
10 Quick Questions
You will recall from Maths Quest 11 Mathematical Methods CAS that a relation is a set of ordered pairs that can be graphed or may be described by a rule. As we have seen, the inverse of a relation can be found by: 1. interchanging the x and ycoordinates of an ordered pair; for example, (−1, −2) becomes (−2, −1) 2. reflecting the relation in the line y = x y 3. interchanging x and y in the rule and rearranging y=x 3 the equation to make y the subject; for example, (1, 2) y = x − 1 becomes x = y − 1, giving y = x + 1. 2 (2, 1) (0, 1) The domain of a relation becomes the range of its inverse 1 and the range of a relation becomes the domain of its inverse. (−1, 0) (1, 0) x 0 1 2 3 The diagram at right shows a set of ordered pairs −3 −2 −1 −1 (0, −1) − − − (−2, −1) A = {( 1, 2), (0, 1), (1, 0), (2, 1)}, the line y = x and the −2 inverse B = {(−2, −1), (−1, 0), (0, 1), (1, 2)}. (−1, −2) − −3 The domain of A is { 1, 0, 1, 2} and the range of A is {−2, −1, 0, 1}. The domain of B is {−2, −1, 0, 1} and the range of B is {−1, 0, 1, 2}.
using matrices to describe a reﬂection in the line y = x In chapter 2, we looked at how matrices can be used to find the image of a point that is transformed on a plane, or to find the rule for the graph of a relation that undergoes a transformation or series of transformations.
Chapter 5
inverse functions
229
As the graph of an inverse is the reflection in the line y = x of the original, we can use a matrix to describe this transformation and find the image of a point on f (x), or the rule of the inverse − f 1(x) given f (x). 0 1 x The matrix operation that produces a reflection in the line y = x is . 1 0 y Finding the image of a point using this method is trivial, as we know that the points on the graph of an inverse are found by interchanging the (x, y) values of any point on the original. This can be shown as follows: The graph of a relation passes through the point (3, −4). Find the image of this point on the graph of the inverse. 0 1 3 − 4 − = 1 0 4 3 That is, (3, −4) maps to (−4, 3). A more general application will be to find the rule of the inverse of a relation. This can be done using matrices but in practice this is unnecessary as it simply involves swapping the variables and rearranging the new equation. Again, it is not time effective to employ this method, but it is shown as a demonstration of the relationship of the inverse graph to its original. x x ′ 0 1 x y T = = = y y ′ 1 0 y x That is, (x, y) maps to (y', x' ). Therefore, we can write y' = x and x' = y, which relates to the practice of reversing the variables to find the rule of the inverse. WoRkED ExAMplE 1
Sketch the graph of each of the following relations. State the domain and range of each. a {(−3, −1), (−1, 1), (1, 3), (3, 5)} b y = x2 + 2 Think a
b
WRiTE/DRAW
1
Plot each coordinate pair on a set of axes.
2
The domain is the set of first elements of the ordered pairs.
3
The range is the set of second elements of the ordered pairs.
1
The relation is a parabola with a minimum turning point (0, 2).
2
Sketch the parabola.
3
The domain is the set of real numbers.
4
The range is any real number greater than or equal to 2.
a
y 5 4 3 2 1 −3−2−1 0 −1
b
Domain = {−3, −1, 1, 3} 1 2 3
y
x
Range = {−1, 1, 3, 5}
y = x2 + 2
(0, 2) 0
x
Domain = R Range = [2, ∞)
WoRkED ExAMplE 2
Find the inverse of each relation in worked example 1. Sketch the graph of each inverse relation, stating its domain and range.
230
Maths Quest 12 Mathematical Methods CAS for the Tinspire
Think a
WRiTE/DRAW a {(−3, −1), (−1, 1), (1, 3), (3, 5)}
1
Write the original relation.
2
Interchange the x and y elements to obtain the inverse. Plot these points on a set of axes.
3
The inverse is {(−1, −3), (1, −1), (3, 1), (5, 3)} y 3 2 1 −1 0 1 2 3 4 5 x −1 −2 −3
4 5
b
1 2 3 4 5
6
Domain = {−1, 1, 3, 5}
State the domain, which is the set of first elements of the ordered pair. State the range, which is the set of second elements of the ordered pair. Write the original relation. Interchange x and y in the rule.
Range = {−3, −1, 1, 3} b y = x2 + 2
The inverse is: x = y2 + 2 y2 = x − 2
Subtract 2 from both sides. Take the square root of both sides to make y the subject.
y=± x−2 y
The graph of y = ± x − 2 is a ‘sideways’ parabola with a turning point (2, 0). (use a CAS calculator to verify this graph.)
+ x− 2 y =−
x
2
0
7
Sketch the graph of the relation. State the domain.
Domain = [2, ∞)
8
State the range.
Range = R
Note that the domain of the original relation is the range of its inverse, and the range of the original relation is the domain of its inverse. WoRkED ExAMplE 3
eBook plus
For each relation graphed below, sketch the graph and its inverse on the same axes. Draw in the line y = x. a
b
y
0
(0, −5)
y
(2, 0) x
c
4 (−2, 2)
−2
(−2, 4)
(2, 2)
0
2
y
x
4
Tutorial
int0540 Worked example 3
(1, 4)
(−3, 0) 0
Chapter 5
x
inverse functions
231
Think a
b
c
DRAW
1
Copy the graph shown.
2
Interchange the points (2, 0) and (0, −5) to (0, 2) and (−5, 0), respectively, and mark them on a set of axes.
3
Draw a straight line through each pair of points.
4
Draw the line y = x, noting that the inverse is a reflection of the original function in y = x.
1
Copy the graph shown.
2
Interchange the points (−2, 2), (0, 0), (0, 4) and (2, 2) to (2, −2), (0, 0), (4, 0) and (2, 2) and mark on a set of axes.
3
Draw a circle through the second set of 4 points.
4
Draw the line y = x, noting that the inverse is a reflection of the original relation in y = x.
1 2
3
4
Copy the graph shown.
a
y
(0, 2) (−5, 0)
x
(2, 0)
0
(0, −5)
y=x b
y (0, 4)
(−2, 2)
y=x (2, 2)
2
(4, 0) x
0 −2 c
Interchange the points (−3, 0), (−2, 4), (0, 0) and (1, 4) to (0, −3), (4, −2), (0, 0) and (4, 1) and mark them on the same set of axes. Draw a ‘sideways cubic’ through the second set of points. Draw the line y = x, noting that the inverse is a reflection of the original function in y = x.
(−2, 4)
(2, −2) y 4
(1, 4)
y=x
2 (4, 1) (−3, 0) −2
0 −2 (0, −3)
2
4
x
(4, −2)
Note: This example shows that the graph of the inverse relation can be obtained by reflecting the graph of the original relation in the line y = x. Some specific points can be marked and used as a guide; intercepts are particularly useful. Note also that a relation and its inverse intersect on the line y = x.
How to visualise the graph of an inverse The following activity can help you to quickly see the shape of an inverse from the graph of a relation. Make a sketch of the graph of a relation in the righthand bottom corner of a blank page. This should be 4 or 5 cm square. You will need to use a pen or a dark pencil for best results. Fold the corner in to the centre at 45°. Keeping the page orientation the same, view the axes through the paper and you will see the graph of the inverse!
232
Maths Quest 12 Mathematical Methods CAS for the TINspire
Looking ‘through’ paper
Fold in along this line at 45°
REMEMbER
1. 2. 3. 4.
A relation is a set of ordered pairs which may be described by a rule. The inverse of a relation can be found by interchanging the x and y elements. The graph of a relation and its inverse are reflected in the line y = x. If the rule for a relation is known, then the rule for its inverse is obtained by interchanging x and y in the rule and then making y the subject. 5. The domain of a relation is the same as the range of its inverse and the range of a relation is the same as the domain of its inverse. ExERCiSE
5A eBook plus Digital doc
SkillSHEET 5.1 Domain and range
eBook plus Digital doc
Spreadsheet 041 Function grapher
Relations and their inverses 1 WE 1 Sketch the graph of each of the following relations. State the domain and range of each. a {(0, 1), (1, 2), (3, 2), (3, 5)} b {(−8, 7), (−5, 2), (−2, −1), (1, 1)} c y=x d y = 2x − 5 e 2x + 4y = 8 f y = x2 + 4x 3 g y = x2 − 1 j y= h y = (x + 1)2 i x2 + y2 = 4 x k y=2 l x = −4 m y = 2x3 2 WE 2 Find the inverse of each relation in question 1. Sketch the graph of each inverse relation, stating its domain and range. (Check your graphs using a CAS calculator.)
ExAM Tip The best procedure for determining the range of a function is through sketching a graph for the given domain.
3 MC The graph of a relation passes through the points (0, 1), (1, 2) and (3, 3). The graph of the inverse of this relation must pass through the points: A (2, 1) and (3, 3) B (0, 1) and (3, 3) C (0, 1) only D (1, 2) and (3, 3) E (3, 3) and (4, 10)
Chapter 5
inverse functions
233
4 MC A relation has xintercepts 2 and −3. The yintercepts of the inverse of this relation are: B −2 and −3 C 2 and −3 A −2 and 3 D 2 and 3 E cannot be determined 5 WE 3 For each relation, sketch the graph and its inverse on the same axes. Draw in the line y = x. a
b
y
d
y
−2
0
f
y
x
3
−6
j
k
y
l
y
y
4 3
x
−4
(−1, −1)
4
0
x
−4
m
n
y
−3
3
p
y
x
(−2, 3)
5b
−4
−2
0
2 x
0
x
(1, −2)
Functions and their inverses Recall that a function is a relation which has only one yvalue for each xvalue. The graph of a function can be crossed only once by any vertical line. To find the inverse of a function, we use the same procedures that we used for relations in the previous exercise. 1. The rule for the inverse of a function is obtained by either: (a) interchanging the x and yvalues of its ordered pairs or (b) interchanging x and y in the rule for the function, then making y the subject.
234
y
2 −4
x
x
0
x
0
o
y 0
2
0
x
−1 0
x
−2
(1, 1) 0
y
2 x
0
x
y
x
0
−4
h
y
0
i
2 x
0
g
y −2
0
y
(1, 3)
x
0
e
c
y
Maths Quest 12 Mathematical Methods CAS for the TINspire
y y = f(x)
0
Vertical lines cross the curve only once.
x
2. The graph of the inverse of a function is obtained by either: (a) interchanging the x and yvalues of its ordered pairs, or (b) reflecting the graph of the function through the line y = x, or (c) finding the rule of the inverse and graphing the new function. The domain of a function is the range of its inverse. The range of a function is the domain of its inverse. WoRkED ExAMplE 4
State whether or not each of the following is a function by using the vertical line test. a y = (x + 1)2 − 3 b x2 + y2 = 1 4 +1 c y= x−2 Think a
WRiTE/DRAW
1
Write the equation.
2
State the type of graph.
3
Do a rough sketch and use the vertical line test to determine if the relation is a function.
a y = (x + 1)2 − 3
A parabola with minimum turning point (−1, −3) y
x
0 (−1, −3)
b
4
State whether or not it is a function.
1
Write the equation.
2
State the type of graph.
3
Do a rough sketch and use the vertical line test to determine if the relation is a function.
(0, −2)
y = (x + 1)2 − 3 is a function. b x2 + y2 = 1
A circle, centre (0, 0), radius 1 y (0, 1) (−1, 0)
(1, 0) 0
x
(0, −1)
c
4
State whether or not it is a function.
1
Write the equation.
2
State the type of graph.
x2 + y2 = 1 is not a function. c y=
4 +1 x−2
A hyperbola with asymptotes x = 2 and y=1
Chapter 5
inverse functions
235
3
Do a rough sketch and use the vertical test to determine if the relation is a function.
y (x = 2)
(−2, 0) 0 (0, −1)
4
State whether or not it is a function.
y=
eBook plus
For each of the following functions, sketch, on the same set of axes, the graph of the function, its inverse and the line y = x. State the domain and range for the function and its inverse. a y = x2 − 2x b y = loge (x + 1) a
(y = 1)
4 + 1 is a function. x−2
WoRkED ExAMplE 5
Think
x
Tutorial
int0541 Worked example 5
WRiTE/DRAW a y = x2 − 2x
1
Write the original function.
2
Change it to turning point form by completing the square.
= x2 − 2x + 12 − 12 = (x2 − 2x + 1) − 1 = (x − 1)2 − 1
3
Determine the turning point of the parabola.
This is a parabola with turning point (1, −1).
4
Find the yintercept by letting x equal 0 and substituting in the original equation.
If x = 0, y = 02 − 2(0) y=0 The yintercept is 0.
5
Find the xintercepts by letting y equal 0.
If y = 0, x2 − 2x = 0
6
Solve the equation by factorising the lefthand side.
x(x − 2) = 0 x = 0 and x = 2 The xintercepts are 0 and 2.
7
Sketch the graph of the function.
y
0 −1
x 1 2 (1, −1) y = x2 − 2x
8
Interchange x and y to find the equation of the inverse.
x = (y − 1)2 − 1
9
Rearrange to make y the subject.
x + 1 = (y − 1)2 y −1= ± x +1 y = 1 ± x + 1 which is a sideways parabola.
236
Maths Quest 12 Mathematical Methods CAS for the Tinspire
10
Interchange the x and ycoordinates of the turning point and intercepts in the original equation to find the turning point and intercepts of the inverse.
11
On the original axes, sketch the graph of the inverse and the line y = x.
(1, −1) becomes (−1, 1), (2, 0) becomes (0, 2) and the origin remains the same. y
y=x y = x2 – 2x
x y=1± x+1
b
12
State the domain and range of the function.
The domain is R and the range is [−1, ∞).
13
State the domain and range of the inverse.
The domain of the inverse is [−1, ∞) and the range is R. b y = loge (x + 1)
1
Write the function.
2
Find the vertical asymptote using the fact that loge (0) is undefined.
Vertical asymptote occurs where x+1=0 x = −1 is the vertical asymptote.
3
Find the xintercept using loge (1) = 0.
xintercept occurs where y = 0 or x+1=1 x = 0 so the xintercept is 0.
4
Sketch the graph of the function.
x = −1
y
x
0
−1
y = loge (x + 1)
5
Interchange x and y to find the equation of the inverse.
x = loge (y + 1)
6
Rearrange to make y the subject.
ex = y + 1 y = ex − 1
7
Find the horizontal asymptote and intercepts from the original function.
x = −1 is the original vertical asymptote, so y = −1 is the horizontal asymptote for the inverse. (0, 0) is on both graphs.
Chapter 5 Inverse functions
237
8
On the original axes, sketch the graph of the inverse and the line y = x.
x=
−1
y
y = ex − 1
y=x
y = loge (x + 1) x y = −1 9
State the domain and range of the function.
The domain is (−1, ∞) and the range is R.
10
State the domain and range of the inverse.
The domain of the inverse is R and the range is (−1, ∞).
Worked Example 6
If f(x) = ln (x + 1) + 1, use a CAS calculator to: a find f 1(x) b draw the graph of f(x) and its inverse f 1(x). Think a
1 2 3
4
b
238
WRITE
Let y = f (x). Interchange x and y. To make y the subject, press: • Menu b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve(ln(y + 1) + 1 = x,y) (This means to solve the equation with respect to y). Then press ENTER ·.
a y = ln (x + 1) + 1
x = ln (y + 1) + 1

Write the answer in the form f 1(x). 
To draw the graphs of f (x) and f 1(x), open a Graphs page. Complete the function entry line as: f 1(x) = ln(x + 1) + 1 Then press ENTER ·. Complete the function entry line as: f 2(x) = ex  1  1 Then press ENTER ·.
f 1(x) = ex  1  1 b
Maths Quest 12 Mathematical Methods CAS for the TINspire
REMEMbER
1. A function is a relation which has only one yvalue for each xvalue. The graph of a function can be crossed only once by a vertical line. 2. The graph of the inverse is obtained by: (a) interchanging the x and ycoordinates in the ordered pairs (b) reflecting the graph in the line y = x (c) finding the rule of the inverse and then graphing the inverse. 3. A horizontal asymptote y = a is the vertical asymptote x = a of the inverse, and vice versa. 4. The domain of a function is the same as the range of its inverse. The range of a function is the same as the domain of its inverse. ExERCiSE
5b
Functions and their inverses 1 WE 4 State whether or not each of the following is a function by using the vertical line test. a y = 2x − 1 d y=
−
1 −3 (2 x + 1)2
Digital doc
SkillSHEET 5.2 Matching graphs with equations
c y=
e x2 + y2 = 4
f y = 3e(x −2) + 1
h y= x +2
g y = loge (x + 1) − 2 eBook plus
2 Sketch the graph of each of the following functions and state the domain and range of each. (Verify that it is correct using a CAS calculator.) a (−4, −2), (−2, 0), (0, 1), (2, 4), (3, 6)} c f (x) = 5 − 2x d f (x) = x2 − 9 4 f f ( x) = g f (x) = x2 + 8x x
b 3x + 4y = 12 e f (x) = (x + 2)2 x3 h f ( x) = 2
i f (x) = 2ex
k f ( x) = 4 − x 2
j f (x) = loge (2x)
3 WE 5 Find the inverse of each function in question 2. Sketch the graph and state the domain and range of each inverse. (Verify using a CAS calculator.)
ExAM Tip When the inverse function is asked for, the domain must be given. Students will be penalised in the future if the domain is left out. If only the rule for the inverse function is asked for, the domain does not have to be given.
4 For each function graphed below: i copy the graph of each function and sketch its inverse on the same axes ii state the domain and range of f (x) iii state the domain and range of the inverse of f (x). eBook plus Digital doc
Spreadsheet 059 Inverse graphs
4 −1 x+3
b y = 3(x − 1)2 + 2
a
b
y
c
y
y=x
f(x)
[Assessment report 2007]
f(x)
f(x)
y
4
1 0
2
x
0
x 0
y=x
x
y=x
Chapter 5
inverse functions
239
d
e
y
f
y
f(x)
y x=1
3
f(x)
3 f(x)
y=x
g
y=x
y=x y
h
f(x)
y=x
f(x) −5
1
5x
0
x
0
−4
x
0
y y=x
(−4, 1) –3
i
y
f(x)
y=x
3
x
2
3 x
0
−3
x
0
0
−5
j
y
k
y=x
0
2
y x = −3
f(x) 6
−3 −2
eBook plus Digital doc
WorkSHEET 5.1
3
−2 y=x
Questions 5 to 8 relate to the following function. f : R + → R, f ( x ) =
(3, 4)
x
0
−2
5 MC The range of f (x) is: A R+ B R
y 4
f(x)
2 x
l
y=x
−
0
3
x D R−
E (−∞, 0]
6 MC The domain of the inverse of f (x) is: A R− B R C R+
D [0, ∞)
E (−∞, 0]
7 MC The range of the inverse of f (x) is: C [0, ∞) B R A R−
D (−∞, 0]
E R+
C [0, ∞)
x
8 MC The inverse of f (x) can be defined by: A y = x2, where x ∈ [0, ∞) B y = x2, where x ∈ R− 2 + D y = x , where x ∈ R E y = x, x where x ∈ [0, ∞)
C y = x2 where x ∈ R
9 WE 6 For each of the following functions, fully define the inverse. a f : [−2, ∞) → R, f (x) = (x − 2)2 − 3 b f : R → R, f (x) = 3e x − 1 + 2
5C
inverse functions A onetoone function is a function where for each xvalue there is only one yvalue and vice versa. The graph of a onetoone function can be crossed only once by any vertical or horizontal line. A function that is not onetoone is manytoone.
240
Maths Quest 12 Mathematical Methods CAS for the Tinspire
A function will have an inverse that is also a function if and only if it is a onetoone function. − If a function, f , is onetoone, then its inverse function is denoted by f 1. Furthermore: − dom f 1 = ran f −1 ran f = dom f −1 The graph of f is obtained by reflecting the graph of f through the line y = x. (Note that if f − crosses the line y = x at any point, f and f 1 will intersect at that point.) The maximal domain of a y function is the largest domain y = f(x) for which its rule is defined. If a function is given without its domain specified, then it is understood that the maximal or implied domain is Horizontal lines x 0 intended. cross the curve only once.
Vertical lines cross the curve only once. eBook plus
WoRkED ExAMplE 7
1 . x+2 State whether or not it is a onetoone function, and hence determine if the inverse function exists. If the inverse function exists: on the same set of axes, sketch f(x), its inverse and the line y = x − use a CAS calculator to find the points of intersection of f(x) and f 1(x) find the maximal domain and range of the original function find the domain and range of the inverse function.
Tutorial
Consider the function f ( x ) = a b c d e
WRiTE/DRAW
Think a
int0542 Worked example 7
a
1 , a basic hyperbola x+2 translated 2 units to the left. f ( x) =
1
Write the function.
2
Find the vertical asymptote by making the denominator of the fraction equal to 0.
Vertical asymptote occurs where x+2=0 x = −2
3
Find the horizontal asymptote remembering there is no vertical translation.
Horizontal asymptote occurs where y = 0.
4
Sketch the graph of f (x) and use vertical and horizontal line tests to decide whether or not f (x) is a onetoone function.
x = −2
y (0, 1–2)
−2
x
0
1 f(x) = x + 2 5
State whether or not the inverse function exists.
−
f (x) is a onetoone function, so f 1(x) exists.
Chapter 5
inverse functions
241
b On the same axes sketch f (x), f
−1
b The horizontal asymptote is y = −2.
(x) and y = x.
The vertical asymptote is x = 0. 1 0, 2 becomes 1 , 0
( )
( ) 2
The equation is y = 1x − 2. x = −2 1) (0, — 2 1 y = —— x+2
y
y=x
0 1 , 0) (— 2
1 x− 2 y= — x y = −2
c The points of intersection are (−2.414,
c Find the points of intersection.
−2.414)
and (0.414, 0.414). Note that these points both lie on the line y = x.
d The domain is R\{−2} and the range is
d Find the domain and range of f (x).
R \{0}. e The domain of the inverse is the range of the original
e The domain of the inverse is R \{0} and
the range is R \{−2}.
function. The range of the inverse is the domain of the original function.
Functions are either onetoone or manytoone. If a function, f, is not onetoone but is manytoone, it is possible to restrict its domain so that the limited domain is onetoone and hence the inverse function exists. WoRkED ExAMplE 8 −
State the largest negative domain of f(x), shown in the figure, so that f 1(x) exists. y f(x) (0, 9)
−3
0
Think
242
1
Decide whether f (x) is a onetoone function.
2
The parabola can be divided through its axis of symmetry, x = 3, so that 2 onetoone functions are formed.
Maths Quest 12 Mathematical Methods CAS for the Tinspire
x
WRiTE/DRAW
f (x) is not onetoone, because horizontal lines cut the parabola in 2 places.
3
State the domain of the left part of the parabola that is onetoone.
f (x) is onetoone if the domain is restricted to (−∞, −3].
4
State the domain of the right part of the parabola that is onetoone.
f (x) is also onetoone if the domain is restricted to [−3, ∞).
5
State the largest negative domain of f (x) so that − f 1(x) exists.
The largest negative domain for f 1(x) to exist is (−∞, −3].
−
REMEMbER −
A function f (x) has an inverse function f 1(x) if and only if the function is onetoone. − The domain of a function f (x) is the range of its inverse f 1(x). − The range of a function f (x) is the domain of its inverse f 1(x). −1 The graph of f (x) is obtained by reflecting f (x) in the line y = x. The maximal or implied domain is used when the domain of a function is not specified. It is the largest domain for which the rule can be defined. 6. If a onetoone function intersects with its inverse, the points of intersection lie on the line y = x. 7. If a function is not onetoone it is possible to restrict the domain so that the new function is onetoone and hence an inverse function exists.
1. 2. 3. 4. 5.
ExERCiSE
5C
inverse functions 1
WE 7 Consider the following functions. For each function: i state whether or not it is a onetoone function and hence determine if the inverse function exists. If the inverse function exists: ii on the same set of axes, sketch f (x), its inverse and the line y = x − iii use a CAS calculator to find the points of intersection of f (x) and f 1(x) iv find the maximal domain and range of the original function v find the domain and range of the inverse function. a f (x) = 4x + 1 b f (x) = 6x c f (x) = 5 d f (x) = x2 + 2 e f (x) = (x − 3)2 f f (x) = (x + 1)3 2 g f ( x) = i f (x) = x2 − 6x + 3 h f ( x ) = − 16 − x 2 x 4x − 2 k f (x) = 2 loge (x − 1) j f (x) = e
2 Find: i which of the functions below − has an inverse function f 1(x) −1 ii f (x), if it exists.
ExAM Tip When asked for the domain of an − inverse, simply stating that the domain of f 1 is the same as the range of f was not sufficient to gain the mark. Students were expected to explicitly state the domain as a set. [Assessment report 2006]
a f (x) = 4x
b f ( x) =
1 −2 x2
c f (x) = 5 − x2
d f (x) = (x − 1)2
e f ( x) =
x3 2
f f (x) = (x + 5)2 − 28
Chapter 5
inverse functions
243
g f ( x) = x − 2 j f (x) = 5 − ex − 2
h f ( x ) = 16 − x 2 k f (x) = 3e−x − 2
i f (x) = 2ex + 1 l f (x) = loge (3x)
m f (x) = 2 loge (x − 4)
n f (x) = 1 + 2 ln (x)
o f (x) = 3 − loge (2x + 3)
3 Copy and complete the following table. Function
Inverse of the function
Domain
Range
Domain
Range
a
R
R−
b
[1, ∞)
R
c
[−3,
d
[0, 3] R+
R
e
[−10, ∞)
R
f
3]
R−
(0, ∞)
g h
[−5, 5]
[0, 8] R+
R
4 MC Consider the function in the figure shown at right. a The function would be onetoone if the domain were restricted to: A [−3, 5] B [2, 10] C [1, 3] D [0, 4] E [0, 2.2] b The largest domain that restricts it to a onetoone function is: B [0, ∞) A (−∞, 4] C R+ − − E ( ∞, 2] D R
y
4
2 + 1. 5 MC Consider the function f ( x ) = ( x − 3)2 a The function would be onetoone if the domain were restricted to: A (−3, ∞] B [−3, ∞) C (3, ∞) D (1, ∞) E [3, ∞) b The inverse would be a function if the domain of f (x) were: A [0, 4] B (1, 4] C [2, 4] D [0, 3) E [1, ∞) 6 MC Consider the function f : S → R, f (x) = x2 + 1. − a The maximal domain, S, for f 1(x) to exist is: + C [1, ∞) A R B R− + − E (−∞, 1] D R ∪ {0} or R ∪ {0} − b For f 1 to exist, the largest possible positive set, S, is: C [0, ∞) D R+ A R B R− 7 WE8 exists. a
2
b
y x
f(x)
c
y f(x)
(−3, 3)
y f(x)
−3
Maths Quest 12 Mathematical Methods CAS for the Tinspire
0
x
x
E [1, ∞)
For each function graphed below, state the largest possible domain of f so that f
0
244
0
−1 0
x
−1
d
y 0
f(x) 2
e
y
f
f(x)
y 9
x
f(x)
1
g
h
y
−9
x
0
i
y
y
f(x) 0 1 −3 −7
5D
5
x=5
x
0
x
5
0
9 x f(x)
9x
0
f(x)
Restricting functions As we have seen in the previous exercise, functions which are not onetoone can have their domains restricted so that they become onetoone. As a result their inverses will then be − functions, that is, f 1 will exist.
WoRkED ExAMplE 9
eBook plus
Consider the function fully defined as follows: f : [ − 3, 3] → R, f ( x ) = 9 − x 2 − a Find the largest possible positive domain so that f 1 exists. −1 b Use this restricted domain to fully define f (x).
1
int0543 Worked example 9
WRiTE
Think a
Tutorial
a Let y = f (x)
Write the function.
y = 9 − x2 y2 = 9 − x2
2
Square both sides.
3
Add x2 to both sides.
x2 + y2 = 9
4
State the shape and key features of the graph.
This is the equation of a circle with centre (0, 0) and radius 3.
5
Describe f (x), remembering that a positive square root implies the upper part of a semicircle. Sketch the graph of f (x).
The graph of f ( x ) = 9 − x 2 is an upper semicircle.
6
y 3
−3 7
State the possible domains if f (x) is onetoone.
8
State the largest positive domain for f 1(x) to exist.
−
0
3
x
f (x) is onetoone if the domain is [−3, 0] or [0, 3]. The largest positive domain for f exist is [0, 3].
Chapter 5
−1
to
inverse functions
245
b
1
b If dom f = [0, 3], then ran f = [0, 3].
Find the range of f (x). −1
−1
= ran f.
dom f
−1
= ran f = [0, 3]
ran f
−1
= dom f = [0, 3]
2
Find the domain of f
3
Find the range of f
4
Write the function.
y = 9 − x2
5
Find the inverse by interchanging x and y.
Inverse is x = 9 − y 2
6
Make y the subject.
7
Select the appropriate inverse function.
8
State the function f 1.
−1
using dom f
using ran f
−1
= dom f.
x2 = 9 − y2 x2 + y2 = 9 y2 = 9 − x2 y = ± 9 − x2 Since ran f
−
−1
= [0, 3], use y = 9 − x 2 .
Therefore the inverse of f with the largest positive domain is: − − f 1: [0, 3] → R, f 1 ( x ) = 9 − x 2
WoRkED ExAMplE 10
If f: S → R, f(x) = 3(x − 1)2 − 2: − a find the largest positive set, S, for which f 1(x) exists −1 b fully define f and sketch its graph. WRiTE/DRAW
Think a
a y = 3(x − 1)2 − 2
1
Write the function.
2
State the shape and key features of the graph.
3
Sketch the graph of f (x).
A parabola with turning point (1, −2), yintercept (0, 1) and xintercepts (0.184, 0) and (1.816, 0) y
(0, 1) (0.184, 0) (1.816, 0) x 0 (1, −2)
b
4
State whether or not it is a onetoone function.
5
State the possible domains if f (x) is to be onetoone.
6
State the largest positive domain, set S, for f to exist. Find the range of f (x).
1 2 3
246
Find the domain of f Write the function.
−1
using dom f
−1
−1
= ran f.
Maths Quest 12 Mathematical Methods CAS for the Tinspire
The parabola is not onetoone so the domain must be restricted. The domain could be (−∞, 1] or [1, ∞). S = [1, ∞) b The range is [−2, ∞). −
The domain of f 1 is [−2, ∞). y = 3(x − 1)2 − 2
4
x = 3(y − 1)2 − 2 x + 2 = 3(y − 1)2 x+2 = ( y − 1)2 3
Interchange x and y and make y the subject.
x+2 Reject the negative 3 solution because the domain is [−2, ∞) x+2 y = 1+ 3 y −1=
5
−
−
f 1:[−2, ∞) → R, f 1 ( x ) = 1 +
Fully define the inverse function.
x+2 3
key feature of inverse functions A feature of inverse functions is that by taking the inverse function of a function, or vice versa, the independent variable (x) is always obtained. That is: − − f 1 [ f (x)] = f [ f 1(x)] = x. The following example illustrates how this works. WoRkED ExAMplE 11
eBook plus
For f : R → R, f(x) = x3: − a find f 1(x) and state its range − − − − b find f [ f 1(x)] and f 1[f(x)] and show that f [ f 1(x)] = f 1[ f(x)] = x. Think a
1 2 3
b
int0544 Worked example 11
WRiTE
Write the function. Interchange x and y. Make y the subject by writing both sides to the power of 13 and simplifying.
a f(x) = y = x3
Its inverse is x = y3. 1
Write the rule for f 1(x).
5
Find the range of f 1(x), remembering that ran − f 1 is the same as dom f.
6
Write your answer in the same form as the question.
1
Find f [ f 1(x)] by rewriting f 1(x) as x 3 .
1
y = x3
2
Find f ( x 3 ) by replacing x by x 3 in f(x) = x3.
3
Simplify.
4
Find f
−1
5
Find f
−1
6
Simplify.
dom f = R − ran f 1 = R f
1
−
b
1
1
1
−
f 1 ( x) = x 3
−
−
1
(y3 ) 3 = x 3
−
4
7
Tutorial
−1
1
−
: R → R, f 1 ( x ) = x 3 1
−
f [ f 1 ( x )] = f ( x 3 ) 1 = x3
3
=x − f 1[ f ( x )] =
x3.
[f(x)] by rewriting f(x) as
(x3) by replacing x by x3 in f 1 ( x ) = x 3.
−
1
−
f 1 (x3 ) 1
= (x3 ) 3 =x
−1
Compare f [ f (x)] to f
−1
[ f(x)].
− f [ f 1 (x ( x )] =
f −1[ f ( x)] x )] = x
Chapter 5
inverse functions
247
WoRkED ExAMplE 12
eBook plus
a Sketch the graph of f(x) = x2 − 3x + 3, showing the turning point and relevant Tutorial intercept(s). int0545 b Find the rule of the inverse by an algebraic method and sketch this graph Worked example 12 on the same set of axes together with the line y = x. c Is the inverse a function? d The inverse is a reﬂection in the line y = x of the original function f(x). Use this information to find any points of intersection between the original curve and its inverse. − e Find the maximum value of a for f: (−∞, a] → R, f(x) = x2 − 3x + 3 so that f 1(x) exists. WRiTE/DRAW
Think a
use a CAS calculator to help in drawing a graph of f (x) including the relevant points.
a
y 4 3 2 1
f(x) = x2 − 3x + 3
(1.5, 0.75)
0 b
c
248
−
1 2 3 4
x
b x = y2 − 3y + 3
1
To find the equation of f 1(x), let y = x2 − 3x + 3. Interchange x and y.
2
To make y the subject, on a Calculator page, press: • MeNu b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve(y2 − 3y + 3 = x,y) Then press eNTeR ·.
3
Write the answer in the form of f 1(x).
4
use the calculator to draw the graphs − of f(x), f 1(x) and y = x.
−
The inverse is a onetomany relation and therefore is not a function.
f 1 ( x) = ± −
4x − 3 + 3 2
y 4 3 2 1
f(x) = x2 − 3x + 3 y=x
0
1 2 3 4 5 6 7
x
c The graph of the inverse does not pass
the vertical line test, as it is a onetomany relation, and therefore it is not a function.
Maths Quest 12 Mathematical Methods CAS for the Tinspire
d
e
The points of intersection between f (x) and − f 1(x) will occur at the intersections of the graph of f (x) and y = x. To locate these points on a Calculator page, press: • MeNu b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve(x2 − 3x + 3 = x, x) Then press eNTeR ·. The points of intersection are at (1, 1) and (3, 3). Draw the graphs of f(x) and y = x.
d
For the lefthand branch of the parabola, we need f (x) = x2 − 3x + 3, since the domain is (−∞, a). It is a onetoone function. Locate the turning point of the graph and the xcoordinate will be the value of a. To locate the minimum of f (x) = x2 − 3x + 3 on a Calculator page, press: • Menu b • 4:Calculus 4 • 7:Function Minimum 7 Complete the entry line as: f Min(x2 − 3x + 3, x) Then press eNTeR ·. 3 The xvalue of the turning point is 2 , which 3 implies that a = 2 .
e
f(x) = x2 − 3x + 3
y 4 3 2 1
y=x
(1.5, 0.75)
1 2 3 4
0
x
The points of intersection are at (1, 1) and (3, 3).
3
f : ( − ∞, 2 ) → R, f ( x ) = x 2 − 3 x + 3 a=
3 2
REMEMbER
1. Functions that are not onetoone can have the domain restricted so that they become onetoone. Then their inverses will be functions. 2. A fully defined function is written in the form f : [domain] → R, f (x) = rule. − 3. The graph of f 1(x) is a reflection of f (x) in the line y = x. − 4. The points of intersection of f (x) and f 1(x) lie on the line y = x. 5. By taking the inverse function of a function or vice versa, the independent variable (x) − − is always obtained: f 1[f (x)] = f [f 1(x)] = x. ExERCiSE
5D eBook plus Digital doc
SkillSHEET 5.3 Function notation
Restricting functions 1 WE 9a Find the largest possible domain(s) of the following functions so that the inverse function exists. (use a CAS calculator to sketch the graph if necessary.) a f : R → R, f (x) = x2 + 3 b f : R → R, f (x) = 3x2 − 1 1 c f : R → R, f (x) = (x + 3)2 − 2 d f : R \{0} → R, f ( x ) = 2 − 3 x 1 1 f f : R \ {2} → R, f ( x ) = e f : R \{4} → R, f ( x ) = +1 2 x−2 ( x − 4)
Chapter 5
inverse functions
249
g f : [− 5, 5] → R R,, f ( x) x) = i
−
25 − x 2
R,, f ( x) x) = 1 − x 2 h f : [− 1, 1] → R j f : R → R, f (x) = x2 − 2x + 5
f : [4, ∞) → R R,, f ( x) x) = x − 4
k f : R → R, f (x) = e x + 2
l f : (5, ∞) → R, f (x) = 2 loge (x − 5) −
2 WE 9b For each function in question 1 use the restricted domain to fully define f 1(x). (If there are two possible domains, use the one which is to the right.) 3 MC use the function f : [0, ∞) → R, f (x) = x2 to answer the following questions. − a Any points of intersection of f (x) and f 1(x) must lie on the line: A y=0 B y=x C y = 2x D y=x+1 E x=0 − b f 1(x) is correctly defined by: − − − − A f 1 : [0, ∞) → R, f 1(x) = x B f 1 : [0, ∞) → R, f 1(x) = −x −
−
−
−
R, f 1 ( x) x = x D f 1 : [0, ∞) → R, f 1 ( x ) = − x C f 1 : [0, ∞) → R, 1 − − R, f 1 ( x) x = 2 E f 1 : (0, ∞] → R, x − c The range of f 1(x) is: −∞, 0] D [0, ∞) A ( B R C R+ E R− −1 d The graphs of f and f intersect at the point(s): A (0, 0) and (1, 1) B (1, 1) only C (0, 0) and (1, −1) 1 D (0, 0) and (3, 3) E (1, 1) and (2, 4 ) 4 Copy each of the following graphs, then on the same set of axes sketch the graphs of y = x and − y = f 1(x). b a y y f(x)
3 5
c
f(x)
0
6
f(x)
y
x
−1
d
x
0
y 3 f(x)
x
0
e
f
y 1
250
0
x
3 y
f(x) x
0
f(x)
−1
0
x
Maths Quest 12 Mathematical Methods CAS for the Tinspire
(2, −5)
g
h
y
y 2
4 f(x)
0
1
x
0
3
x
5
f(x) −4
5 WE 10 If f : S → R, f ( x ) = 3 + x − 1, find: − a the largest set, S, for which f 1(x) exists −1 b f and sketch its graph. 6 If f : [−2, a] → R, f (x) = (x − 4)2 − 5, find: a the largest possible value of a so that f will have an inverse which is a function − b f 1 and sketch its graph. − 1 7 WE 11 For f : R+ → R, f ( x ) = x − a find f 1 and state its range −1 b find f [ f (x)] − c find f 1[ f (x)] − − d show that f [ f 1(x)] = f 1[ f (x)] = x. 1 : 8 For f : R → R, f ( x ) = 2 x +2 a sketch the graph of f (x) − b state the largest positive domain for f (x) so that f 1(x) exists −1 c sketch the graph f (x) on the same set of axes as f (x), using the positive domain − − d show that f [ f 1(x)] = f 1[ f (x)] = x. 9 WE 12 If g : [b, 8] → R, g(x) = 9 − x2, find: a the smallest value of b so that g(x) has an inverse that is a function − b g1 − c the range of g 1(x). 10 Given that g( x ) = 3 + 4 − x 2 , fully define two inverse functions, − g 1, using maximal domains.
Chapter 5
eBook plus Digital doc
WorkSHEET 5.2
inverse functions
251
Summary Functions and their inverses
• A function is a relation which has only one yvalue for each xvalue. The graph of a function can be crossed only once by a vertical line. • A onetoone function is a function which has only one xvalue for each yvalue. The graph of a onetoone function can be crossed only once by any vertical or horizontal line. • A function which is not onetoone is manytoone. • The rule for the inverse of a relation, or a function, can be obtained by either: 1. interchanging the first and second elements of the ordered pairs, or 2. interchanging x and y in the rule and making y the subject. • Graphing the inverse — The graph of the inverse of a relation, or a function, can y be obtained by either: −1(x) f 1. interchanging the first and second elements of the ordered pairs of the relation or function, or 2 (−5, 2) 2. reflecting the graph of the relation or function through the line y = x, or x 0 3. using the rule of the relation or function to find the rule of the inverse and −5 f(x) then graphing the inverse. y=x • The domain and range of a function and its inverse are interchanged. (2, −5) • The graphs of a function and its inverse intersect on the line y = x. Inverse functions
• • • • •
−
The inverse of a onetoone function, f (x), is also a function and is denoted by f 1(x). − dom f 1 = ran f −1 ran f = dom f − − f 1[ f (x)] = f [ f 1(x)] = x The implied (or maximal) domain of a function is the largest domain for which the function has meaning. When the domain is not mentioned, use the maximal domain. Restricting functions
−
• A function, f (x), which is not onetoone can have its domain restricted so that f 1(x) exists. − • The points of intersection of f (x) and f 1(x) lie on the line y = x. • A function is fully described using the notation f : X → Y, f (x) = rule where X is the domain and Y is the codomain. − • The graph of f 1(x) is a reflection of f (x) in the line y = x. −1 − • f [ f (x)] = f [ f 1(x)] = x
252
Maths Quest 12 Mathematical Methods CAS for the TINspire
ChApTER REViEW c
ShoRT AnSWER
1 Sketch the graph of the inverse of each of the following relations. y a
y
8 x
0
2
f(x) (4, −4) x
0
−3
–3
b
y (−4, 4)
2 −6
−2
0
x
2 Sketch the graph of each of the following functions and state its domain and range. a y = − 36 − x 2
b f (x) = 2e x − 2
3 using the graph of the functions given below sketch the graph of the inverse of each function. y a x = −2 1 −2
−1 0
b
f(x)
x
y f(x)
0
(−1, −4)
x
4 State the largest possible domain for each of the following functions to be onetoone. a f (x) = 3 − loge (x + 4) b g(x) = (x − 2)2 1 c f ( x) = 5 − x2 5 For the following functions: i Sketch their graphs and, by inspection, define them as onetoone or manytoone functions. ii For the functions defined in i as manytoone, define the maximal domain for which the inverse is a function, stating their domains in formal function notation. If there is more than one option, choose the ‘righthand’ option. iii Find the rules of their inverses. iv For all of the onetoone functions from i and the redefined functions from iii, sketch all these inverse functions. a f (x) = 2x − 1 b f (x) = 2(x − 1)3 + 1 c f (x) = x2 + x d f ( x) = 2 − x − 3 e f (x) = 3ex + 1 2 −1 f f ( x) = 3( x − 2) − 1 +1 g f ( x) = ( x + 3)2 6 a Sketch the graph of f (x) = loge (x − 2) + 1, marking the intercepts and asymptotes and stating the domain and the range. b Find the rule for the inverse function. − c If f 1(m) = 3, find the value of m. − d Draw the graphs of y = x and f 1(x) on the same axes as f (x) = loge (x − 2) + 1, marking the point (m, 3) and checking that it is on − f 1(x). State the domain and the range. e Mark the image of (m, 3) on the original graph.
Chapter 5
inverse functions
253
7 If g : (−∞, b) → R, g(x) = x2 + 5x − 1, find the − largest value of b for g 1 to exist.
A
B
y
8 Part of the graph of the function f : R → R, − f (x) = a + be x is shown below.
−1
y
0
x
0
−1
y y=1 0
1
y
C
x
3 The graph of the function with the equation y = f (x) is shown below. y 1
2
The graph of the inverse function f be: y
2, 4, −2 −2, 4 0, 0
B
D
y
0
1
x
1
x
−1
is most likely to y 1 0
x
−1 0
C
−4
x
0
1 A relation has an xintercept of −4 and yintercept of 2. The inverse of this relation has an xintercept and yintercept, respectively, of: −4,
x
x
0 −1
A
MulTiplE ChoiCE
1
y
E
[© VCAA 2007]
[© VCAA 2008]
x
1
0
The line with the equation y = 1 is an asymptote and the graph passes through the origin. a explain why a = 1 and b = −1. Let h be the function h: [0, 2] → R, h(x) = f (x). b State the range of h using exact values. − c Find the inverse function h 1. d Sketch and label the graph of the inverse − function h 1 on the graph axes shown above. 9 Let f : R → R, f (x) = e2x − 1. a Find the rule and domain of the inverse − function f 1. − b Sketch the graph of y = f (f 1(x)) for its maximal domain. ax − c Find f (−f 1(2x)) in the form where a, b bx +c and c are real constants.
y
D
0
A B C D E
x
x
y
−1 0
x
2 The graph that best represents the inverse of the relation shown in the figure below is: E
y
−1
0
x
y
0
[© VCAA 2005]
254
Maths Quest 12 Mathematical Methods CAS for the Tinspire
4 The relation x2 + (y + 1)2 = 16 has an inverse that is: A a onetoone function B a manytoone relation C a manytomany relation D a onetomany function E a onetoone relation −
intercept of the inverse function f 1(x), 5 The xintercept where f ( x ) = 3 − x − 3 , is closest to: A 1.27 B 6 C −3 D −1.27 E −6 6 The yintercept of the inverse function f 2 + 3, is closest to: where f ( x ) = ( x − 2) 4 A 3 B no xintercept −4 C 3 D 2 E −2
−1 (x),
7 Asymptotes exist on the graph of the inverse 2 − + 3, at: function f 1(x), where f ( x ) = ( x − 2) A x = 1 only B y = 1 only C x = 3, y = 2 D x = 2, y = 1 E x = 3, y = −2 4 8 The inverse of the relation y = 2 is: x 2 2 B y= A y=± x x −2 2 D y= C y= x x −2 E y= x 9 A function has a domain of [0, ∞) and a range of R+. The domain of its inverse must be: A R+ B R D [0, ∞) C R+ E (−∞, 0] 1 10 The implied domain of the function, f ( x ) = 2 x −1 is: A R C R \{0} E R+
B R \{1} D R \{−1, 1}
Questions 11 to 13 can be answered by considering the function f (x) = 4 ln (x + 3) − 2. 11 The maximal domain for f (x) is: A [3, ∞) B (−∞, −3) − D (−3, ∞) C R \{ 3} E R+ −
12 dom f 1 is equal to: A R+ C R E (−3, ∞) 13 The rule for f
−1
B [0, ∞) D (−∞, −3)
is completely described by:
−
−
−
−
x
A f 1 : R → R, f 1 ( x ) = e 4 − 3 1 1 B f : R → R, f ( x ) = e −
x+2 4
−3
−
1 − − R, f 1 ( x) x) = e C f : ( ∞, 3) → R, −
−
1 1 D f : R → R, f ( x ) = e
E
−
−
f 1 : R + → R, f 1 ( x ) =
x+4 3
x+4 2
−3
−2
x−4 e 2
−3
14 At which one of the following points is it possible for a function and its inverse to intersect? A (2, 2) B (1, 2) C (2, 1) D (0, 1) − − E ( 1, 2) −
15 The value of f [ f 1 (− 13 )] is equal to: A −3 1 B 3 C x 1 D 3 E 3 −
16 The function shown in the figure is onetoone if the domain is restricted to: A B C D E
y
[2, ∞) [−3, ∞) [−3, 2] (−∞, 2] [0, ∞)
f(x) −3
0
2
x
17 If f : S → R, f (x) = x2 − 10x + 18, then the largest − possible set, S, for f 1 to exist is: A B C D E
(−∞, 0] [0, ∞) [0, 5] [5, 100] (−∞, 5]
Chapter 5
inverse functions
255
ExTEnDED RESponSE
1 using a CAS calculator: a Find the rule for the inverse of the function f (x) = 0.213x2 + 1.127x − 2.124. b explain why this inverse is not a function. c Find the value of a, correct to 2 decimal points, for [a, ∞), the maximal domain of f (x) so that the inverse is a function. Let this new function be g(x). − d Find the domain and range of f 1(x). −1 e Solve g(x) = g (x) for x, correct to 2 decimal places. 2 The function, f (x), in the figure is a parabola with a turning point at (0, 4). a Find the function which describes f (x). b Find the coordinates of the point, A. c State the domain and range of f (x). d Sketch the graph of f (x) and its inverse on the same set of axes. e What shape is enclosed between f (x) and its inverse? f i State the domain for which the inverse of f (x) is a function. ii Give the rule for the inverse of f (x) over this domain. g i State the domain for which the inverse of f (x) is not a function. ii Give the rule for the inverse of f (x) over this domain. 3 The graph of f (x) = exx2 is shown at right. a use a calculator to find the turning points, rounding answers to 2 decimal places as appropriate. b State the domain and range of f (x). c On the same set of axes, sketch f (x), its inverse and the line y = x. Mark all turning points and asymptotes. d State the domain and range of the inverse of f (x). e use a CAS calculator to solve exx2 = x to find the points of intersection of f (x) and its inverse. − f Find the largest possible negative domain of f (x) so that f 1(x) exists (that is, the inverse is a function). g Find the domain of the inverse function.
y 4
y=x
A −2
0
2
x
–2
y
f(x) = exx2 x
4 a use a CAS calculator to draw the graph of f (x) = 6 loge x − 3. − b Find the value of a for (a, ∞), the maximal domain of f (x), such that f 1(x) exists. Let this new function be g(x). − c Solve g(x) = g 1(x) for x, correct to 2 decimal places. − d Consider the function f (x) = a loge (x − 3), where a > 0. Let f 1(x) = h(x). Given that x a 1 a f ′( x ) = and h ′( x ) = e , find the value of a (correct to 2 decimal places) for which the graph of a x−3 f (x) and its inverse intersect only once.
eBook plus Digital doc
Test Yourself Chapter 5
256
Maths Quest 12 Mathematical Methods CAS for the Tinspire
eBook plus
ACTiViTiES
Chapter opener Digital doc
• 10 Quick Questions: Warm up with ten quick questions on inverse functions. (page 229) 5A
Relations and their inverses
Interactivity int0248:
• Inverses: use the interactivity to consolidate your understanding of graphs of the inverses of relations. (page 229) Tutorial
• WE 3 int0540: Watch a worked example on how to sketch relations and their inverses. (page 231) Digital docs
• SkillSHeeT 5.1: Practise identifying domain and range. (page 233) • Spreadsheet 041: Investigate graphs of functions. (page 233)
5D
Restricting functions
Tutorials
• WE 5 int0541: Watch a worked example on sketching functions and their relations. (page 236)
• WE 9 int0543: Watch a worked example on defining inverse functions. (page 245) • WE 11 int0544: Watch a worked example on identifying that the composite functions − − f [ f 1(x)] = f 1[ f (x)] = x. (page 247) • WE 12 int0545: Watch a worked example on restricting a function to define an inverse function. (page 248)
Digital docs
Digital docs
5B
Functions and their inverses
Tutorial
• SkillSHeeT 5.2: Practise making graphs with equations. (page 239) • Spreadsheet 059: Investigate inverse graphs. (page 239) • WorkSHeeT 5.1: Sketch relations, functions and their inverses. (page 240) 5C Tutorial
Inverse functions
• WE 7 int0542: Watch a worked example on sketching graphs of inverse functions. (page 241)
• SkillSHeeT 5.3: Practise using function notation. (page 249) • WorkSHeeT 5.2: Recognise types of functions and sketch polynomials and power functions. (page 251) Chapter review Digital doc
• Test Yourself: Take the endofchapter test to test your progress. (page 256) To access eBookPLUS activities, log on to www.jacplus.com.au
Chapter 5
inverse functions
257
6 Circular (trigonometric) functions areaS oF STudy
• Graphs and identification of key features of graphs of the circular functions: y = sin (x), y = cos (x), y = tan (x) • Transformation from y = f (x) to y = Af (n(x + b)) + c, where A, n, b and c ∈ R, and f is one of the circular functions specified above and the relation between the graphs of the original and transformed functions for functions such as: y = −10 sin (π (x − c)) − 5, c ∈ R • Graphs of sum, difference, product and composite functions of f and g, where f and g are functions such as y = sin ((x) + 2x 2x and y = cos cos (2 (2x) • Graphical and numerical solutions of trigonometric equations
6a 6b 6c 6D 6E 6F
Revision of radians and the unit circle Symmetry and exact values Trigonometric equations Trigonometric graphs Graphs of the tangent function Finding equations of trigonometric graphs 6G Trigonometric modelling 6H Further graphs 6I Trigonometric functions with an increasing trend
• Recognition of the general form of possible models for data presented in graphical or tabular form, using circular functions • General solutions of equations such as 1 cos ( x ) + cos cos (3 x ) = 2 , x ∈ R and the specification of exact solutions or numerical solutions, as appropriate, within a restricted domain • Applications of simple combinations of functions and interpretation of features of the graphs of these functions in modelling practical situations, for example, y = ax + b + m sin (nx) as a possible pattern for economic growth cycles
eBook plus Digital doc
nit c
1
P
nit
revision of basic concepts
y
1u
revision of radians and the unit circle
1u
6a
10 Quick Questions
Angles are measured in degrees or radians. O S x To define a radian we can use a circle which has a radius of one unit. This circle is called the unit circle. If we take a piece of string which is the same length as the radius and place it along the circumference of the circle from S to P to form an arc, then A unit circle the angle formed by joining S and P to O, the centre of the circle, measures one radian. The radius of the circle can be any length and can still be regarded as a unit. As long as the arc is the same length as the radius, the angle will always measure one radian. In general, therefore, a radian is the angle formed at the centre of any circle by radii meeting an arc which is the same length as the radius of the circle. Note the following.
258
maths Quest 12 mathematical methods CaS for the Tinspire
y 1. One radian is written as 1c (or 1 radian can be written as 1). A 2. The circumference of a circle is 2π r units in length. P 3. If the radius is one unit, as in the case of the unit circle, then the r r circumference is 2π units, and the angle at the centre of the circle c is 2π radians. 1 r 4. 2π radians = 360° B S x 5. The length of the semicircle from S through A to B is half the circumference and is π units. 6. π radians = 180° 7. An arc length of r units subtends an angle of 1 radian. A radian 8. An arc length of 2π r units subtends an angle of 2π radians. πr 2π r 9. An arc length of a quarter of a circle is units that is, and subtends an angle of 2 4 π radians. 2
Finding the number of degrees in one radian π c = 180° 180° 1c = = 57.296° (correct to 3 decimal places) π
Since we have
Converting radians to degrees Radians are converted to degrees using the following equation. 180° 1c = π Worked Example 1
Convert the following to degrees, giving the answer correct to 2 decimal places. a 2c b 6.3c c 9π
10
Think a
1 2
b
Write
180 Multiply the number of radians by . π Simplify where possible.
180° π 360° = π = 114.59°, correct to 2 decimal places.
a 2c = 2 ×
3
Write the answer correct to 2 decimal places.
1
b On a Calculator page, type: 6.3 Note: The radian sign does not need to be entered if the calculator is already in radian mode. To convert 6.3c to degrees, press Catalog k, press D to get to the items beginning with the letter D. Then select ¢DD. Then press ENTER ·.
Chapter 6 Circular (trigonometric) functions
259
2
Then press ENTER ·. Then write the answer.
6.3c = 360.96° correct to 2 decimal places. c
1
Multiply the number of radians 180 by . π
2
Simplify by cancelling.
c c 9π = 9π × 180°
10
10
π
= 162°
Converting degrees to radians Degrees are converted to radians using the following equation. 180° = π c 1° = Worked Example 2
Convert the following to radians. a 2° b 36.35° c 150° Think a
260
1
Write
On a Calculator page, type: 2 The calculator is in radian mode so you will need to type the degree symbol before converting to radians. You will find the degree symbol in the Maths expression template by pressing Ctrl / Catalog k.
Maths Quest 12 Mathematical Methods CAS for the TINspire
πc 180
2
Complete the entry line as: 2° Then press ENTER ·. Then write the answer.
π or 0.035c correct to 3 decimal places. 90 π b 36.35° = 36.35 × 180 2° =
b
1
Multiply the number of degrees by π . 180
2
Simplify.
= 0.634c
Note: In this example it is not appropriate to leave your answer in exact form. c
1
Multiply the number of degrees by π . 180
2
Simplify, leaving your answer in exact form.
c
150° = 150 ×
π 180
150π 180 5π = 6 =
Special cases (degrees to radians) Note the following special cases. 180° = π π 90° = 2 π 60° = 3 π 45° = 4 π 30° = 6
Divide both sides by 2. Divide both sides by 3. Divide both sides by 4. Divide both sides by 6.
Basic definitions of sine, cosine and tangent Sine and cosine In the unit circle the vertical distance PR is defined as sine (θ ) or sin (θ ) and the horizontal distance OR is defined as cosine (θ ) or cos (θ ). The coordinates of the point P are (cos (θ ), sin (θ )) where θ can be in radians or degrees. The xcoordinate of P is cos (θ ) and the ycoordinate of P is sin (θ ).
y P (cos (θ ), sin (θ )) sin (θ) θ x O cos (θ) R
sin (q ) and cos (q )
Chapter 6 Circular (trigonometric) functions
261
Special cases (sin, cos) Note the following special cases for sin and cos. sin (0) = 0 π sin = 1 2 sin (π) = 0
y
3π sin = − 1 2
π
cos (0) = 1 π cos = 0 2 cos (π) = −1
π – 2
sin (π–2 ) cos (π ) cos (0) 0
sin (2π) = 0 sin (0°) = 0 sin (90°) = 1 sin (180°) = 0 sin (270°) = −1 sin (360°) = 0
3π ) sin (— 2
3π cos = 0 2
x
cos (2π) = 1 cos (0°) = 1 cos (90°) = 0 cos (180°) = −1 cos (270°) = 0 cos (360°) = 1
3π — 2
Special cases
y
Tangent
T
Using the unit circle, the vertical distance TS is defined as tan θ. TS is the tangent to the circle which intersects with the xaxis and ∠TOS = θ. Using Pythagoras’ theorem in triangle OPR (figure below), PR2 + OR2 = OP2. 2 So, sin (θ ) + cos2 (θ ) = 1 From the diagram below, ∆OPR is similar to ∆OTS (angle, angle, angle).
Special cases (tan)
x
T
it
P un
sin (θ ) tan (θ ) = cos (θ )
S
tan (q )
y
1
TS PR Therefore: = OS OR tan (θ ) sin (θ ) = 1 cos (θ )
O
tan (θ )
θ
O
tan (θ ) sin (θ )
θ cos (θ ) R 1 unit Identities
S x
Note the following special cases for tan. tan (0) = 0
π tan is undefined 2
tan (0°) = 0
tan (90°) is undefined
tan (π) = 0
3π tan is undefined 2
tan (2π) = 0
tan (180°) = 0
tan (270°) is undefined tan (360°) = 0 sin (90o ) 1 = , which is undefined. It can be seen that tan (90°) is undefined because tan (90o ) = cos (90o ) 0 REMEMBER
1. Radians: (a) 1c = the size of the angle formed where the length of an arc is equal to the radius of the circle. (b) π c = 180° (c) Angles are in radians unless a degree symbol is shown.
262
Maths Quest 12 Mathematical Methods CAS for the TINspire
180 . π π 3. To change degrees to radians, multiply by . 180 4. Identities:
2. To change radians to degrees, multiply by
(b) tan (θ ) =
(a) sin2 (θ ) + cos2 (θ ) = 1
exerCiSe
6a eBook plus Digital doc
Spreadsheet 142 Unit circle
eBook plus
revision of radians and the unit circle 1 We1 Convert the following to degrees, giving answers correct to 2 decimal places. a 3c 7π c e 20
b 5c c 4.8c d 2.56c 5π c 5π c 3π c h f g 4 6 10 2 We2 Convert the following to radians. Give exact answers for a , b , c and d . Write other answers correct to 2 decimal places.
Digital doc
SkillSHEET 6.1 Changing degrees to radians
sin (θ ) cos (θ )
a 5°
b 15°
c 120°
d 130°
g 235° h 260° i 310° f 78.82° 3 Evaluate using a calculator. Give answers correct to 3 decimal places.
e 63.9° j 350°
a sin (0.4)
b sin (0.8)
c cos (1.4)
d cos (1.7)
e tan (2.9)
f tan (2.4)
g sin (75°)
h sin (68°)
i cos (160°)
j cos (185°)
k tan (265°)
l tan (240°)
4
Evaluate the following. a sin (0) 3π e tan 2
b sin (π) π f tan 2
c cos (2π)
d cos (π)
g sin (90°)
h sin (360°)
i cos (180°)
j cos (0°)
k tan (270°)
l tan (240°)
5
Evaluate without using your calculator. a sin2 (20°) + cos2 (20°)
b cos2 (50°) + sin2 (50°) π π e sin 2 + cos 2 2 2
d sin2 (2.5) + cos2 (2.5) g 2 sin2 (α) + 2 cos2 (α)
c sin2 (π) + cos2 (π) θ θ f sin 2 + cos 2 2 2
h 5 sin2 (β) + 5 cos2 (β)
6 Write the following in order from smallest to largest. a sin (35°), sin (70°), sin (120°), sin (150°), sin (240°) b cos (0.2), cos (1.5), cos (3.34), cos (5.3), cos (6.3) 8
15
7 If sin (θ ) = 17 and cos (θ ) = 17 , find tan (θ ). 8 If sin (A) = 0.6 and cos (A) = 0.8, find tan (A). Draw a triangle marking in the position of angle A and possible lengths of the sides.
π radians is equal to: 3 a 0° b 30° c 45°
eBook plus Digital doc
9 mC
SkillSHEET 6.2 Tangent ratios
D 60°
E 90°
Chapter 6
Circular (trigonometric) functions
263
10 MC The expression 1 − sin2 (α) is equal to: c cos (α) D tan (α) a 1 b cos2 (α)
E tan2 (α)
11 The temperature T °C inside a shop t hours after 2 am is given by π T = 15 − 3 cos t 12 Calculate the exact temperature after 4 hours and the temperature to the nearest tenth of a degree at 9.00 am.
6B
Symmetry and exact values Exact values Using the equilateral triangle (of side length 2 units) shown at right, the following exact values can be found. π sin (30°) = sin = 6
1 2
π cos (30°) = cos = 6 π tan (30°) = tan = 6
3 2 1
=
3
3 3
30° 2
π sin (60°) = sin = 3
3 2
π cos (60°) = cos = 3
1 2
π tan (60°) = tan = 3 3
60° 1 1 Exact values of sine, cosine and tangent of 30° and 60°
Using the right isosceles triangle shown, the following exact values can be found. π sin (45°) = sin = 4
1 2
π cos (45°) = cos = 4
=
1 2
=
2 2
2
3
45° 1
2
45° 1 Exact values of sine, cosine and tangent of 45°
2 2
π tan (45°) = tan = 1 4
The unit circle and symmetry properties The unit circle is symmetrical so that the magnitude of sine, cosine and tangent at the angles shown are the same in each quadrant but the sign varies. y
π–
θ
S
θ
θ
θ θ
θ θ
π+
A
T
C
x
2π –
Symmetrical properties
In the first quadrant sin, cos and tan are all positive. In the second quadrant only sin is positive.
264
Maths Quest 12 Mathematical Methods CAS for the TINspire
θ
In the third quadrant only tan is positive. In the fourth quadrant only cos is positive. This can be remembered as All Students To Class (ASTC). 1st quadrant
2nd quadrant
3rd quadrant
4th quadrant −sin
sin (2π − θ ) = −sin (θ )
sin (θ )
sin (π − θ ) = sin (θ )
sin (π + θ ) =
cos (θ )
cos (π − θ ) = −cos (θ )
cos (π + θ ) = −cos (θ )
cos (2π − θ ) = cos (θ )
tan (π + θ ) = tan (θ )
tan (2π − θ ) = −tan (θ )
tan (θ )
tan (π − θ ) =
−tan
(θ )
(θ )
Worked example 3
eBook plus
Without using a calculator, find the value of: a sin (150°) Think a
b
Tutorial
5π b cos . 4
int0546 Worked example 3
WriTe a sin (150°) = sin (180 − 30)°
1
Find the equivalent first quadrant angle.
2
As 150° is in the 2nd quadrant, sine is positive.
= sin (30°)
3
Write the exact value.
=
1
Find the equivalent first quadrant angle using 5π 4π π = + . 4 4 4
2
Decide on the sign required. As it is in the 3rd quadrant, cosine is negative.
π = − cos 4
3
Write the exact value.
=
b cos
1 2
5π π = cos π + 4 4
− 2 2
Angles are not restricted to values between 0 and 2π; that is, the domain is not restricted to [0, 2π]. If an angle is greater than 2π radians, it is necessary to subtract multiples of 2π so that the angle is within one turn of the unit circle. Each 2π radians is a complete turn of the circle. Worked example 4
If sin (x) = 0.6, cos (x) = 0.8, and x is in the first quadrant, find: a sin (3π − x) b cos (4π + x). Think a
1
WriTe
Write 3π as 2π + π — that is, one complete cycle and then the angle π − x.
a sin (3π − x) = sin (2π + π − x)
= sin (π − x)
x
2
y
x
= sin (x)
As it is in the 2nd quadrant, sine is positive.
Chapter 6
Circular (trigonometric) functions
265
b
= 0.6
3
Substitute the given value.
1
4π = 2 × 2π — that is, two complete cycles and then the angle x.
b cos (4π + x)
y x
2
As it is in the 1st quadrant, cosine is positive.
= cos (x)
3
Substitute the given value.
= 0.8
Worked example 5
eBook plus
π < θ < π calculate cos (θ ) and hence find tan (θ ). 2
If sin (θ ) = 12 and 13 Think
x
Tutorial
int0547 Worked example 5
WriTe
Method 1 1
2 3 4 5
In the given triangle, find the length of the third 132 = 52 + 122 side by using Pythagoras’ theorem. Let θ′ be the first quadrant angle corresponding to θ. adjacent 5 cos (θ ′) = cos (θ ′) = 13 hypotenuse
θ is in the second quadrant, therefore cos (θ ) is negative. opposite tan (θ ) = adjacent θ is in the second quadrant, therefore tan (θ ) is negative.
cos (θ ) = tan (θ ′) = tan (θ ) =
13 θ'
−5 13 12 5 − 12 5
Method 2 1
Use the rule sin2 (θ ) + cos2 (θ ) = 1.
2
Rearrange. 12 13
cos2 (θ ) = 1 − sin2 (θ ) for sin (θ ).
3
Substitute
4
Evaluate.
5
Take the square root of both sides, remembering that the answer could be positive or negative.
6
cos is negative in the 2nd quadrant.
7
Use tan (θ ) = tan (θ ).
sin2 (θ ) + cos2 (θ ) = 1 144
= 1 − 169 25
= 169
sin (θ ) to find the value of cos (θ )
cos (θ ) = ± 13 5
Take cos (θ ) = tan (θ ) = tan (θ ) =
266
maths Quest 12 mathematical methods CaS for the Tinspire
−5 13 12 13 5 13 − 12 5
as
π <θ <π 2
12 5
Method 3 1
On a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: 12 π solve (sin (θ ) = , θ )  < θ < π 13 2 Then press ENTER ·.
2
Complete the entry lines as: − 12 cos θ = π − sin 1 13 − 12 tan θ = π − sin 1 13 Press ENTER · after each entry. Write the answers.
Given sin (θ ) = 12 , 13 cos (θ ) = tan (θ ) =
π < θ < π, 2
−5 13 − 12 5
Negative angles For negative angles, move in a clockwise direction. In the diagram, RQ = −PR, and OR = OR, so T1S = −TS An alternative way to find tan (−θ ) is: tan
(− θ ) (−θ )
(− θ )
− sin
sin (θ ) − = = = tan (θ ) cos (− θ ) cos (θ )
y P
sin (θ ) θ O –θ
R S sin (– θ) Q
−sin
sin = (θ ) cos (−θ ) = cos (θ ) tan (−θ ) = −tan (θ )
The diagram shows 0 < θ <
T
x
T1
Negative angles
π ; however, these relationships are true for all values of θ. 2
sin (−150°) = −sin (150°) cos (−190°) = cos (190°) tan (−280°) = −tan (280°)
Chapter 6 Circular (trigonometric) functions
267
Worked Example 6
Find the exact value of: a sin (−135°) b cos (−240°) c tan (−330°). Think a
b
c
Write
1
sin (−135°) = −sin (135°)
2
135° is in the 2nd quadrant.
= −sin (180° − 45°)
3
Sine is positive in the 2nd quadrant.
= −sin (45°)
4
Give the exact value.
=
1
cos (−240°) = cos (240°)
2
240° is in the 3rd quadrant.
= cos (180° + 60°)
3
Cosine is negative in the 3rd quadrant.
= −cos (60°)
4
Give the exact value.
=
1
tan (−330°) = tan (30°)
2
Give the exact value.
2
3
268
x
−1 2
c tan (−330°) = tan (30°)
=
1
=
3 3
− 4π
a sin = 3
4π is in the 3rd quadrant. 3 Sine is negative in the 3rd quadrant.
Maths Quest 12 Mathematical Methods CAS for the TINspire
y
3
T −330°
Write
sin (−θ ) = −sin (θ )
y
–240°
Method 1: Technologyfree 1
x
–135°
b cos (−240°) = cos (240°)
Find the exact value of: − 4π − 5π a sin b tan . 3 6
a
135°
− 2 2
Worked Example 7
Think
y
a sin (−135°) = −sin (135°)
− sin
4π 3
π = − sin π + 3 π = − − sin 3 π = sin 3
S x
b
4
Give the exact value.
1
tan (−θ ) = −tan (θ )
2
5π is in the 2nd quadrant. 6
3
Tangent is negative in the 2nd quadrant.
4
Give the exact value.
= − 5π
b tan 6 =
3 2
5π 6
− tan
π = − tan π − 6 π = − − tan 6 π = tan 6 =
1
=
3 3
− 4π sin = 3
3 2
− 5π tan = 6
3 3
3
Method 2: Technologyenabled 1
On a Calculator page, complete the entry lines as: − 4π a sin 3 b
− 5π tan 6 Press ENTER · after each entry.
2
Write the answers.
Complementary angles
π Complementary angles add to 90° or radians. Therefore, 30° and 60° are complementary angles. 2 π π π In other words and are complementary angles, and θ and − θ are also complementary 2 6 3 angles.
The sine of an angle is equal to the cosine of its complement. Therefore, sin (60°) = cos (30°). We say that sine and cosine are complementary functions. The complement of the tangent of an angle is the cotangent or cot — that is, tangent and cotangent are complementary functions (as well as reciprocal functions). 1 cot (θ ) = tan (θ )
Chapter 6 Circular (trigonometric) functions
269
1st quadrant
2nd quadrant
3rd quadrant
π π sin − θ = cos (θ ) sin + θ = cos (θ ) 2 2
4th quadrant
3π 3π sin − θ = − cos (θ ) sin + θ = − cos (θ ) 2 2
3π 3π π π cos − θ = sin (θ ) cos + θ = − sin (θ ) cos − θ = − sin (θ ) cos + θ = sin (θ ) 2 2 2 2 3π π π tan − θ = cot (θ ) tan + θ = − cot (θ ) tan − θ = cot (θ ) 2 2 2
3π tan + θ = − cot (θ ) 2
Worked example 8
eBook plus
If sin (θ ) = 0.4 and tan (α) = 0.6, find: π 3π + α . a cos + θ b tan 2 2
Tutorial
int0548 Worked example 8
Think a
b
WriTe
π + θ = − sin (θ ) 2
a cos
1
Cosine and sine are complementary functions and cosine is negative in the second quadrant.
2
Substitute the given value for sin (θ ).
1
Tangent and cotangent are complementary functions and tangent is negative in the 4th quadrant.
= −0.4 3π
+ α = b tan 2 =
− cot (
α)
−1
tan (α ) −1
2
Substitute the given value for tan (α).
=
3
Calculate.
= −1.667, correct to 3 decimal places
0.6
rememBer
1. Exact values can be determined by using an equilateral triangle and a right isosceles triangle as shown.
30° 3
45° 2 60° 1
270
maths Quest 12 mathematical methods CaS for the Tinspire
2
1
1
45°
0°
30° π 6
45° π 4
60° π 3
90° π 2
180°
π
270° 3π 2
360°
θ
0
sin (θ )
0
1 2
2 2
3 2
1
0
−1
0
cos (θ )
1
3 2
2 2
1 2
0
−1
0
1
tan (θ )
0
3 3
1
Undef.
0
Undef.
0
3
2π
2. All angles in the unit circle are positive in the 1st quadrant, sine is positive in the 2nd quadrant, tangent is positive in the 3rd quadrant and cosine is positive in the 4th quadrant. Symmetry properties: sin (π − θ ) = sin (θ )
sin (π + θ ) = −sin (θ )
sin (2π − θ ) = −sin (θ )
cos (π − θ ) = −cos (θ )
cos (π + θ ) = −cos (θ )
cos (2π − θ ) = cos (θ )
tan (π + θ ) = tan (θ )
tan (2π − θ ) = −tan (θ )
tan (π − θ ) =
−tan
(θ )
3. Negative angles sin (−θ ) = −sin (θ ), cos (−θ ) = cos (θ ), tan (−θ ) = −tan (θ ) 4. Complementary angles π (a) Complementary angles add to 90o or radians. 2 (b) The sine of an angle is equal to the cosine of its complement. (c) The complement of the tangent of an angle is the cotangent.
exerCiSe
6B
π sin − θ 2 = cos (θ )
π sin + θ 2 = cos (θ )
3π sin − θ 2 − = cos (θ )
3π sin + θ 2 − = cos (θ )
π cos − θ 2 = sin (θ )
π cos + θ 2 = −sin (θ )
3π cos − θ 2 = −sin (θ )
3π cos + θ 2 = sin (θ )
π tan − θ 2 = cot (θ )
π tan + θ 2 = −cot (θ )
3π tan − θ 2 = cot (θ )
3π tan + θ 2 = −cot (θ )
Symmetry and exact values Note: Give answers as surds with rational denominators.
eBook plus
1 We3a
Without using a calculator, find the exact values of the following.
a sin (120°)
b cos (135°)
c tan (330°)
SkillSHEET 6.3
d cos (225°)
e sin (210°)
f tan (150°)
Rationalising the denominator
g sin (315°)
h cos (300°)
i tan (225°)
j cos (390°)
k sin (405°)
l tan (420°)
Digital doc
Chapter 6
Circular (trigonometric) functions
271
2 We3b eBook plus Digital doc
Spreadsheet 142 Unit circle
Find the exact values of the following.
3π a sin 4
5π b cos 6
2π c tan 3
4π 3
5π e sin 4
7π f tan 6
g sin
11π 6
5π h cos 3
7π i tan 4
9π j cos 4
13π k sin 6
7π l tan 6
d cos
3 We4 If sin (x) = 0.3, cos (a) = 0.5, tan (b) = 2.4 and x, a and b are in the first quadrant, find the value of the following. b cos (π − a) c tan (2π − b) d cos (π − x) a sin (π − x) e sin (π − a) f tan (π + b) g sin (2π − x) h cos (2π − a) i tan (π − b) j cos (2π + x) k sin (2π + a) l tan (2π + b) n cos (3π + a) o tan (3π − b) m sin (3π − x) If sin (α ) =
4
7 25
and cos(α ) =
24 , 25
find tan (α) and show that sin2 (α) + cos2 (α) = 1.
π < x < π , find cos (x) and hence find tan (x). 2 −1 3π , find sin (x) and tan (x). b If cos ( x ) = , and π < x < 2 2 If sin ( x ) = 12, and
5 a We5
c If sin ( x ) =
−
3 2
, and
3π < x < 2π , find cos (x) and tan (x). 2
3π , find sin (x) and cos (x). 2 Find the exact value of the following.
d If tan ( x ) = 3 , and π < x < 6 We6 a e i m
sin (−30°) sin (−120°) tan (−240°) sin (−420°)
b f j n
cos (−45°) tan (−135°) cos (−330°) cos (−390°)
c g k o
tan (−60°) sin (−225°) sin (−315°) tan (−405°)
d cos (−150°) h cos (−210°) l tan (−300°)
Evaluate (sin α + cos α)2 + (sin α − cos α)2.
7 8 We7
Find the exact values of: −π a sin b cos 3 6 −π
9
− 7π 6
− 3π
d cos 4
− 2π e sin 3
− 5π f tan 6
g sin
− 5π h cos 4
− 4π i tan 3
− 5π j cos 3
−13π k sin 6
− 9π l tan 4
−π −π Show that cos2 + sin 2 = 1. 4 4
10 We8
If sin (θ ) = 0.3, cos (x) = 0.7 and tan (α) = 0.4, find the value of the following.
π a sin − x 2
272
−π c tan 4
3π + θ b cos 2
maths Quest 12 mathematical methods CaS for the Tinspire
π c tan − α 2
π + θ 2
d cos
3π π − x e sin f tan + α 2 2 3π 3π − α + θ i tan j cos 2 2 π 11 mC If x = , 3 sin (2x) is equal to: 12 3 3 3 3 2 c a b 2 2 2
π
3π − θ h cos 2 3π + α l tan 2
g sin + x 2
3π − x k sin 2
D
1 2
E
π 2
π 12 mC The expression 1 − sin 2 − θ is equal to: 2
13
π a cos − θ 2
π b sin − θ 2
π D cos2 − θ 2
E 0
π c sin 2 − θ 2
A weight on a spring moves so that its speed, v cm per second, is given by the formula πt v = 10 + 2 sin 6 a Find the initial speed. b Find the speed of the weight after 5 seconds. c What is the greatest speed that the weight can reach?
eBook plus Digital doc
SkillSHEET 6.4 Problem solving using trigonometry
14 The height, H, in metres, that sea water reaches up a particular tree trunk at a Caribbean resort is governed πt by the equation H = 0.4 cos + 0.5, 12 where t is the number of hours past midnight. Find the height of water up the trunk at: a midnight b 8 am c 8 pm.
6C
Trigonometric equations From the general equation sin (x) = a, we can find an infinite number of solutions. An example of this general equation is: 3 sin ( x ) = . 2 3 π π One of the solutions is x = , because sin = . 3 2 3 π 3 However, we also know that sin π − = because sine 3 2 is positive in the second quadrant.
y
sin (π – –3π )
–π 3
–π 3
sin ( –3π ) x
π 2π and . 3 3 (There are no solutions in the third and fourth quadrants because here, sine is negative.) For this equation there are two solutions between 0 and 2π. They are
Chapter 6
Circular (trigonometric) functions
273
To find a greater number of solutions we can go around the unit circle as many times as we wish, finding new solutions each time. Since
π sin 2π + = 3
3 2
π and sin 3π − = 3
3 2
π 2π 7π 8π , , . and 3 3 3 3 We can also go in a negative direction. In the domain [−π, π] there are only 2π π 2 solutions: and . 3 3 in the domain [0, 4π] there are 4 solutions:
Worked Example 9
Find all solutions to the equation cos ( x ) =
− 2 2
in the domain [0, 2π].
Think
Write
Method 1: Technologyfree − 2 2
1
Write the question.
cos ( x ) =
2
Find the equivalent angle in the first quadrant, ignoring the sign.
Basic angle is
3
In the 2nd and 3rd quadrants cos (x) is negative. So write the appropriate values of x in these quadrants.
2nd quadrant: π x=π − 4 3rd quadrant: π x=π + 4 3π 5π x= , 4 4
4
Simplify.
Method 2: Technologyenabled 1
On a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: −
2
, x )  0 ≤ x ≤ 2π 2 Then press ENTER ·. solve(cos( x ) =
2
274
Write the answer.
x=
Maths Quest 12 Mathematical Methods CAS for the TINspire
3π 5π , 4 4
π . 4
y
3π — 4
x 5π — 4
Worked Example 10
Find all solutions to the equation sin (α) = 0.7 in the domain [0, 4π]. Give your answers correct to 4 decimal places. Think
Write
Method 1: Technologyfree 1
Find the equivalent angle in the 1st quadrant, ignoring the sign.
The basic angle is 0.7754.
2
Find in which quadrants sin (α) is positive.
sin (α) is positive in the 1st and 2nd quadrants.
3
Find the values of α in the 1st and 2nd quadrants.
α = 0.7754, π − 0.7754 = 0.7754, 2.3662
4
Since the domain is [0, 4π], we need to go around the circle twice, so add 2π to each of the first two solutions.
α = 0.7754, 2.3662, 0.7754 + 2π, 2.3662 + 2π
5
Simplify giving answers correct to 4 decimal places.
α = 0.7754, 2.3662, 7.0586, 8.6494, correct to 4 decimal places.
Method 2: Technologyenabled 1
On a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve(sin (a) = 0.7,a)  0 ≤ a ≤ 4p Then press ENTER ·.
2
Write the answer.
If sin (a) = 0.7, 0 ≤ a ≤ 4p, then a = 0.7754 or 2.3662 or 7.0586 or 8.6494, correct to 4 decimal places.
Worked Example 11
Calculate all solutions to the equation 2 sin ( 2θ ) = Think
−
3 in the domain 0° ≤ x ≤ 360°. Write
1
Since the equation contains 2θ rather than θ, multiply both end points of the domain by 2.
0° ≤ θ ≤ 360° 0° ≤ 2θ ≤ 720°
2
Simplify the trigonometric equation.
2 sin (2θ ) = − 3 sin (2θ ) =
− 3 2
Chapter 6 Circular (trigonometric) functions
275
3
Find the ‘first quadrant (basic) angle’, ignoring the negative sign.
Basic angle = 60°
4
Sine is negative in the 3rd and 4th quadrants. Find the angles in these quadrants equivalent to the basic angle.
2θ = 180° + 60°, 360° − 60° 2θ = 240°, 300°
5
As the domain has been extended to 720°, we need to go around the circle two times, so add 360° to each of the two initial values.
2θ = 240°, 300°, 240° + 360°, 300° + 360° 2θ = 240°, 300°, 600°, 660°
6
Simplify.
θ = 120°, 150°, 300°, 330°
Worked Example 12
Calculate the sum of the solutions between 0 and 2π for the equation sin (3x) = cos (3x). Think
Write
1
Adjust the domain as shown.
0 ≤ x ≤ 2π 0 ≤ 3x ≤ 6π
2
Divide both sides by cos (3x).
3
Find the basic angle.
4
Solve for x between 0 and 6π. Note that tan is positive in the 1st and 3rd quadrants.
5
Cancel where possible.
sin (3x) = cos (3x) tan (3x) = 1 π Basic angle = 4 π π π π π π 3 x = , π + , 2π + , 3π + , 4π + , 5π + 4 4 4 4 4 4 π 5π 9π 13π 17π 21π = , , , , , 4 4 4 4 4 4 π 5π 9π 13π 17π 21π x= , , , , , 12 12 12 12 12 12 π 5π 3π 13π 17π 7π x= , , , , , 12 12 4 12 12 4
6
Check that all answers are between 0 and 24π 8π . They are. 2π (2π = = ) 12 4 Calculate the sum of the solutions.
7
Sum of the solutions π 5π 3π 13π 17π 7π = + + + + + 12 12 4 12 12 4 11π = 2
General solutions of trigonometric equations π 3π and x = . 4 4 However, if a domain is not specified, there are an infinite number of solutions as multiples of π 3π 2π can be added or subtracted indefinitely to and . In this situation a general solution is 4 4 obtained where the solutions are in terms of a parameter, n, where n is an integer (that is, n ∈ Z). The solution to the equation sin ( x ) =
276
1
2
where x ∈ [0, 2π] is x =
Maths Quest 12 Mathematical Methods CAS for the TINspire
π π becomes x = 2 nπ + , where n ∈ Z. 4 4 3π 3π The general solution for the second quadrant solution x = becomes x = 2 nπ + , 4 4 where n ∈ Z. 3π Note that the general solution x = 2 nπ + can be expressed as 4 π π x = 2 nπ + π − = (2 n + 1)π − . 4 4 Substituting different integer values of n will give specific solutions, as shown in the table below. π 3π n x = 2 nπ + , n ∈ Z x = 2 nπ + , n ∈Z 4 4 3π − 5π π − 7π −1 x = − 2π + = x = − 2π + = 4 4 4 4 3π 3π π π 0 x=0+ = x=0+ = 4 4 4 4 π 9π 3π 11π 1 x = 2π + = x = 2π + = 4 4 4 4 π 17π 3π 19π 2 and so on x = 4π + = x = 4π + = and so on 4 4 4 4 The general solution for the first quadrant solution x =
In general: − − • If sin (x) = a, then x = 2nπ + sin 1 (a) and x = (2n + 1)π − sin 1 (a), where a ∈[−1, 1] and n ∈ Z. − • If cos (x) = a, then x = 2nπ ± cos 1 (a), where a ∈[−1, 1] and n ∈ Z. −1 • If tan (x) = a, then x = nπ + tan (a), where a ∈ R and n ∈ Z. Worked example 13
eBook plus
Find the general solution of the equation 2 cos ( x ) − 1 = 0. Hence, find all the solutions for x ∈ [−2π, 2π]. Think
int0549 Worked example 13
WriTe
1
Write down the general solution for cos (x) = a.
2
Substitute a = − 1 cos 1 2
Tutorial
1 2
into the general equation and evaluate
recognising that it is an exact angle.
3
Write the two separate solutions and specify n ∈ Z.
4
Note the answer from step 2 could be further simplified by combining the two terms. A CAS calculator will give the answer in this form.
5
Substitute n = −1, n = 0 and n = 1 into each of the general solutions.
6
Write down the solutions for x ∈ [−2π, 2π].
−
x = 2nπ ± cos 1 (a) x = 2 nπ ± cos
−1
( ) 1
2
π x = 2 nπ ± 4 π π x = 2 nπ + and x = 2 nπ − , n ∈ Z. 4 4 8nπ ± π (8n ± 1)π ,n∈Z x= = 4 4 − 7π 9π and x = 4 4 − π π n = 0: x = and x = 4 4 π 7π π 9π n = 1: x = 2π − = and x = 2π + = 4 4 4 4 − 7π − π π 7π x= , , , 4 4 4 4
n = −1 : x =
Chapter 6
−
Circular (trigonometric) functions
277
Worked Example 14
Find the general solution of the equation 2 sin ( 2 x ) = domain 0 ≤ x ≤ 2π. Think
278
−
3 and hence find all solutions for x in the
Write
1
On a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve(2 sin (2 x ) = − 3 , x ) Then press ENTER ·.
2
Write the two separate solutions and specify n ∈ Z.
Solving 2 sin (2x) = 3 gives (6 n − 1) π (3n + 2) π x= and x = , n ∈ Z. 6 3
3
Substitute n = 0 and n = 1 and n = 2 into each of the general solutions, to find the solutions in the domain 0 ≤ x ≤ 2p.
n = 0: x =
4
This answer can be verified using a CAS calculator. Complete the entry line as: − solve(2 sin (2 x ) = 3 , x )  0 ≤ x ≤ 2π Then press ENTER ·.
−
−π
2π 6 3 5π 5π n = 1: x = and x = 6 3 11 π 8π n = 2: x = and x = 6 3 2π 5π 5π 11π , , , For 0 ≤ x ≤ 2p, x = 3 3 6 6
Maths Quest 12 Mathematical Methods CAS for the TINspire
and x =
Worked Example 15
Find the general solution of the equation sin (3x) = cos (3x) and hence find all solutions for x in the domain 0 ≤ x ≤ 2π. Think 1
Divide both sides by cos (3x).
Write
sin (3 x ) = cos (3 x ) sin (3 x ) =1 cos (3 x ) tan (3x) = 1 
2
Write down the general solution for tan (3x) = a. 3x = np + tan 1 (a)
3
Substitute a = 1 into the general equation and evaluate tan 1 (1), recognising that it is an exact angle. Divide each side by 3 to solve for x.
4
To find the solutions in the domain 0 ≤ x ≤ 2p, substitute n = 0, 1, 2 . . . into each of the general solutions.
5
Write down the simplified solutions for 0 ≤ x ≤ 2p.
6
This answer can be verified using a CAS calculator. Note: 1. The general formula here is found using − 3π the basic angle of , which means 4 that values of n start at 1 rather than 0. 2. Remember to scroll to the right to view the full set of solutions over the given domain.

3x = np + tan 1 (1) π 3 x = nπ + 4 x=
(4 n + 1) π , n ∈Z 12
n= 0: x =
π 12
n = 1: x =
5π 12
n = 2: x =
9π 12
n = 3: x =
13 π 12
n= 4: x =
17 π 12
n = 5: x =
21π 12
x=
π 5π 3π 13π 17π 7π , , , , , 12 12 4 12 12 4
Chapter 6 Circular (trigonometric) functions
279
Worked Example 16
Find the solutions of the equation cos ( x ) + cos ( 3 x ) = 12 , x ∈[ − 2π , 2π ] . Think
Write
1
On a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: 1 solve(cos ( x ) + cos (3 x ) = , x ) − 2π ≤ x ≤ 2π 2 Then press ENTER ·. An exact solution is not available.
2
A numerical solution will be appropriate and can be found by pressing Ctrl / ENTER ·.
3
Write the solutions.
Solving cos( x ) + cos(3 x ) =
1 2
for
x ∈ [2p, 2p] gives x =  5.654 87, 2.0944, 0.628 319, 0.628 319, 2.0944, 5.654 87
REMEMBER
1. Given a general equation such as sin (x) = a, there can be an infinite number of solutions. The domain is usually restricted and it is important to find all values for x within this domain. 2. If the domain is given in radians, then the solution(s) to x should be in radians. If the domain is given in degrees, then the solution(s) to x should be in degrees. 3. Adjust the domain to match what has been done to the angle in the question.
280
Maths Quest 12 Mathematical Methods CAS for the TINspire
4. Sine is positive in the 1st and 2nd quadrants, cosine is positive in the 1st and 4th quadrants and tangent is positive in the 1st and 3rd quadrants. 5. If sin (x) = a, then the general solution is x = sin−1 (a) + 2nπ, n ∈ Z. 6. If cos (x) = a, then the general solution is x = cos−1 (a) + 2nπ, n ∈ Z. 7. If tan (x) = a, then the general solution is x = tan−1 (a) + nπ, n ∈ Z. 8. Find all the solutions within the specified domain, by substituting integer values for n into the general solution. 9. If given cos2 (x) = a, then you will need to find values in all 4 quadrants (as x will have both positive and negative values). 10. If the equation is of the form sin (ax) = k cos (ax), divide both sides by cos (ax) to change the equation to tan (ax) = k.
exerCiSe
6C
Trigonometric equations Find all solutions to the equations below in the domain [0, 2π].
1 We9
2
eBook plus Digital doc
Spreadsheet 135 Trig equations
a cos (θ ) = 0
b sin (θ ) =
d sin (θ ) = −1
e cos (θ ) =
−1 2
c
cos (θ ) =
c
sin (θ ) =
1 2
− 3 2
Find all the values of θ between 0° and 360° for which: a sin (θ ) = 1
b cos(θ ) =
1 2
d cos (θ ) = −1
e sin (θ ) =
1
3 2
2
3 We10 For each equation below, find all the values of x between 0 and 4π. Give answers correct to 4 decimal places. a cos (x) = −0.6591 b sin (x) = 0.9104 c cos (x) = 0.48 d sin (x) = −0.371 4 Find all the values of x between 0° and 360° for which: a sin (x) = 0.2686 b cos (x) = −0.7421 − c sin (x) = 0.5432 d cos (x) = 0.1937 Give answers correct to 2 decimal places. 5
Find the solutions to the following equations in the domain 0 ≤ x ≤ 2π. a 2 sin (x) = 1 c 2 sin ( x ) =
−
b 3 cos (x) = 0 d
3
2 cos( x ) = 1
6 We11 Find the solutions to the following equations in the domain 0° ≤ x ≤ 360°. Give exact answers where possible, otherwise give answers correct to 2 decimal places. a cos (2x) = 1 b 2 sin (2x) = −1 c 2 cos (3 x ) = e sin (3x) = g 4 sin
−
2
−0.1254
( x ) = 0.913 1 2
d 2 sin (3 x ) = 3 f 3 cos (2x) = 0.5787 h
2 cos( x ) = − 0.2751
Chapter 6
Circular (trigonometric) functions
281
7 Find all the solutions between 0 and 2π to the following equations. Give exact answers where possible, otherwise give answers correct to 4 decimal places. x 2
a 4 sin (x) + 2 = 6
b 3 cos (x) − 3 = 0
c cos
x d sin + 5 = 5.32 3
e 2 sin (3x) − 5 = −4
f
g 2 cos (2 x ) + 3 = 0
h
1 3
1
+ 4 = 4.21
2 cos (3 x ) + 2 = 3
sin 2 x − 1 = − 0.8039
8 We12 Calculate the sum of the solutions between 0 and 2π for each of the following equations. Give exact answers for questions a to d. Otherwise give answers correct to 4 decimal places. a sin (x) = cos (x) b sin (2x) = cos (2x) c sin (2 x ) = 3 cos (2 x )
eBook plus Digital doc
WorkSHEET 6.1
f sin (x) + 3 cos (x) = 0 e sin (3x) + 2 cos (3x) = 0 d 3 sin (3 x ) = cos(3 x ) 9 A particle moves in a straight line so that its distance, x metres, from a point O is given by the equation x = 3 + 4 sin (2t), where t is the time in seconds after the particle begins to move. a Find the distance from O when the particle begins to move. b Find the time when the particle first reaches O. Give your answer correct to 2 decimal places. 10 We13 Find the general solution of the following equations. Hence, find all solutions for −2π ≤ x ≤ 2π for each equation. a 2 cos ( x ) − 3 = 0
b tan ( x ) =
1
2 sin ( x ) − 1 = 0
c
3
11 We14 Find the general solution of the equation 2 sin (2x) − 1 = 0. Hence, find all solutions for −π ≤ x ≤ π. 12 Find the general solution of the equation 2 cos (3x) − 1 = 0. Hence, find all solutions for −π ≤ x ≤ π. 13 We15 Find the general solutions for each of the following equations. Hence, find all solutions for x ∈ [0, 2π]. a
3 sin ( x ) = cos ( x )
b sin (2x) = cos (2x)
c
14 We16 Find the solutions of the equation sin (2 x ) + sin (3 x ) =
6d
3 ,x 2
−
3 sin (3 x ) = cos (3 x )
∈[− π , π ] .
Trigonometric graphs Graphs of the sine and cosine functions The graph of the function f (x) = sin (x) is drawn below. It is drawn over the domain [−2π, 2π]. The graph repeats itself every 2π radians. We say that it has a period of 2π. Half the distance between the maximum and minimum values is 1 so we say y that the amplitude is 1. 1 It is possible to take any value of x for the function 0.5 f (x) = sin (x), so the domain of the whole function is R. The range is [−1, 1]. The graph is shown at right. 3π 3π −π −–π 0 –π π — −2π − — 2π x 2
2
−0.5
2
−1 f(x) = sin (x),
282
maths Quest 12 mathematical methods CaS for the Tinspire
−2π
≤ x ≤ 2π
2
The graph of the function f (x) = cos (x) is the same shape as the sine graph, but the graph has been translated to a different position. The period is also 2π. The amplitude is 1, the domain is R and the range is [−1, 1]. The graph is shown at right.
y 1 0.5 3π − π–2 0 −2π − — 2 −π −0.5
–π 2
π
3π — 2
2π x
−1 f(x) = cos (x), 2p ≤ x ≤ 2p
Dilation If we change the amplitude, the distance between the maximum value and the minimum value also changes. y The graph of f (x) = 2 sin (x) is shown at right. The amplitude is 2 2. The period is still 2π. The domain is R and the range is [−2, 2]. 1 This graph is a dilation of the basic graph of f (x) = sin (x) by a factor of 2 from the xaxis. It has been stretched vertically. 0 –π 3π — π 2π x 2 2 Generally, if f (x) = a sin (x) or f (x) = a cos (x), a is the dilation −1 factor in the direction of the yaxis. The amplitude of the −2 graph is a . f(x) = 2 sin (x), 0 ≤ x ≤ 2p If we change the coefficient of x, the period of the graph changes. The graph of f (x) = cos (2x) from 0 to 2π is shown at right. The amplitude is 1 and the period is π. This can be y found by dividing 2π by the coefficient of x. 1 2π In this case = π . The domain is R. The range is [−1, 1]. 0.5 2 The xintercepts can be found by solving the equation 0 –π –π — 3π π — 5π — 3π — 7π 2π x 4 2 2 4 4 4 cos (2x) = 0. −0.5 π 3π 5π 7π 2 x = , , , −1 2 2 2 2 f(x) = cos (2x), 0 ≤ x ≤ 2p π 3π 5π 7π so x = , , , 4 4 4 4 1 This graph is a dilation of the basic graph of f (x) = cos (x) by a factor of 2 from the yaxis. The period has been halved or the graph has been ‘squashed up’. The decimal approximation for these solutions can be found using a graphics calculator. The graphics calculator can also be used to check the number and approximate value of the solutions when solving trigonometric equations. 1 Generally, if f (x) = sin (nx) or f (x) = cos (nx), the graph is dilated by a factor of from the n 2π y yaxis. The period of the graph is . n 4 The graph of f (x) = 4 sin (3x) is shown at right. 3 2
It is drawn from 0 to 2π. The amplitude is 4 and the 1 2π 4π 5π –π — — — 3 3 3 3 2π − 3π 0 period is . The domain is R. The range is [ 4, 4]. π –π — 2π x 2 2 −1 3 −2 The xintercepts are found by solving 4 sin (3x) = 0. −3 So sin (3x) = 0 −4 3x = 0, π, 2π, 3π, 4π, 5π, 6π f (x) = 4 sin (3x), 0 ≤ x ≤ 2p 4π 5π π 2π x = 0, , , π, , , 2π 3 3 3 3 This is an example of a sine graph dilated in both x and y directions. It has a dilation factor of 1 4 from the xaxis, and of 3 from the yaxis.
Chapter 6 Circular (trigonometric) functions
283
Worked Example 17
State the period and the amplitude of the graphs of each of the following functions. a y = 2 sin
( x) 1 4
b y =
−1 cos 3
(2 x ) .
Think a
b
Write
1
Write the equation.
2
Find the period of the graph, using the formula.
3
State the amplitude — it is the coefficient in front of the sine function.
1
Write the equation.
2
Find the period of the graph.
a y = 2 sin
( x) 1 4
2π 1 ,n = 4 n 2π 4 So period = 1 = 2π × = 8π 1 4 Period =
Amplitude = 2 b y=
−1 cos (2 x ) 3
Period = So period =
3
Amplitude =
State the amplitude. Remember that the amplitude is always positive, so we need only the magnitude of the coefficient.
2π , n=2 n 2π =π 2 1 3
Worked Example 18 1 Sketch the graph of y = 2 sin (3θ ) for one complete cycle stating the amplitude, period and range.
Think
Write/draw
Method 1: Technologyfree
284
1
Write the equation.
2
Find the period of the graph.
3
State the amplitude.
1
y = 2 sin (3θ ) 2π ,n=3 n 2π So period = 3 Period =
Amplitude =
Maths Quest 12 Mathematical Methods CAS for the TINspire
1 2
4
5
Sketch the basic sine shape (one full cycle) 2π 1 with the period of and the amplitude of 2 . 3 1 The scale on the xaxis should be 4 of a period (to reflect the 4 quadrants), 2π 2π π so it is ÷ 4= = . 3 12 6
The minimum value of the function is
−1 2
and
y 1– 2
0
π– 6
π– 3
π– 2
−1
1
2— π 3
− 1– 2
Range : [ 2 , 2 ]
1
the maximum value is 2 , so state the range. Method 2: Technologyenabled On a Graphs page, complete the function entry line as: 1 2π f 1( x ) = sin (3 x)  0 ≤ x ≤ 2 3 Then press ENTER ·.
Reflection If the coefficient of the function is negative, the graph is turned upside down, that is, reflected in the xaxis. This does not alter the amplitude, which is always positive. • y = f (x) is the image of f (x) when reflected in the xaxis. y • y = f (x) is the image of f (x) when reflected in the yaxis. 4 3 The graph of f (x) = −4 sin (3x) is shown at right. You 2 will notice that it is the graph of f (x) = 4 sin (3x) turned 1 3π –π — upside down. (The graph of f (x) = 4 sin (3x) is shown 2 2 5π 4π 2π 0 on page 283.) –π — — π — 2π x 3 3 3 3 −1 −2 −3 −4
f (x) = 4 sin (3x), 0 ≤ x ≤ 2p
2π . The domain is R. The range is [−4, 4]. 3 4π 5π π 2π The xintercepts are 0, , , π, , , 2π . 3 3 3 3 1 This graph has a dilation factor of 4 from the xaxis and of 3 from the yaxis. If the function f (x) = 4 sin (3x) is reflected in the xaxis, the result is f (x) = −4 sin (3x). If we reflected the graph of f (x) = 4 sin (3x) in the yaxis, the result would still be f (x) = −4 sin (3x). The amplitude is 4 and the period is
Chapter 6 Circular (trigonometric) functions
285
If we reflect f (x) = 2 cos (3x) in the xaxis, the result is f (x) = −2 cos (3x), but if we reflect it in the yaxis, the graph does not change. This is because f (x) = 2 cos (3x) is symmetrical about the yaxis. Check this on a CAS calculator. y 4 3
Translation If we add a constant to the function, the graph is moved up or down and is said to be translated parallel to the yaxis. The number that is being added becomes the median value of the function. The graph of f (x) = 3 cos (2x) + 1 is shown at right. Compared to f (x) = 3 cos (2x), the graph is shifted 1 unit up. The amplitude is 3, the period is π, the domain is R and the range is [−2, 4]. The xintercepts are found by solving 3 cos (2x) + 1 = 0. So cos 2 x =
−1 3
2 1 −1
0
–π 2
π
3π — 2
2π x
−2 f (x) = 3 cos (2x) + 1, 0 ≤ x ≤ 2p
(cosine is negative in quadrants 2 and 3).
The basic angle is inverse cosine of
1 3
= 1.231c .
2x = π − 1.231, π + 1.231, 3π − 1.231, 3π + 1.231 2x = 1.911, 4.373, 8.194, 10.656 ⇒ x = 0.955, 2.186, 4.097, 5.328 If we add a constant to x, the graph is translated parallel to the xaxis, that is, left or right. π The graph of f ( x ) = sin x − is the graph of 4
y 1 0.5
0 –π 5π π −π π— 2π x 4 4 units to the right. −0.5 4 −1 π The graph of f ( x ) = sin x + is the graph of 4 π f ( x ) = sin x + , − π ≤ x ≤ 2π π 4 f (x) = sin (x) translated units to the left. 4 This graph is a translation of the π basic graph of the function The graph of f ( x ) = sin x + is shown at right. 4 π f ( x ) = sin( x ) units parallel to It is drawn between [−π, 2π]. 4 The amplitude is 1, the period is 2π, the domain is R and the xaxis in a negative direction. the range is [−1, 1]. The yintercept occurs when x = 0. 2 π . So y = sin 0 + = 4 2 π The xintercepts occur when sin x + = 0. 4 − π 3π 7π π So x + = 0, π , 2π ⇒ x = , , 4 4 4 4 π 3π Note: The graph of sin (x) is the same as the graph of cos ( x − ) or cos ( x + ). That is, the 2 2 π 3π sine graph can be turned into the cosine graph by a translation of units left or units 2 2 right. The cosine graph can be turned into the sine graph by the opposite translations.
f (x) = sin (x) translated
If we take a general trigonometric function f (x) = a sin n(x − b) + c, it has an amplitude 2π of a , a period of , a horizontal translation of b units and a vertical translation of n c units.
286
Maths Quest 12 Mathematical Methods CAS for the TINspire
y
π The graph of f ( x ) = 2 cos 3 x − − 1 is shown at right for 3 0 ≤ x ≤ 2π. 2π The amplitude is 2, the period is , the domain is R and the 3 range is [−3, 1].
1 0
–π 3
−1
2π — 3
π
4π — 3
5π — 3
2π x
−2
This graph is a dilation of the basic graph of f (x) = cos (x) by a 1 factor of 2 from the xaxis and a factor of 3 from the yaxis, π along with a translation of units to the right and 1 unit down. 3
−3
Worked Example 19
π Sketch the graph of y = 5 cos x + + 5 for 0 ≤ x ≤ 2π, and state the period and amplitude. 4 Think
Write/draw
π y = 5 cos x + + 5 4
1
Write the equation.
2
Determine the period. (The coefficient of x is 1.)
Period =
3
State the amplitude.
Amplitude = 5
4
Using a pencil, sketch a basic cosine shape with the period and amplitude above.
2π = 2π 1
y 5
y = 5 cos (x)
0
π
–π 2
2π x
3π — 2
−5 5
Apply the horizontal translation of +
π (the + sign 4
y –π 4
means to the left). Again, use pencil.
–π 4
0 –π
−π–4
–π 2
4
Apply the vertical translation of +5 (+ means up).
y
π
3π — 4
5π — 3π — 7π — 4 2 4 2π
x
–π 4
y = 5 cos (x + –4π ) + 5
10
+5
+5
5 +5 +5
−π–4
y = 5 cos (x)
–π 4
−5 6
–π 4
–π 5 y = 5 cos (x + 4 )
0
y = 5 cos (x + –4π )
3π — 4 –π 2
π +5
3π — 7π — 2 4
2π x
−5
Chapter 6 Circular (trigonometric) functions
287
7
Erase the pencilled stages.
y 10
y = 5 cos (x + –4π ) + 5
0
2π x
3π — 4
rememBer
1. Graphs of the form y = a sin n(x − b) + c and y = a cos n(x − b) + c are transformations of y = sin (x) and y = cos (x). 2. a is the amplitude and is a dilation from the xaxis. If a is negative, the amplitude is still positive but the graph is reflected in the xaxis. 3. c is the vertical translation (the translation parallel to the yaxis). If c is positive, the graph is translated c units up, and if c is negative, the graph is translated c units down. 4. The range is [c − a, c + a]. 2π . 5. The period is n 6. The factor n is the horizontal dilation where the graph has been dilated by a 1 factor of from the yaxis. n 7. The value b is the horizontal translation (the translation parallel to the xaxis). If b is positive, the graph is translated b units to the right, and if b is negative, the graph is translated b units to the left.
exerCiSe
6d
Trigonometric graphs 1 We17
eBook plus Digital doc
SkillSHEET 6.5 Period and amplitude of sine and cosine graphs
eBook plus Digital doc
Spreadsheet 124 Sine graphs
State the period and amplitude of the graphs of each of the following.
a y = cos (x) 1 3
d y = cos ( x ) g y = 3 sin
1 2
c y = 4 sin (x)
e y = 2 cos (3x)
f y = 3 sin (2x)
h y = 2 cos
( x) 1 3
i
y=
−1 3
cos (2 x )
j y = −4 sin (3x) 2 We18 Sketch the graphs of each of the following for one complete cycle and state the amplitude, the period and the range. 2
a y = 3 cos(θ )
b y = 4 sin (θ )
d y = 2 cos (3θ )
e y = 2 cos (3θ )
g y = 4 sin
288
( x)
b y = sin (x)
( θ) 1 2
1
h y = 3 cos
maths Quest 12 mathematical methods CaS for the Tinspire
( θ) 1 3
c y = 3 sin (2θ ) f y=
−1 3
sin (2θ )
eBook plus
3
Sketch the graph of the function f:R → R where f (x) = 4 cos (3θ ) for 0 ≤ θ ≤ 2π. State the amplitude, period and range.
4
Sketch the graph of the function f:R → R where y = 2 sin (2x) for −π ≤ x ≤ π. State the amplitude, period and range.
5
From the basic graphs of y = sin (x) and y = cos (x), state the horizontal translation and the vertical translation for each of the following. π π a y = sin x + + 3 b y = cos x − + 1 3 2
Digital doc
Spreadsheet 012 Cosine graphs
π c y = 3 cos x − − 2 4
π d y = 2 sin x + − 1 3
6 Sketch the graphs of the following for one complete cycle stating the amplitude, the period and the range. a y = sin (x) + 1 b y = cos (x) − 1 c y = 2 cos (x) − 2 d y = 2 sin (x) + 3 e y = sin (3x) − 1 f y = cos (2x) + 1 1
h y = 2 sin (2 x ) + 3
g y = 3 cos (3x) − 2 j y = 2 cos
( x) − 1
y = 3 sin
i
( x) + 4 1 2
1 3
7 We19 Sketch the graphs of the following for 0 ≤ θ ≤ 2π. State the period and amplitude. π π π a y = sin θ − b y = cos θ + c y = 3 cos θ − 3 4 2
π d y = 2 sin θ − 4 π g y = cos 3 θ − + 1 6
π e y = 2 sin 2 θ + 2 π h y = 2 sin 2 θ − − 2 2
π f y = 3 cos 3 θ + 3 π i y = 2 sin θ − − 1 4
1
j y = cos 2 (θ − π ) + 1 8
Write down the amplitude, period and range of the following graphs. a
b
y 4
y 4
2 3π −π 0 –π – −π – −π −— 4 2 4 4 −2
2 –π 2
3π π — 4
x
0 −2
−4
c
2π
3π
4π x
−4
d
y 2
y 1
1 0 −1
π
0.5 –π 2
π
3π — 2
2π
5π — 2
3π x
−0.5
−2
0 π 2π 3π 4π 5π 6π 7π 8π 9π x
−1
Chapter 6
Circular (trigonometric) functions
289
y
e
y
f
4 3 2 1 −π
g
π x
0
−π–2 −1 0 −2
π
–π 2
3π — 2
x −7
h
y 1.5 1 0.5 −0.5 −1 −1.5
9
3
y 1 0.5
0
–π 2
π
3π — 2
0
2π x
−0.5
–π 2
π
2π
3π — 2
5π — 2
3π x
−1
State the maximum and minimum values for each of the following. a y = cos (x) b y = sin (x) c y = 3 sin (x) d y = 2 cos (x) e y = 2 cos (3x) f y = 3 sin (2x) g y = 4 sin (2x) + 1 h y = cos (4x) − 2 i y = 2 cos (x − π) − 3 j y = 4 sin 3 (x + π) + 1
1
k y = 2 sin 3 ( x + 2π ) + 2
1
l y = 3 cos 2 ( x − 2π ) − 4
10 Sketch the graphs of the following over the domain [0, 2π] and state the period, amplitude and range. a y = −cos (x) b y = −sin (x) c y = −2 sin (x) d y = −3 cos (x) e y = −3 cos (2x) f y = −sin (3x) g y = 1 − 4 sin (x) j y = 2 − sin (x − 2π)
h y = −2 cos (2x) − 2
i y =
−1 2
cos 3 ( x + π ) + 1
π to the left, what is the new equation? 3 π 12 If the graph of y = 2 cos (3x) − 2 is translated to the right and 3 units up, what is the new 4 equation? π 13 If the graph of y = 3 sin (x − π) + 1 is translated to the left and 3 units down, what is the new 3 equation? 11 If the graph of y = sin (x) + 1 is translated
Water level
14 The level of the water in the Banksia River was measured at hourly intervals from midnight and the results recorded. The graph below shows the results. Find: y a the amplitude 3 b the period 2 c the maximum height of the river d the minimum height of the river 1 e at what times the river has maximum height 0 2 4 6 8 10 12 14 16 18 20 22 24 x f at what times the river has a minimum Hours height g the equation of the curve which is of the form y = A sin a(x) + B.
290
Maths Quest 12 Mathematical Methods CAS for the TINspire
6E
Graphs of the tangent function The graph of the function f (x) = tan (x) is shown at right. − x = —2π It is drawn over the domain [−2π, 2π]. The graph repeats y − x = 3—2π itself every π radians, so its period is π. There are vertical x = –2π x = 3—2π 3 asymptotes through half the period, 2 3π π that is, at x = ± and ± , so the function is not 2 2 1 defined at these points. Generally, the vertical 0 –π π 3—π −2π − 3—2π −π − –2π 2π x π 2 2 asymptotes are given by the equation x = (2k + 1) , −1 2 −2 where k ∈ Z (that is, k = 0, ±1, ±2 …). Hence, the domain π −3 of the tangent function is R \ {x : x = (2k + 1) , k ∈ Z}. 2 Unlike sine and cosine functions, the tangent graph does not have an amplitude; it extends infinitely up and down,so its range is R.
Dilation
y = 2 tan (x) y = tan (x)
y
Compared to f (x) = tan (x), the graph of f (x) = a tan (x) is dilated by a factor of a from the xaxis. Vertical dilation does not affect the period, domain or range, or the position of the asymptotes or the xintercepts. In fact, its effect can be seen only when two graphs are sketched on the same set of axes. The diagram at right shows the graphs of f (x) = tan (x) and f (x) = 2 tan (x) over the domain [0, 2π]. The graph of f (x) = 2 tan (x) is the dilation of the basic tangent graph by a factor of 2 from the xaxis. It has been stretched vertically. Compared to f (x) = tan (x), the graph of f (x) = tan (nx) is dilated 1 by a factor of from the yaxis. Horizontal dilation y n x = π–4 3 affects the period and domain of the graph, as well as the position of the asymptotes and xintercepts. 2 π The period of f (x) = tan (nx) is and the general equation 1 n π of the asymptotes is x = (2k + 1) ,where k ∈ Z. 0 π– π– 2n −1 4 2 π The domain changes to R \ {x : x = (2k + 1) , k ∈ Z} −2 2n accordingly. (The range is not affected by the value of n.) −3 At right is the graph of f (x) = tan (2x). This graph is a dilation of the basic f (x) = tan (x) graph by a factor of 12 π from the yaxis; its period is . As the period has been 2 halved, the resultant graph is ‘compressed’ horizontally.
3 2 1 0 −1
π 3—2π
–π 2
2π
x
−2 −3
x = –2π x = 3—2π
x = 3—4π x = 5—4π x = 7—4π
π
3— π 4
5— π 4
y
3— π 7— π 2 4
2π x
x = –2π x = 3—2π
3 2 1
Reflection If the coefficient a in f (x) = a tan (x) is negative, the graph is reflected in the xaxis. Reflecting the graph does not affect its period, domain or range, or the position of the asymptotes. At right is the graph of f (x) = −2 tan (x). This graph is a reflection in the xaxis of the graph f (x) = 2 tan (x), shown previously.
0 −1
–π 2
π
3— π 2
2π
x
−2 −3
Chapter 6 Circular (trigonometric) functions
291
Translation y If a constant is added to the function, the graph is translated x = –2π x = 3—2π vertically, that is, up or down, parallel to the yaxis. Thus, the 3 graph of f (x) = tan (x) + c represents a translation of the graph 2 f (x) = tan (x) by c units in the y direction. If c > 0, the graph is 1 shifted up, and if c < 0, it is shifted down. Vertical translation does not affect the period, domain or range, or the position of 0 –π π 3—π 2π x 2 2 the asymptotes. The axial intercepts, however, will change. −1 The diagram at right shows the graph of y = tan (x) + 2. It is −2 translated 2 units up, compared to the basic graph of y = tan (x). −3 If a constant is added to x, the graph is translated horizontally, that is, right or left, parallel to the xaxis. Thus, the graph of f (x) = tan (x − b), y x = –4π x = 5—4π represents a translation of the graph f (x) = tan (x) 3 by b units in the x direction. If b > 0, the graph is 2 shifted to the right, and if b < 0, it is shifted to the left. Horizontal translation has no effect on the 1 period or range, but it does affect the domain and 3— π π 5— π π 7— π the position of the asymptotes and the axial 2π x –π –π 3— − –π 0 4 4 2 4 4 2 4 −1 intercepts. π −2 The diagram at right shows the graph of y = tan ( x + ). 4 π translated units to the left, compared to the basic 4 graph of y = tan (x). Having considered each transformation individually, we can now summarise them as follows. Compared to the basic graph of y = tan (x), the graph of y = a tan [n(x − b)] + c is: – dilated by a factor of a from the xaxis 1 π – dilated by a factor of from the yaxis (and hence has a period of ) n n – reflected in the xaxis if a < 0 – translated c units vertically (up if c > 0 and down if c < 0) – translated b units horizontally (to the right if b > 0 and to the left if b < 0).
Worked Example 20
x State the period and sketch the graph of y = 2 tan , showing one full cycle. 4 Think
Write/draw
1
Write the equation.
x y = 2 tan 4
2
Find the period.
Period =
π ; n= n
So, period =
292
Maths Quest 12 Mathematical Methods CAS for the TINspire
π 1 4
1 4
=π ×
4 = 4π 1
3
Sketch the tangent shape with a period of 4π. The graph is dilated by a factor of 2 parallel to the yaxis, so in comparison to the graph of y = tan (x), each ycoordinate will be doubled. The vertical asymptote goes through half of the period, so it is at x = 2π.
y
x = 2π
3 2 1 0 −1
π
2π
3π
4π
x
−2 −3
Worked example 21
eBook plus
π State the period and sketch the graph of y = − tan 2 ( x − ) + 1 4 for 0 ≤ x ≤ π. Think
Tutorial
int0550 Worked example 21
WriTe/draW
1
Write the equation.
π y = − tan 2 x − + 1 4
2
Find the period.
Period =
3
Using pencil, sketch the basic tangent shape with π a period of . 2 (Since 0 ≤ x ≤ π, we need to show two cycles.) Remember that the asymptotes are at the middle of each cycle (that is, halfway through the period).
y
π π ; n = 2. so period = n 2 x = π–4 x = 3—4π
3 2 1 0 −1
–π 4
–π 2
3— π 4
π x
−2 −3 4
Since we need a negative tangent graph, reflect the graph in the xaxis.
y
x = π–4 x = 3—4π
3 2 1 0 −1
–π 4
–π 2
3— π 4
π x
−2 −3
Chapter 6
Circular (trigonometric) functions
293
5
Translate the graph
π units to the right. 4
y 3
x = π–4 x = π–2
2
x = 3—4π
1 0 −1
–π 4
–π 2
3— π 4
π
x
−2 −3 6
y
Translate the graph 1 unit up.
x = π–2 x = π
3 2 1 0 −1
–π 4
–π 2
3— π 4
π
x
−2 −3
7
Erase the pencilled stages to see the final graph.
y
x = π–2 x = π
3 2 1 0 −1
–π 4
–π 2
3— π 4
π
x
−2 −3
REMEMBER
1. The graph of y = tan (x) has a period of π and does − − y x = 3—2π x = —2π x = –2π x = 3—2π not have an amplitude. 3 2. The equations of the vertical asymptotes of 2 π y = tan (x) are given by x = (2k + 1) , where 1 2 k ∈ Z, that is, k = 0, ±1, ±2 . . . 0 –π π 3—π −2π − 3—2π −π − –2π 2π x 2 2 3. The domain of the tangent function is −1 π R \ {x : x = (2k + 1) , k ∈ Z} and its range is R. −2 2
294
Maths Quest 12 Mathematical Methods CAS for the TINspire
4. Compared to the basic graph of y = tan (x), the graph of y = a tan [n(x − b)] + c is: – dilated by a factor of a from the xaxis 1 π – dilated by a factor of from the yaxis (and hence has period of ) n n – reflected in the xaxis if a < 0 – translated c units vertically (up if c > 0 and down if c < 0) – translated b units horizontally (to the right if b > 0 and to the left if b < 0). exerCiSe
6e
Graphs of the tangent function 1
We20 State the period and sketch the graphs of each of the following, showing one full cycle. a y = 4 tan (x) b y = tan (2x) c y = −tan (3x) x 1 f y = − 2 tan d y = 2 tan (4x) e y = − tan 4 x 3 x − h y = −5 tan (2x) g y = 3 tan i y = 21 tan (4 x ) 2
( )
j
y = 13 tan
( x) 1 2
2 For each of the following, state: i the period ii the transformations, compared to the basic graph y = tan (x). a y = 2 tan (3x) d y = − tan
( x) − 2 1 2
π 1 g y = 4 tan 3 x + 6
b y = −tan (4x) + 1 π e y = − 5 tan x + 2
c y = 3 tan (2x) − 4 π f y = tan 2 x − 4
x 1 h y = 6 − 2 tan 3
π 1 i y = tan 4 x − − 3 12
π j y = 2 tan x + + 5 3 3
We21 State the period and sketch the graphs of each of the following for 0 ≤ x ≤ π.
a y = tan (x) + 2
π d y = 2 tan x − 2 x 1 g y = 2 tan − 1 2 j y=
6F
1 3
b y = tan (2x) − 3
π c y = − tan x + 4
1 e y = − tan 4 ( x − π ) π h y = − 2 tan 2 x + 8
x 1 f y = − 3 tan + 1 2 6 −3 π tan 4 x + + 1 i y= 4 12
1 π tan 2 x − − 2 2
Finding equations of trigonometric graphs
eBook plus Interactivity
Sometimes it is necessary to be able to find the equation of a trigonometric function from a graph. The following worked examples illustrate how this can be done.
Chapter 6
int0251 Finding equations of trigonometric graphs
Circular (trigonometric) functions
295
Worked Example 22
The equation of the following graph is in the form y = a sin (nx). Find the values of a and n. Hence, find the equation of the function.
y 2 1 0 −1
π
2π
3π
4π
x
−2
Think
Write
1
State the amplitude of the graph.
Amplitude = 2
2
The coefficient a represents the amplitude. Since the graph is not reflected in the xaxis, a is positive.
The graph is not reflected, so a = 2.
3
State the period of the graph (it is the length of one full curve).
Period = 4π.
4
Use the formula for the period to find the value of n.
Period = n=
5
Substitute the values of a and n into y = a sin (nx) to find the equation of the function.
2π 2π , so = 4π ; n n
2π = 4π
1 2
y = a sin (nx); a = 2, n = ∴ y = 2 sin
1 2
( x) 1 2
Worked Example 23
The graph shown is a trigonometric function of the form y = a sin (nx) + c. Find the values of a, n and c. Hence, find the equation of the function.
y 2 1 −1 0 –4π −2 −3 −4
Think 1
2
3
296
The amplitude (a) is half the distance between the maximum and minimum values. The period is the interval from one point on the graph to the next point where the graph begins to 2π repeat itself. The period is . n The line through the centre of the graph is y = −1, so the graph has been translated down 1 unit.
3π — 4
π
5π — 4
Write
y = a sin (nx) + c 1
Amplitude a = 2 (2 + 4) = 3 2π = π , so n = 2. The period is n
c = −1 So the equation is y = 3 sin (2x) − 1.
Maths Quest 12 Mathematical Methods CAS for the TINspire
7π — 4
2π x
Worked Example 24
This graph is a trigonometric function of the form y = c + a cos (nx). Find the values of a, n and c. Hence, write the equation of the function. y 3 2 1 0 −1
1
2
3
Think
The amplitude (a) is half the distance between the maximum and minimum values.
2
We know that the graph is a cosine graph so it must be inverted; that is, a is negative. The period is the amount of time taken to complete the pattern once.
4 5
5
6 x
Write
1
3
4
c represents the vertical translation. This graph has been translated up one unit. The equation of this trigonometric function is in the form y = c + a cos (nx).
y = c + a cos (nx) 1 Amplitude a = 2 (3 + 1) =2 a = −2 Period =
2π =6 n
2π =n 6 π so n = 3 c=1
y = c + a cos ( nx )
π 3 π y = 1 − 2 cos x 3 c = 1, a = − 2, n =
REMEMBER
1. Trigonometric functions can be expressed in the form of y = a sin [n(x − b)] + c or y = a cos [n(x − b)] + c. 2. The amplitude a can be found by halving the distance between the maximum and minimum values. 3. The period is the interval from one point on the graph to the next point where the 2π graph begins to repeat itself. The period is . n 4. The vertical shift is c. 5. The horizontal shift is b. Exercise
6F
Finding equations of trigonometric graphs 1 WE22 The equations of the following graphs are of the form y = a sin (nx). Find the values of a and n. Hence, find the equation of each function.
Chapter 6 Circular (trigonometric) functions
297
a
eBook plus Digital doc
Spreadsheet 124 Sine graphs
Digital doc
y 2 1
0
−1 −2 −3
eBook plus
b
y 3 2 1 –π 4
–π 2
0 1 2 3 4 5 6 7 8 9 10 11 12 x −1 −2
π x
3π — 4
Spreadsheet 012 Cosine graphs
2 The equations of the following graphs are of the form y = a cos (nx). Find the values of a and n. Hence, find the equation of each function. b a y y 2
2
1
1
0 −1
–π 8
–π 4
3π — 8
–π 2
x
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 x −1 −2
−2
3
We23 The equations of the following graphs are of the form y = a sin (nx) + c. Find the values of a, n, and c and hence write the equation of the function. b a y y
1
1.5
0 1 2 3 4 5 6 7 8x −1 −2
1.0 0.5 0
–π 2
π
−3 −4 −5
2π x
3π — 2
4 The equations of the following graphs are of the form y = a cos [n(x − ε)]. Find the values of a, n and ε. Hence, write the equation of the function. b a y y 5
4 2 0 −2
–π 4
–π 2
3π — 4
π x
0
−4
–π 4
–π 2
3π — 4
π x
−5
5 The equations of the following graphs are of the form y = a sin [n(x + ε)] + c. Find the values of a, n, ε and c. Hence, write the equation of the function. b a y y 1 0 −1
–π 6
–π –π 3 2
2π — 5π — 3 6
π
7π — 4π — 3π — 5π 11 — —π 6 3 2 3 6
2π x
−2 −3
298
maths Quest 12 mathematical methods CaS for the Tinspire
5 4 3 2 1
0 −1
–π 2
π
3π — 2
2π
x
6
We24 The equations of the following graphs are of the form y = c + a cos (nx). Find the values of a, n and c. Hence, write the equation of the function.
a
b
y 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 x −1
7
8
y 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12x –1
mC If the amplitude is 2, the period is 6 and there is a vertical translation of −2, then the
equation of the form y = a sin (nx) + b is: a y = 2 sin (6x) − 2
b y = 6 sin (2x) − 2
π D y = 2 sin 3
π E y = 2 sin 6
x − 2
π c y = 2 − 2 sin x 3
x − 2
mC If the period is π, the range is [−2, 4], and the horizontal
eBook plus
π Digital doc , the equation for the trigonometric function of the WorkSHEET 6.2 4 form y = a cos [n(x + ε)] + b is: π π π a y = 3 − 2 cos π x + + 1 b y = 3 cos 2 x + − 1 c y = 2 cos 3 x + + 1 4 4 4 translation is
π D y = 2 cos 3 x + + 1 4
6G
π E y = 3 cos 2 x − + 1 4
Trigonometric modelling In real life there are many examples of periodic behaviour. Sine and cosine functions such as y = a sin [n(x − b)] + c and y = a cos [n(x − b)] + c are often used to model this behaviour.
eBook plus eLesson
eles0092 Trigonometric modelling
Worked example 25
While out in his trawler John North, a fisherman, notes that the height of the tide in the harbour can be found by using the equation: π h = 5 + 2 cos t , 6 where h metres is the height of the tide and t is the number of hours after midnight. a What is the height of the high tide and when does it occur in the first 24 hours? b What is the difference in height between high and low tides? c Sketch the graph of h for 0 ≤ t ≤ 24. d John North knows that his trawler needs a depth of at least 6 metres to enter the harbour. Between what hours is he able to bring his boat back into the harbour?
Chapter 6
Circular (trigonometric) functions
299
Think
Write
Method 1: Technologyfree a
1
Write the given equation.
2
For high tide, find the maximum value of h.
π t 6
a h = 5 + 2 cos
For maximum h, π cos t = 1 6 So h = 5 + 2 × 1 = 7 Alternatively, maximum value = median + amplitude so h = 5 + 2 = 7
b
3
Find when high tide occurs.
1
Find the minimum value of h.
π cos t = 1, so 6 π t = 0, 2π , 4π , . . . 6 t = 0, 12, 24, . . . A high tide of height 7 m occurs at midnight, noon the next day, and midnight the next night. b For minimum h,
π cos t = −1 6
2
c
So h = 5 + 2 × −1 = 3 Alternatively, min. value = median − amplitude so h = 5 − 2 = 3. The difference between high and low tides is 7 − 3 = 4 metres.
Find the difference between high and low tides.
Use the information from the previous page to sketch the graph.
c
h 6 4 2 0 2 4 6 8 1012 14 16 18 202224 t
d
1
2
300
Find t using the equation when h = 6.
Write the answer in words.
d When h = 6,
π 5 + 2 cos t = 6 6 π 2 cos t = 1 6 π 1 cos t = 2 6 π π 5π 7π 11π t= , , , .... 6 3 3 3 3 t = 2, 10, 14, 22, . . . From the graph we can see that John North can bring his boat back into harbour before 2 am, between 10 am and 2 pm and between 10 pm and 2 am the next morning.
Maths Quest 12 Mathematical Methods CAS for the TINspire
Method 2: Technologyenabled 1
On a Graphs page, complete the function entry lines as: π f 1( x ) = 5 + 2 cos x 6 f 2( x ) = 6 Press ENTER · after each entry. Select an appropriate window.
2
To find the points of intersection, press: • MENU b • 7:Points & Lines 7 • 3:Intersection Point(s) 3 Use the NavPad to move the cursor to the first point of intersection and press Click x twice and the points of intersection will appear.
t = 2, 10, 14, 22, . . . rememBer
1. Read the question carefully to find out what is being asked. 2. Answer the questions in words where appropriate. 3. Ensure that graphs are sketched over the given domain. exerCiSe
6G eBook plus Digital doc
Spreadsheet 041 Function grapher
Trigonometric modelling 1
We25 Competition is severe, so Fred Greenseas decides that he will catch more fish in an inlet several kilometres east of the place where John North fishes. There is a sandbar at the entrance to the inlet and the depth of water in metres on the sandbar is modelled by the πt function d (t ) = 6 + 2.5 sin where t is the number of hours after 12 noon. 6 a What is the greatest depth of the water on the sandbar and when does it first occur? b How many hours pass before there is once again the maximum depth of water on the sandbar? c What is the least amount of water on the sandbar? d Sketch the graph of d for 0 ≤ t ≤ 24. e Fred Greenseas needs a depth of 7.25 metres to cross the sandbar. Between what hours is he able to enter and leave the inlet?
Chapter 6
Circular (trigonometric) functions
301
2 A student wanting to catch fish to sell at a local market on Sunday has discovered that more fish are in the water at the end of the pier when the depth of water is greater than 8.5 metres. π The depth of the water (in metres) is given by d = 7 + 3 sin t , where t hours is the 6 number of hours after midnight on Friday. a b c d e
What is the maximum and minimum depth of the water at the end of the pier? Sketch a graph of d against t from midnight on Friday until midday on Sunday. When does the water first reach maximum depth? Between what hours should the student be on the pier in order to catch the most fish? If the student can fish for only two hours at a time, when should she fish in order to sell the freshest fish at the market from 10.00 am on Sunday morning?
3 The mean daily maximum temperature in Tarabon, an experimental town in a glass dome, is π modelled by the function T (m) = 18 + 7 cos m , where T is in degrees Celsius and m is the 6 number of months after 1 January 2007. a What was the mean daily maximum temperature in March 2007, and in August 2007? b What is the highest mean daily maximum temperature in Tarabon? In which months does it occur? c What would the mean daily maximum temperature be in February 2008? d If the pattern continued, how many months would pass before the mean daily maximum temperature would be the same again as it was in February 2008? 4 The height above the ground of the middle of a skipping rope as it is being turned in a child’s game is found by using the equation h = a sin (nt) + c, where t is the number of seconds after the rope has begun to turn. During the game, the maximum height the rope reaches is 1.8 metres and it takes 2 seconds for the rope to complete a full turn. a Find the values of a, n and c and hence write the equation of h in terms of t. b Sketch the graph of h against t for 0 ≤ t ≤ 5. c After how much time from the beginning of the turn will the rope be 25 cm above the ground? Give your answer correct to the nearest tenth of a second. y
5 The graph at right shows the path of a small lizard as it runs over a hill. a Calculate the height of the hill, in metres, and hence find the amplitude of the trigonometric function. b If the ground was flat, how far would the lizard run to reach the same spot on the other side of the hill? Hence, find the period of the function. 6 When Sloane and Michael were riding on a Ferris wheel, they realised that as the wheel turned clockwise, the height of their seat in metres after t minutes could be modelled by the function h(t) = a − b cos (nt). The graph of h against t is shown for the whole ride. a State the values of a, b and n. b Write down the rule for h(t). c How many times do Sloane and Michael reach the highest point during their ride and how far above the ground is it? d How high are they when the ride begins to move? e How many minutes pass before they are at this point again?
302
Maths Quest 12 Mathematical Methods CAS for the TINspire
h
h = 34
30 25 20 15 10 5 0
0
3
6
h=2 9 12 15 t
y = 1 − cos ( –4π x)
x
f How far above the ground are they after 1 minute? g If the ride began when the boys were at the height found in part f , what would the function become? h Draw a graph of the new function for the first 6 minutes of the ride. 7 On a summer day the hourly temperature, which can be approximated to a cosine curve, was recorded. The maximum temperature was 30 °C and occurred at 3.00 pm. The minimum temperature was 10 °C and occurred at 3.00 am. The temperature was first recorded at 12 midnight, then every hour for 24 hours. a What is the amplitude of the function? b What is the period of the function? c What is the middle value of the function? d How far has this middle value been translated upwards from the xaxis? e When do the maximum and minimum temperatures occur? f Using the above, write an equation that will model this function. g Check the accuracy of your work by using your equation to find the temperature after 3, 9, 15 and 21 hours. h What is the temperature at midnight? Give your answer correct to the nearest degree. Check that your answer makes sense.
6H
Further graphs Earlier in the book we discussed in detail how to graph the sum and difference of functions. In this section we will further apply these techniques to trigonometric functions.
Graphing the sum and difference of functions The sum (and difference) function is defined over the intersection of the domains of the individual functions. That is, dom (f (x) ± g(x)) = dom f (x) ∩ dom g(x). To sketch the graph of the sum function, the method of addition of ordinates can be used. This method involves sketching the graphs of the individual functions on the same set of axes and then adding their ycoordinates. This technique is illustrated in the worked example below.
Worked Example 26
Using addition of ordinates, sketch the graph of y = sin (x) + cos (x) for the domain [0, 2π]. Think
Draw
Method 1: Technologyfree 1
Sketch the graphs of y = sin (x) and y = cos (x) on the same set of axes over the required domain. (Note that both y = sin (x) and y = cos (x) are defined over [0, 2π], so the sum function is defined over the same domain.)
y 2 1 0 −1
–π 4
–π 2
3— π 4
π
5— π 4
3— π 2
7— π 4
2π x
−2
Chapter 6 Circular (trigonometric) functions
303
2
Moving from left to right, add the ycoordinates of the two graphs and plot the resultant points. The π π 3π 5π 3π 7π π, , , ‘key’points are at x = 0, , , 4 2 4 4 2 4 and 2π.
y 2 1 0 −1
–π 4
–π 2
3— π 4
π
5— π 4
3— π 2
7— π 4
–π 4
–π 2
3— π 4
π
5— π 4
3— π 2
7— π 4
–π 4
–π 2
3— π 4
π
5— π 4
3— π 2
7— π 4
2π x
−2 3
Join the points with a smooth curve.
y 2 1 0 −1
2π
x
−2 4
Erase y = sin (x) and (y) = cos (x) to see the final graph.
y 2 1 0 −1
2π x
−2
Method 2: Technologyenabled On a Graphs page, complete the function entry lines as: f 1(x) = sin (x) f 2(x) = cos (x) f 3(x) = f 1(x) + f 2(x) Press ENTER · after each entry. Select an appropriate window.
Note that a difference function can be treated as a sum function where the second additive is negative. For instance, the function y = sin (x) − cos (x) can be viewed as y = sin (x) + (−cos (x)). So to obtain the graph of y = sin (x) − cos (x), we can sketch the graphs of y = sin (x) and y = −cos (x) on the same set of axes and add their respective ycoordinates.
Graphing modulus (absolute value) functions As was discussed previously, to sketch the graph of the modulus function y =  f (x), sketch the graph of y = f (x) and then reflect all sections of the graph that are below the xaxis in the xaxis. The technique is shown in the worked example below.
304
Maths Quest 12 Mathematical Methods CAS for the TINspire
Worked Example 27
Sketch the graph of y = 3 cos (2x) over the domain [0, 2π]. Think
Draw
Method 1: Technologyfree 1
First sketch the graph of y = 3 cos (2x). The amplitude of this graph is 3 and its period is 2π = π . Since the domain is [0, 2π], we need 2 to sketch two full cycles.
y 3 2 1 0 −1
–π 4
–π 2
3— π 4
π
5— π 4
3— π 2
7— π 4
–π 4
–π 2
3— π 4
π
5— π 4
3— π 2
7— π 4
–π 4
–π 2
3— π 4
π
5— π 4
3— π 2
7— π 4
2π x
−2 −3 2
Reflect all sections of the graph that are below the xaxis in the xaxis.
y 3 2 1 0 −1
2π x
−2 −3 3
Erase the sections below the xaxis to see the final graph.
y 3 2 1 0 −1
2π x
−2 −3
Method 2: Technologyenabled On a Graphs page, complete the function entry line as: f 1(x) = 3 cos (2x) Then press ENTER ·. Select an appropriate window.
Chapter 6 Circular (trigonometric) functions
305
Graphing product functions The product function is defined over the intersection of the domains of the individual functions. That is, dom (f (x) g(x)) = dom f (x) ∩ dom g(x). Some features of the product function can be established by using the following rules: 1. The xintercepts of f (x) g(x) occur where either f (x) or g(x) have their xintercepts. 2. f (x) g(x) is above the xaxis where f (x) and g(x) are either both positive, or both negative. 3. f (x) g(x) is below the xaxis where one of the functions f (x) or g(x) is positive and the other is negative. The general shape of the graph and the coordinates of the turning points can be found using a CAS calculator. Worked Example 28
Find the domain and sketch the graph of the product function y = x sin (x). Use a CAS calculator for assistance. Think
306
1
The required function can be viewed as a product of two functions: f (x) = x and g(x) = sin (x). The domain of the product function is equal to the intersection of the domains of the two individual functions.
2
On a Graphs page, complete the function entry line as: f1(x) = x × sin (x) Then press ENTER ·. Select an appropriate window.
3
To locate the maximums and minimums, press: • MENU b • 5:Trace 5 • 1:Graph Trace 1 Trace along the graph to find a local maximum. Press ENTER · to lock in the point. This method can also be used to locate minimums and intercepts.
write/Draw
Let f (x) = x, dom f (x) = R Let g(x) = sin (x), dom g(x) = R dom (f (x) g(x)) = dom f (x) ∩ dom g(x) =R So the domain of y = x sin (x) is R.
Maths Quest 12 Mathematical Methods CAS for the TINspire
Of course the graphs of product functions are not limited to those involving trigonometric functions.
Worked Example 29
If f(x) = 2x and g ( x ) = x + 1, sketch the graph of f ( x) g ( x ) = 2 x x + 1. Think 1
write/Draw
Sketch the graphs of f (x) and g(x).
y
y = 2x y= x+1
(0, 1) x
(−1, 0) 0
2
Find the domain of f(x) and the domain of g(x).
Dom f = R and dom g = [−1, ∞)
3
Find the domain of f(x)g(x).
Dom fg = [−1, ∞)
4
Find the xintercepts of both f and g and hence find the xintercepts of the product fg.
xintercept for f (x) is when x = 0 and f (x) = 0 xintercept for g(x) is when x = −1 and g(x) = 0 Hence, the xintercepts for the product are when x = 0 and x = −1.
5
Find the values of x for which the product is negative.
f (x) is negative and g(x) is positive for x ∈ (−1, 0), so fg is negative for x ∈ (−1, 0).
6
Find the values of x for which the product is positive.
f(x) and g(x) are both positive for x ∈ (0, ∞), so fg is positive for x ∈ (0, ∞).
7
Find the turning point using a graphics calculator. Round the answer to 2 decimal places as appropriate.
The turning point is (
8
Sketch the graph of the product.
y
3
, − 0.77).
y = 2x x + 1
(−1, 0) (0, 0) −3 (— , 3
−2
x
−0.77)
Graphing composite functions A composite function is formed from two functions in the following way. If f (x) = x + 5 and g(x) = 2x are two functions, then we combine the two functions to form the composite function g(f (x)) = 2f (x) = 2(x + 5). That is, f (x) replaces x in the function g(x).
Chapter 6 Circular (trigonometric) functions
307
The composite function reads g of f and can be written g f. Another composite function is f (g(x)) = g(x) + 5 = 2x + 5. In this case, g(x) replaces x in f (x). This composite function reads f of g and can be written f g. For the composite function f (g(x)) to be defined, the range of g must be a subset of the domain of f. Furthermore, if f (g(x)) is defined, the domain of f (g(x)) equals the domain of g(x). Composite functions can be rather complex to graph by hand, so a CAS calculator can be used for assistance when sketching. Worked example 30
eBook plus
For the pair of functions f(x) = cos (x) and g( x ) = x : a show that f(g(x)) is defined b find f(g(x)) and state its domain c sketch the graph of f(g(x)), using a CAS calculator for assistance. Think a
b
c
Tutorial
int0551 Worked example 30
WriTe/draW
For f(g(x)) to exist, the range of g must be a subset of the domain of f. So find both the range of g and the domain of f to show that this condition is observed. 1
Form the composite function f (g(x)) by substituting g(x) into f (x).
2
The domain of f (g(x)) must be the same as the domain of g(x). Since the domain of g(x) is R+ ∪ {0}, so is the domain of f (g(x)).
On a Graphs page, complete the function entry line as: f 1( x ) = cos( x ) Then press ENTER ·. Select an appropriate window.
a f (x) = cos (x); domain of f (x) = R
g( x ) = x : range of g(x) = R+ ∪ {0} Range of g(x) ⊆ domain of f(x) ∴ f (g(x)) is defined
b f ( g( x )) = cos ( x )
Domain of f (g(x)) = R+ ∪ {0}
c
The graphs of composite functions are not limited to those involving trigonometric functions. This is demonstrated in the following example. Worked example 31
For f(x) = x2 − 2 and g(x) = x4: a show that both f(g(x)) and g(f(x)) are defined b find both f(g(x)) and g(f(x)), stating the domain and range of each one c on separate axes, sketch the graphs of f(g(x)) and g(f(x)).
308
maths Quest 12 mathematical methods CaS for the Tinspire
Think
Write/draw
a
Show that ran f ⊆ dom g and that ran g ⊆ dom f.
a Dom f = R; ran f = [−2, ∞).
b
1
Write f(x).
b f (x) = x2 − 2
2
Form the composite function f (g(x)) by substituting g(x) into f (x). Simplify.
f (g(x)) = (x4)2 − 2 = x8 − 2
3
State the domain and range of f (g(x)).
Domain = R, range = [−2, ∞).
4
Form the composite function g(f (x)) by substituting f (x) into g(x).
g(f(x)) = (x2 − 2)4
5
State the domain and range of g(f (x)).
1
Sketch f (g(x)), rounding xintercepts to 2 decimal places.
c
Dom g = R; ran g = [0, ∞]. Range of f ⊆ domain of g, so g(f (x)) exists. Range of g ⊆ domain of f, so f (g(x)) exists.
Domain = R, range = [0, ∞)
c The yintercept is −2, and the xintercepts
are −1.09 and 1.09. y
y = x8 − 2 (−1.09, 0) 0
2
Sketch g(f (x)).
(1.09, 0) x (0, −2)
The yintercept is 16, and the turning point is (2, 0). y (0, 16) y = (x2 − 2)4 0
(2, 0) x
REMEMBER
1. For the sum/difference function, dom (f (x) ± g(x)) = dom f (x) ∩ dom g(x). The graph of the sum/difference function can be obtained by using the addition of ordinates method. 2. The graph of the modulus function, y =  f (x), can be obtained by sketching the graph of y = f (x) and then reflecting all of the sections of the graph that are below the xaxis in the xaxis. 3. For the product function, dom (f (x) g(x)) = dom f (x) ∩ dom g(x). Some features of the graph of the product function are as follows: (i) the xintercepts of f(x) g(x) occur where either f (x) or g(x) have their xintercepts
Chapter 6 Circular (trigonometric) functions
309
(ii) f (x) g(x) is above the xaxis where f (x) and g(x) are either both positive or both negative (iii) f (x) g(x) is below the xaxis where one of the functions f (x) or g(x) is positive and the other is negative. 4. For the composite function f (g(x)) to be defined, the range of g must be a subset of the domain of f. Furthermore, if f (g(x)) is defined, the domain of f (g(x)) equals the domain of g(x). exerCiSe
6h eBook plus
Further graphs 1
Digital doc
SkillSHEET 6.6 Addition of ordinates
We26 Using addition of ordinates, sketch the following graphs for the domain [0, 2π]. a y = sin (x) + cos (2x) b y = cos (x) + sin (2x) c y = 2 sin (x) + cos (x) d y = 2 cos (x) + sin (x) e y = 2 sin (x) + cos (2x) f y = 2 cos (x) + sin (2x) g y = 2 sin (2x) + cos (x) h y = 2 cos (2x) + sin (x)
2 Apply the addition of ordinates method to sketch each of the following graphs over the domain [0, 2π]. Use a CAS calculator to check your answers. a y = sin (x) + x b y = cos (x) − x c y = 3 sin (x) − 2x 1
3
1
d y = cos (2 x ) + 4 x 2
e y = 2 sin (4 x ) − 8 x 3
g y = 4 sin (x) − 5 loge (x + 1)
h y = 3 cos (2 x ) −
f y = tan (x) − 2x2
20 ( x + 2)2
We27 Sketch each of the following graphs over the domain [0, 2π].
a y = sin (2x)
b y = 2 sin (4x)
1 c y = cos x 2
d y = 3 cos (3x)
e y =  2 tan (x)
f y =  4 sin (x) + 2
g y = 1 − 2 cos (2x)
h y = tan (2x) + 3
4 Sketch each of the following graphs over the domain [0, 2π]. Remember to observe the appropriate order of transformations. x a y = 3sin (x) b y = 2 cos (2x) + 1 c y = − 2 cos + 3 2 5
6
We28 Find the domain and sketch the graph of each of the following product functions. Use a CAS calculator for assistance.
a y = 0.5x sin (x)
b y = (x − 1) cos (x)
c y = 3 sin (x) loge (x)
d y = 2 cos ( x ) x
e y = 8 cos (x) sin (x)
f y = (4 − 2x) sin (2x)
g y = (1.2)x cos (x)
h y = sin
d f (x) =  x , g(x)
= x2
−1
g f ( x ) = x , g( x ) = 1 − x
310
2− x
We29 For each of the following functions f (x) and g(x), sketch the graph of f (x)g(x).
a f ( x ) = x, g ( x ) = x + 2
7
x 2
b f (x) = x − 2, g(x) = ex
c f (x) = x − 1, g(x) = loge (x)
e f ( x ) = x , g( x ) =
f f (x) = 1 − x2, g(x) = ex
3
−
x+2
h f (x) = x2, g(x) = loge (x)
We30 For each of the following pairs of functions:
i show that f (g(x)) is defined ii find f (g(x)) and state its domain iii sketch the graph of f (g(x)), using a CAS calculator for assistance.
maths Quest 12 mathematical methods CaS for the Tinspire
a f (x) = cos (x) and g(x) = loge (x) x2 c f ( x ) = 2 sin ( x ) and g( x ) = 4 e f (x) = x2 and g(x) = sin (x) g f ( x ) = cos
x 4
b f ( x ) = sin (2 x ) and g( x ) = x d f ( x ) = x + 2 and g( x ) = 2 cos ( x ) f f (x) = 2x and g(x) = cos (2x)
1
and g( x ) = 2 x 2
h f ( x ) = 2 sin ( x ) + 1 and g( x ) = x − 3
8 WE31 For each of the following pairs of functions f (x) and g(x): i state whether f (g(x)) and g(f (x)) are defined ii for the composite functions that are defined, find f (g(x)) and g(f (x)), stating the domain and range of each one iii on separate axes, sketch the graphs of f (g(x)) and g(f (x)) that are defined.
6I
a f (x) = x − 2, g(x) = ex
b f (x) =  x g(x) = x2 − 1
c f (x) = 1 − x2, g(x) = ex
d f ( x ) = x , g( x ) = sin ( x )
e f (x) = ex, g(x) = cos (x)
f f (x) = loge x, g(x) = sin (x)
Trigonometric functions with an increasing trend Consider a trigonometric function where there is an increasing trend, for example, economic growth cycles, tidal heights due to global warming or increasing seasonal populations. These situations can be modelled by a function of the form y = ax + b + m sin (nx) where ax + b represents the increasing trend line and m sin (nx) represents the seasonal variation.
Worked Example 32
Consider a remote island where global warming has caused the temperature to increase by 0.1 degree each month. The mean daily temperature is modelled by the function π T ( m) = 16 + 0.1 m + 6 cos m , where T is the temperature in degrees Celsius and m is the number 6 of months after January 2008. a Sketch a graph of the function for a five year period from January 2008, using a CAS calculator for assistance. b Find the mean daily temperature for March 2009 c When will the mean daily temperature first reach 23 degrees? Think
Write
a On a Graphs page, complete the function entry
line as: π f 1( x ) = 16 + 0.1x + 6 cos x 6 Then press ENTER ·. Select an appropriate window: XMin: 0 XMax: 60 YMin: 10 YMax: 30
Chapter 6 Circular (trigonometric) functions
311
b
March 2009 occurs when m = 14. On a Calculator page, complete the entry line as: f1(14) Then press ENTER ·.
The mean daily temperature in March 2009 would be 20.4 degrees Celsius. c
1
2
π To solve 23 = 16 + 0.1m + 6 cos m , 6 for m, return to the Graphs page. Complete the function entry line as: f 2(x) = 23 Then press ENTER ·.
Press: • MENU b • 7:Points & Lines 7 • 3:Intersection Point(s) 3 Select each function and then press ENTER ·. The points of intersection will appear.
m = 11.5631 Hence, the first time the temperature reaches 23 degrees Celsius will be during the 12th month after January 2008. That is, during January 2009. REMEMBER
Trigonometric functions with an increasing trend can be modelled by a function of the form y = ax + b + m sin (nx) where ax + b represents the increasing trend line and m sin (nx) represents the seasonal variation.
312
Maths Quest 12 Mathematical Methods CAS for the TINspire
Exercise
6I
Trigonometric functions with an increasing trend 1 WE32 A fisherman finds himself stranded on an island, where the mean daily temperature is increasing as a result of global warming. He finds that the temperature can be modelled by the π function T (m) = 12 + 0.2m + 5 cos m , where T is the temperature in degrees Celsius and 6 m is the number of months after January 2008. a Sketch a graph of the function for a 2year period beginning with January 2008. b Find the mean daily temperature for December 2009. c When will the temperature first reach 18 degrees? 2 In a region of country Victoria, a study shows that increased wheat production causes the π mouse population to increase according to the function M = 15 000 + 100t − 4000 cos t , 6 where M is the number of mice and t is the number of months after July 2008. a How many mice are being added to the average population per month? b Draw a graph of this situation for a 5year period beginning with July 2008. c How many mice would you expect to be in the region in December 2008? d When would the mice population first reach 20 000? 3 The value of a particular stock on the market follows a trigonometric model, and inflation causes the stock’s value to have an overall upward trend. The value of the stock can be π represented by the equation V (t ) = 20 + 0.02t + 5 sin t , where V is the value of the stock 6 in dollars and t is the number of months after January 2006. a What is the inflation rate per month? b What was the initial value of the stock? c What will be the stock’s value after 6 months? d When will its value first reach $25.50?
Chapter 6 Circular (trigonometric) functions
313
Summary Revision of radians and the unit circle
• 1c = the size of the angle formed where the length of an arc is equal to the radius of the circle. • πc = 180° • Angles are in radians unless a degree symbol is shown. 180 • To change radians to degrees, multiply by . π π • To change degrees to radians, multiply by . 180 • Identities 1. sin2 (θ ) + cos2 (θ ) = 1 sin (θ ) 2. tan (θ ) = cos (θ )
Symmetry and exact values
• Exact values can be determined by using the equilateral triangle and the right isosceles triangle shown at right.
θ
π
3 30°
2
45°
1
45°
60° 1 π
π
π
2
1 3π 2
0° (0)
30° 6
45° 4
60° 3
90° 2
180° (π)
sin (θ )
0
1 2
2 2
3 2
1
0
−1
0
cos (θ )
1
3 2
2 2
1 2
0
−1
0
1
tan (θ )
0
3 3
1
Undef.
0
Undef.
0
3
360° (2π)
270°
• The unit circle is symmetrical so that the magnitude of sine, cosine and tangent are the same in each quadrant but the sign varies. All functions (sine, cosine and tangent) are positive in the 1st quadrant, sine is positive in the 2nd quadrant, tangent is positive in the 3rd quadrant and cosine is positive in the 4th quadrant. sin (π − θ ) = sin (θ )
sin (π + θ ) = −sin (θ )
sin (2π − θ ) = −sin (θ )
cos (π − θ ) = −cos (θ )
cos (π + θ ) = −cos (θ )
cos (2π − θ ) = cos (θ )
tan (π + θ ) = tan (θ )
tan (2π − θ ) = −tan (θ )
tan (π − θ ) =
−tan
(θ )
• Negative angles sin (−θ ) = −sin (θ ), cos (−θ ) = cos (θ ), tan (−θ ) = −tan (θ ) π • Complementary angles add to 90o or radians. 2 • The sine of an angle is equal to the cosine of its complement. • The complement of the tangent of an angle is the cotangent.
( − θ ) = cos (θ ) cos ( − θ ) = sin (θ ) tan ( − θ ) = cot (θ ) sin
π 2
π 2
π 2
314
( + θ ) = cos (θ ) cos ( + θ ) = sin (θ ) tan ( + θ ) = cot (θ ) sin
π 2
π 2
−
π 2
−
( cos ( tan ( sin
Maths Quest 12 Mathematical Methods CAS for the TINspire
3π 2 3π 2 3π 2
) − θ ) = sin (θ ) − θ ) = cot (θ )
− θ = − cos (θ ) −
sin
(
3π 2
( tan ( cos
)
+ θ = − cos (θ )
3π 2
3π 2
)
+ θ = sin (θ )
)
+ θ = − cot (θ )
Trigonometric equations
• Given a general equation such as sin x = a, there can be an infinite number of solutions. The domain is usually restricted and it is important to find all values for x within this domain. • If the domain is given in radians, then the solution(s) to x should be in radians. If the domain is given in degrees, then the solution(s) to x should be in degrees. • Adjust the domain to match what has been done to the angle in the question. • Sine is positive in the 1st and 2nd quadrants, cosine is positive in the 1st and 4th quadrants and tangent is positive in the 1st and 3rd quadrants. • If sin (x) = a then the general solution is x = sin 1 (a) + 2np, n ∈ Z. • If cos (x) = a then the general solution is x = cos 1 (a) + 2np, n ∈ Z. 1 • If tan (x) = a then the general solution is x = tan (a) + np, n ∈ Z. • Find all the solutions within the specified domain by substituting integer values for n into the general solution. • If the equation is of the form sin (ax) = k cos (ax), divide both sides by cos (ax) to change the equation to tan (ax) = k. Trigonometric graphs
Sine and cosine graphs • Graphs of the form y = a sin [n(x − b)] + c and y = a cos [n(x − b)] + c are transformations of y = sin (x) and y = cos (x). • The amplitude a is a dilation from the xaxis. If a is negative, the amplitude is still positive but the graph is a reflection in the xaxis. • The vertical translation c is a translation parallel to the yaxis. If c is positive, the graph is translated c units up, and if c is negative, the graph is translated c units down. • c represents the median value of the function. • The range is [c −  a , c +  a ]. 2π • The period is . n 1 • The factor n is the horizontal dilation where the graph has been dilated by a factor of from the yaxis. n • The value b is the horizontal translation or a translation parallel to the xaxis. If b is positive, the graph is translated b units to the right and if b is negative, the graph is translated b units to the left. Tangent graphs − − y x = 3—2π x = —2π x = –2π x = 3—2π The graph of y = tan (x) has the following properties. 3 • It has no amplitude. 2 • The period is π. − − 1 • There are xintercepts at x = . . ., 2π, π, 0, π, 2π, . . . − 3π − π π 3π , , , ,... • There are vertical asymptotes at x = . . ., 0 –π π 3—π −2π − 3—2π −π − –2π 2π x 2 2 2 2 2 2 −1 • The range is R. −2 The graph of y = tan (nx) has the following properties. • It has no amplitude. −3 π • The period is . n kπ • There are xintercepts at x = ± where k = 0, 1, 2, . . . n (2k + 1)π • There are vertical asymptotes at x = ± where k = 0, 1, 2, . . . 2n • The range is R. • Compared to the basic graph of y = tan (x), the graph of y = a tan [n(x − b)] + c is: – dilated by the factor of a from the xaxis 1 π – dilated by the factor of from the yaxis (and hence has period of ) n n
Chapter 6 Circular (trigonometric) functions
315
– reflected in the xaxis if a < 0 – translated b units horizontally (to the right if b > 0 and to the left if b < 0) – translated c units vertically (up if c > 0 and down if c < 0). Further graphs
• For the graph of the sum/difference function, dom(f (x) ± g(x)) = dom f (x) ∩ dom g(x). The graph of the sum/ difference function can be obtained by using the addition of ordinates method. • The graph of the modulus function y =  f (x) can be obtained by sketching the graph of y = f (x) and then reflecting all of the sections of the graph that are below the xaxis in the xaxis. • For the product function, dom(f (x)g(x)) = dom f (x) ∩ dom g(x). Some features of the graph of the product function are as follows: – the xintercepts of f (x)g(x) occur where either f (x) or g(x) have their xintercepts – f (x)g(x) is above the xaxis where f (x) and g(x) are either both positive or both negative – f (x)g(x) is below the xaxis where one of the functions f (x) or g(x) is positive and the other is negative. • For the composite function f (g(x)) to be defined, the range of g must be a subset of the domain of f. Furthermore, if f (g(x)) is defined, the domain of f (g(x)) equals the domain of g(x). Trigonometric functions with an increasing trend
• These situations can be modelled by a function of the form y = ax + b + m sin (nx), where ax + b represents the increasing trend line and m sin (nx) is the seasonal variation.
316
Maths Quest 12 Mathematical Methods CAS for the TINspire
chapter review Short answer
1 Convert the following into radians. a 60° b 30° d 90° e 360° g 150° h 300° j 120° k 210°
c f i l
45° 270° 225° 315°
2 Find the value of: 2π a cos 3 5π d sin 4 g cos (135°) j tan (225°)
π b sin 4
7π c tan 6
e tan (2π)
f sin (120°)
h tan (30°)
i cos (315°)
9 The depth, d metres, of water in a shallow bay at πt t hours after 9 am is given by d = 5 + 3 sin , 6 for 0 ≤ t ≤ 24 hours. a Sketch the graph of d for 0 ≤ t ≤ 24 hours. b At what times will the depth of water in the bay be 6.5 metres? c Particular water sports cannot run in the bay when the depth of water is less than 6.5 metres. At what times of the day (not night) will the water sports be able to run?
3 If sin (x) = 0.85 and x is in the first quadrant, find: b sin (π + x) a sin (π − x) c sin (2π + x) d sin (4π − x) 4 Find the general solution to the equation 2 cos ( x ) − 2 = 0. 5 Find the general solution for 3 sin (2 x ) = cos(2 x ) and hence all solutions for x between 0 and 2π. π 6 Consider the graph of y = 3 sin x − + 2. 4 a State the translations required to form this graph from y = sin (x). b State the amplitude and period of the transformed trigonometric function. c Sketch the graph of the transformed function over the domain 0 ≤ x ≤ 2π. 7 State the period and sketch the graph of y = tan (2x) + 2. 8 The equation of the following graph is of the form y = a sin (nx) + b. Find the values of a, n and b, and hence find the equation of the function. y 3 2 1
−1 −2 −3 −4
2
3
4
x
10 Sketch the graph of y =  2 sin (2x)  over the domain [0, 2π]. 11 On the same set of axes and over the domain [0, 2π], sketch the graphs of: a y = 2x b y = sin (x) c y = sin (x) + 2x. 12 On the same set of axes and over the domain [0, 2π], sketch the graphs of: a y = x b y = sin (x) c y = x sin (x). 13 For the pair of functions f (x) = sin (x) and g(x) = x2 + 2: a show that f (g(x)) is defined b find f (g(x)) c state the domain and range of f (g(x)). 14 For the function π f : [− π , π ] → R, f ( x ) = 5 cos (2( x + )): 3 a Write down the amplitude and period of the function.
Chapter 6 Circular (trigonometric) functions
317
b Sketch the graph of the function f. Label intercepts with their coordinates. Label end points of the graph with their coordinates. Exam tip Make sure that you use the period and
amplitude found in part a to help sketch the graph. The graph should be a smooth symmetrical curve with all intercepts (x and y) clearly labelled with coordinates. [Assessment report 1 2006]
[© VCAA 2006]
Multiple choice
1 What is 320° expressed in radians? A 2π B 5π C 8π D 4π 9 18 9 9 c 13π 2 What is expressed in degrees? 6 A 390° B 420° C 150° D 120°
E 16π 9
E 330°
5π 3 The expression sin equals: 3 A
1 2
B
−1 2
C
3 2
D
− 3 2
E 1
4 A trigonometric function is given by f : R → R, f (x) = 3 cos (2x + π) − 1 The amplitude, period, and range of f are respectively: C 3, π, [−4, 2] B 2, π, R A 3, π , [− 4, 2] 2 π − E 3, π D 2, , [ 1, 3] 2
8 The domain and range of the function π f : [0, π] → R, f ( x ) = 4 sin 3 x − + 2 is: 3
B [0, π], [−2, 6] C R, R A [0, π], R −π D [ , 2π ] [− 4, 4] E [0, π], [2, 4] 3 9 Using addition of ordinates for the graph of the difference function y = 2 cos (x) − 3 sin (x), the value π of y at x = is: 3 −1 1+ 3 3 2+3 3 A B C 3 2 2 D
1− 3 3 2
1
6 The function f : [−π, π] → R, f (x) = sin (2x) + cos (2x) has the following xintercepts: − π − 5π 3π 7π − π 3π − π 3π , , , A , B , C 4 4 4 4 2 4 8 8 − π − 5π 3π 7π −π π , , , , D E 8 8 8 8 4 4 7 The solution of the equation 4 sin ( x ) = − 2 3 3π between π and is: 2 A π B π C 4π 6 3 3 7 π 2 π D E 3 3
318
2−3 3 2
x 10 The period of the graph of y = tan is: 3 π B π C 2π D 3π A 3
E 6π
11 The graph of the function f : [−1.5, 1.5] → R, f (x) = a + b cos (π nx) is shown below. The values of a, b, and n are respectively: y
max = 3.5
3 2 1 −1.5
−1.0
−0.5
0 −1
0.5
1.0 1.5 min = −1.5
x
−2
5 When f (x) = 0 the following function f : [0, 2π] → R, f (x) = 2 sin 2 ( x + π ) + 1 has: A 0 solutions B 1 solution C 2 solutions D 3 solutions E 4 solutions
E
A 1, D 1,
−5 2 −5 3
2
2
5
,3
B −1, 3 , 2
,3
E
−1,
−5 3
2
C 1, 5, 3
,2
12 If 0 ≤ c ≤ π, the xintercepts of the function f : [0, 2π] → R, f (x) = b sin (x − c) are: B π, 2π C c, π − c A π − c, 2π − c D π + c, 2π − c E c, π + c 13 The position of a particle from a fixed point O is given by the equation x = 2 − 2 sin (π t). If 0 ≤ t ≤ 2, the particle is at the point O when t equals: E 6π A 0 B 2 C 0.5 E 6π 14 The smallest positive value of x for which tan (2x) = 1 is: π π π A 0 B C D 8 4 2
Maths Quest 12 Mathematical Methods CAS for the TINspire
E p
[© VCAA 2006]
Extended response
1 At the South Pole in midsummer on the planet Marus, the red sun of its solar system does not set. It dips towards the horizon until its lower rim just touches it, then rises until its lowest point is at an angle of D(t)° to the horizontal before sinking again. It continues in this pattern. The angle above the horizontal can be modelled by the following relation: D(t) = a − b sin n(t + c) where t is the time in hours after midnight and a, b, c and n are positive constants. The graph of D(t) for 24 hours is shown on the axes in the figure below. D(t) 10 8 6 4 2 0
2 4 6 8 10 12 14 16 18 20 22 24
t
a State the values of a, b, c and n and hence write the rule for D(t). b What would be the angle above the horizon at 6.00 am and at 9.00 pm? Give an exact answer where possible, otherwise give your answer correct to 2 decimal places. c Use your graph to find at what times the angle to the horizontal is 8°. When does the rim of the sun reach this angle again? d By using an appropriate equation, check your answer and account for any difference in your two solutions. e If a spot on the surface of the sun is 5° above the horizon at midnight, what would be the relation G(t) that models its path? 2 Nathan and Rachel are competing in the National Ballroom Dancing Championships. The judges are evenly spaced around the circular dance floor, standing just outside the edge. As Nathan and Rachel waltz around the circular floor, their distance (in metres) from judge Maya can be described by the function π d = 10.5 − 9 cos t , where t is time (in seconds) from the beginning of the dance. 30 a How far is the couple from judge Maya when they start dancing? b What is the couple’s maximum distance from the judge? c Assuming that, while dancing, Rachel and Nathan trace a perfect circle, what is its diameter? d How long does it take for the couple to complete one full circle around the dance floor? e What is the couple’s average speed (in m/s)? Give your answer i in exact form and ii correct to 2 decimal places. π f If the duration of the waltz is 2.5 minutes, draw the graph of d = 10.5 − 9 cos t over the domain, 30 showing the full length of this dance. g Judge Joseph is positioned further down the dance floor, so that Nathan and Rachel are closest to him 6 seconds after the waltz begins. Write the equation describing the couple’s distance from judge Joseph at any time, t, from the beginning of the dance. h How far is the couple from judge Joseph when they finish the waltz? 3 Tasmania Jones is attempting to recover the lost Zambeji diamond. The diamond is buried at a point 4 km into Death Gorge, which is infested with savage insects. In order to recover the diamond, Tasmania will need to run into the gorge, dig up the diamond and return the same way that he came. The concentration of insects in the gorge is a continuous function of time. The concentration, C, of insects per square metre is given by π (t − 8) + 2)2 − 1000, 8 ≤ t ≤ 16 1000 (cos C (t ) = 2 m 0 ≤ t < 8 or 16 < t ≤ 24 where t is the number of hours after midnight and m is a real constant.
Chapter 6 Circular (trigonometric) functions
319
a What is the value of m? b Sketch the graph of C for 0 ≤ t ≤ 24. c What is the minimum concentration of insects and at what value(s) of t does that occur? The insects infecting the gorge are known to be deadly if their concentration is more than 1250 insects per square metre. d At what time after midnight does the concentration of insects first stop being deadly? e During a 24hour period, what is the total length of time for which the concentration of insects is less than 1250 insects per square metre?
eBook plus Digital doc
Test Yourself
exam Tip
• Be aware of the word ‘continuous’ when working out what value(s) to substitute to find m. • When drawing the graph, clearly show minimum turning points and be aware of the domain stated in the question.
Chapter 6
[Assessment report 2 2007]
[© Vcaa 2007]
320
maths Quest 12 mathematical methods CaS for the Tinspire
eBook plus
aCTiviTieS
chapter opener Digital doc
• 10 Quick Questions: Warm up with ten quick questions on circular functions. (page 258) 6a
Revision of radians and the unit circle
Digital docs
• Spreadsheet 142: Investigate the unit circle. (page 263) • SkillSHEET 6.1: Practise changing degrees to radians. (page 263) • SkillSHEET 6.2 : Practise working with tangent ratios. (page 263) 6b
Symmetry and exact values
Tutorials
• We 3 int0546: Watch a worked example on exact values using degrees and radians. (page 265) • We 5 int0547: Watch a worked example on determining trigonometric ratios. (page 266) • We 8 int0548: Watch a worked example on complementary angle formulas. (page 270) Digital docs
• SkillSHEET 6.3: Practise rationalising the denominator. (page 271) • Spreadsheet 142: Investigate the unit circle. (page 272) • SkillSHEET 6.4: Practise problem solving using trigonometry. (page 273) 6c
Trigonometric equations
Tutorial
• We 13 int0549: Watch a worked example on determining the general solution to a trigonometric equation. (page 277) Digital docs
• Spreadsheet 135: Practise solving trigonometric equations. (page 281) • WorkSHEET 6.1: Use exact values, solve trigonometric equations and application questions. (page 282) 6D
Trigonometric graphs
Digital docs
• SkillSHEET 6.5: Practise identifying the period and amplitude of sine and cosine graphs. (page 288) • Spreadsheet 124: Investigate sine graphs. (page 288) • Spreadsheet 012: Investigate cosine graphs. (page 289) 6E Tutorial
Graphs of the tangent function
6F
Finding equations of trigonometric graphs
Interactivity int0251
• Finding equations of trigonometric graphs: Use the interactivity to consolidate your understanding of trigonometric graphs. (page 295) Digital docs
• Spreadsheet 124: Investigate sine graphs. (page 298) • Spreadsheet 012: Investigate cosine graphs. (page 298) • WorkSHEET 6.2: Sketch trigonometric graphs over given domains, including composite and absolute value graphs, and determine the equations of trigonometric graphs. (page 299) 6G
Trigonometric modelling
eLesson eles0092
• Trigonometric modelling: Learn how trigonometry can be used to model sinusoidal patterns. (page 299) Digital doc
• Spreadsheet 041: Investigate graphs of functions. (page 301) 6H
Further graphs
Tutorial
• We 30 int0551: Watch how to sketch a composite function using a CAS calculator. (page 308) Digital doc
• SkillSHEET 6.6: Practise using addition of ordinates to sketch graphs. (page 310) chapter review Digital doc
• Test Yourself: Take the endofchapter test to test your progress. (page 320) To access eBookPLUS activities, log on to www.jacplus.com.au
• We 21 int0550: Watch a worked example on calculating the period of trigonometric functions. (page 293)
Chapter 6
Circular (trigonometric) functions
321
EXAM PRACTICE 2 Short answer
30 minutes
1 Solve the equation log2 (2x  3)  log2 (3) = 1 for x. 1 mark 2 Solve the equation 32x + 3 33 = 9 for x.
1 mark
[2,0]
3 The graph of the function f : → R where f (x) = e x is reflected in the yaxis, translated 2 units to the left then translated 3 units up. (Note the change to the domain.) a Specify the rule of the transformed graph. b Determine its exact range. 2 marks 4 The diagram below shows the graph of y = sin (x) and a second graph formed by transforming sin (x) by carrying out dilation(s) followed by translation(s). y 1
0 π/6
2π
y = sin (x)
x
−1
π
a Write down the transformations (type, direction and quantity) necessary to create the second graph. b Write down the equation of the second graph. 4 marks
5 Find the exact solutions of the equation sin (3x)  cos (3x) = 0 for 0 ≤ x ≤ π
2 marks
6 Specify the range of the function f :[
−π
π π ) → R, f ( x ) = 2 tan(2 x − ) + 1. 4 8 4 ,
2 marks
7 Solve the equation
e2x  e x
= 2 exactly for x. 2 marks
−1,
8 For f : [ 3] → R, where f (x) = x sin (x): a sketch f b determine the minimum value of f (x) c estimate the maximum value of f (x) to 1 decimal place.
322
Chapters 1 TO 6
Multiple choice
12 minutes
Each question is worth 1 mark. 1 If y = 2abx  1 + 5, then x is equal to: y−5 y log a ( − 5) + 1 a 2 +1 2 B A b b y−5 +1 y−5 2 +b C D 2b b y−5 log a ( ) +1 2 E b 2 The solution set of the equation e2x  3e x = 4 over R is: A {−1, 4} B {4} C {loge (4)} D {loge (1), loge (4)} E {loge (1), loge (4)} 3 The graph of the function f :[2, 5] → R, f (x) = loge (x) is reflected in the yaxis, translated 2 units to the left then translated 3 units up. The domain of the new graph is: A [−7, −1] B [−4, −1] −7, −4] C [ D [loge (7), loge ( 4)] E not defined 4 The function f : [1, ∞) → R, where f (x) = x2 + 1 has as its inverse function: A f 1: R → R, where f −1( x ) = ± x − 1 B f 1: [2, ∞] → R, where f −1( x ) = x − 1 1 C f 1: [1, ∞] → R, where f −1 ( x ) = 2 x +1 D f –1: R+ ∪ {0} → R, where f −1( x ) = x − 1 1 1 E f −1: [ , ∞) → R, where f −1( x ) = 2 2 x +1 5 The equation a sin (3θ )  b = 0 has θ = 2 as one solution. Which of the following could also be a solution? A 3π + 6 B π + 2 2π C 2 D +2 3 2π E +6 3
4 marks
Maths Quest 12 Mathematical Methods CAS for the TINspire
7 The graph of y = 3 cos (2x) − 1 is shown below. If a and b are two adjacent xintercepts, b − a is equal to:
6 The graph shown could be the graph of: y 3a
0 −a
y
a
θ
T
b x
0
2π cos θ +a T
T θ +a a 2a cos 2π
b
2π c a (cos θ + 1) T
D 2a sin (
− 2a
2π π (θ − )) + a T 2
π T E a (sin (θ − ) + 1) 2π 2
π 2 c 3a E π − 2a
b π
a
D 2π − a
exTended reSponSe
20 minutes
Bird population
An extensive study has been made of the population of silverbanded lorikeets in an area proposed for a wind farm. The variation in estimated numbers over a number of months is recorded in the graph below. It is suggested that the relationship could be modelled by a sine function of the form P(t) = a sin (b(t + c)) + d, where P is the bird population and t is the time in months since estimates commenced. y 500 400 300 200 100 0
a b c d e f g
1
2
3
4 5 6 Months
7
8
9 x
What is the amplitude of the sine model (to the nearest 10 birds)? 1 mark What is the period (to the nearest month)? 1 mark What is the mean population (to the nearest 10 birds)? 1 mark After how many months is the population a minimum? 1 mark Determine the values of a, b, c and d in the model P(t) = a sin b(t + c) + d. 4 marks What was the initial bird population when the observations commenced? 1 mark A second researcher conducts observations over a 10year period at the same time each year and records the following estimates for the population. Time (months)
Population
0
500
60
370
120
274
She believes that the population can be modelled by an exponential function of the form Q(t) = Ae−kt, where t is the time in months since the first estimate was calculated. Determine the values of A and k. 2 marks h Explain how both researchers’ results are consistent with the model eBook plus P( x )Q( x ) Population = . 3 marks Digital doc 500 Solutions Exam practice 2
exam practice 2
323
7
7a Review — gradient and rates of change 7b Limits and differentiation from first principles 7c The derivative of xn 7d The chain rule 7e The derivative of e x 7F The derivative of loge (x (x) 7G The derivatives of sin ((x), cos ((x) and tan ((x) 7H The product rule 7I The quotient rule 7J Mixed problems on differentiation
Differentiation AREAS OF STUDY
• Graphical treatment of limits, continuity and differentiation of functions of a single real variable • Deducing the graph of a derivative function from the graph of a function • Derivatives of xn, n ∈ Q, ex, ln (x), sin (x), cos (x), tan (x) • Properties of derivatives (af af (x) ± bg(x))′ = af ′(x) ± bg′(x), a, b ∈ R
f ( x) , and g( x ) f (g(x)) where f and g are polynomial functions, exponential, circular, logarithmic or power functions (or combinations of these functions)
• Derivatives of f (x) ± g(x), f (x) × g(x),
such as x 5 + 1 − x 2 , x sin (2 x ), e cos ( x ) 2 and ln ( x + 4) x
eBook eBoo k plus
7A
Review — gradient and rates of change
Digital doc
10 Quick Questions
Gradient is a measure of how one quantity changes with respect to another quantity; in other words, the rate of change of one quantity with respect to another quantity. The gradient (or slope) is the measure of how the vertical distance (the rise) changes with respect to the horizontal distance (the run). Speed is the measure of how distance travelled changes with respect to the time taken; another way of saying this is that speed is the rate of change of distance with respect to time.
Constant rates When the rate of change of one quantity with respect to another quantity does not change, the rate is constant. If the rate of pay for babysitting is fixed at $15 an hour, this is a constant rate. This means that a babysitter would be paid $15 for 1 hour and $150 for 10 hours. A constant rate can be represented graphically by a straight line.
324
Maths Quest 12 Mathematical Methods CAS for the TINspire
y (1, 15)
(0, 0) x
y
Average rates An average rate of change is the rate of change over a period of time. A car travelling at 60 km/h is probably not actually registering that speed at every moment of the distance travelled; more likely, its speed varies above and below 60 km/h. The average rate of change between two points can be represented by the gradient of a straight line joining the two points.
(0, 0)
Instantaneous rates
y
(1, 0) x (1_2, −33_4)
(−2, 3)
If the police want to find how fast a car is travelling, they measure the speed at a particular instant. This is known as instantaneous (–1, 0) (1, 0) rate of change. Graphically, this is found by drawing a tangent x 0 to the curve at a particular point and finding the gradient of the (−3, 0) tangent. If a section of the graph is increasing, the gradient of the tangent is positive, and if a section of the graph is decreasing, the (0.15, −3.08) gradient of the tangent is negative. For example, for the curve y = (x − 1)(x + 1)(x + 3), the gradient is positive when x < −2 and when x > 0.15; the gradient is negative when −2 < x < 0.15; and the gradient is zero when x = −2 and x = 0.15.
Graphing the gradient function from the graph of a function Differentiation is the process of calculating the gradient function from a given function. It can be used to find the gradient of a curve at a particular point and to find the maximum and minimum values of a function. y f(x) Derivatives can be found at a given point if: There is a break in the 1. the function is continuous at that point graph at x = −1(discontinuous), 2. the function is smooth at that point. so gradient does not exist at x = −1. That is, gradient A function is continuous if there are no exists for R\{−1}. ‘breaks’, ‘jumps’ or ‘asymptotes’ on its graph. x 0 −1 The gradient does not exist where the function is not continuous. An example is shown at right. A function is smooth if there are no ‘sharp’ points on its graph. The gradient does not exist where the function is y not smooth. An example is shown at right. f (x) is not smooth at x = 2 so gradient does not exist at x = 2. That is, gradient exists for R\{2}.
0
x
2 f(x)
The gradient of a function exists wherever the function is smooth and continuous. That is, the gradient of a function exists at a point providing only one tangent can be drawn at that point. An example is shown at right. A function is not differentiable where there is a gap, hole or asymptote, a sharp corner or an end point.
y f(x)
2 −2
0
The gradient of f(x) exists for x ∈ R as f(x) is smooth and continuous. x
Chapter 7 Differentiation
325
Gradient function of straight lines In general, for straight lines in the form f (x) = mx + c (or y = mx + c), the rule for the gradient is f ′(x) = m, that is, a horizontal straight line through y = m. WORKED EXAMPLE 1
For the straight line function shown, sketch the graph of its gradient function.
y
f(x)
0
x
1
−2 THINK
WRITE/DRAW
1
Find the gradient of f (x) using m =
2
Sketch the graph of y = gradient.
1 2
rise . run
m=
to represent the
1 2 y 1– 2
f'(x) x
0
Note: The domain of f ′(x) is R as f (x) is smooth and continuous.
Gradient function of quadratic functions The gradient function of a polynomial function is also a polynomial function but the degree is reduced by 1. That is, the gradient function of f (x) = ax2 + bx + c is of the form y = mx + c. So the gradient of a quadratic function is a linear function.
WORKED EXAMPLE 2
Sketch the graph of the gradient function for the quadratic function shown and state its domain. y
0 THINK
326
f(x)
1
x
WRITE/DRAW
1
Find when f ′(x) = 0.
f ′(x) = 0 when x = 1.
2
Find when f ′(x) > 0.
f ′(x) > 0 when x > 1.
3
Find when f ′(x) < 0.
f ′(x) < 0 when x < 1.
Maths Quest 12 Mathematical Methods CAS for the TINspire
4
Sketch f ′(x).
y
f'(x)
0
5
Find where f (x) is smooth and continuous and hence find the domain of f ′(x).
x
1
The domain is R.
Gradient function of cubic functions The gradient function of a cubic is a quadratic function. WORKED EXAMPLE 3
eBook plus
For the cubic function shown, sketch the gradient function and state its domain.
y
Tutorial
f(x)
int0552 Worked example 3
−3 THINK
x
0 1
WRITE/DRAW
1
Find when f ′(x) = 0.
f ′(x) = 0 when x = −3 and x = 1.
2
Find when f ′(x) > 0.
f ′(x) > 0 when x < −3 and x > 1.
3
Find when f ′(x) < 0.
f ′(x) < 0 when −3 < x < 1.
4
Sketch the graph of the gradient function.
y
−3
5
Find the domain by determining where f (x) is smooth and continuous.
0
f'(x)
x
1
The domain is R.
WORKED EXAMPLE 4
For the function f (x) shown, state the domain of the gradient function f ′(x).
y
−1 0 2
x f(x)
Chapter 7
Differentiation
327
THINK
WRITE
The domain = R \{−1, 2}.
The function is smooth and continuous everywhere except at x = −1 (discontinuous) and x = 2 (not smooth).
REMEMBER
1. 2. 3. 4. 5.
6. 7. 8. 9. 10.
EXERCISE
7A eBook plus Digital doc
A function is smooth if there are no sharp points on its graph. A function is continuous if the graph can be drawn without lifting pen from paper. The gradient of a function exists where the function is smooth and continuous. The graph of the gradient function is a graph of the gradients of all the points of the original function. The gradient function of a polynomial function is also a polynomial function but the degree is reduced by 1. For example, the gradient function of f (x) = ax2 + bx + c is of the form y = mx + c. Wherever the gradient of a function is zero, its gradient function will have an xintercept. Wherever the gradient of a function is positive (sloping /), the gradient function graph is above the xaxis. Wherever the gradient of a function is negative (sloping \), the gradient function graph is below the xaxis. The gradient of a horizontal line is 0. The gradient of a vertical line is undefined.
Review — gradient and rates of change 1 WE 1 For each straight line function shown below, sketch the graph of its gradient function. b c a y y y f(x)
SkillSHEET 7.1
3
1
Gradient — positive, negative and zero
x
0
−1
f(x) −2
0
x
1 f(x)
eBook plus Digital doc
SkillSHEET 7.2
d
y 0
Gradient function
e
f(x) 2
y f(x)
x 3 0
−5
328
Maths Quest 12 Mathematical Methods CAS for the TINspire
x
0 −1
x
2 MC a The gradient of the line in the graph at right is: a 1
b 2
d
c 0
1 2
y f(x)
−1
e
2
x
0 1
b The graph of the gradient function in the graph above is represented by which of the diagrams below? b c a y y y f'(x)
2
f'(x)
d
0
x
0 1
e
y 1– 2
x
2
x
0
y x
0
f'(x)
f'(x)
2
1
f'(x) x
0
3 WE 2 a
−2
Sketch the graph of the gradient function for each quadratic function shown below. b c y y y g(x)
g(x)
x
0
x
0
d
g(x)
x
0
−2
y
e
g(x)
y −2
0
3
x
0 −1
x
g(x)
4 MC a The gradient of the function shown in the graph at right is: a always increasing b always decreasing c decreasing then increasing d increasing then decreasing e constant
y
0
x f(x)
Chapter 7
Differentiation
329
b The gradient function for the graph in question 4a is shown by which of the graphs shown below? a
y
b
y
x
0
c
f'(x)
y
x
0
f'(x)
x
0
f'(x)
d
e
y
y
f'(x)
x
0
x
0 f'(x)
5 WE 3 a
For each cubic function f (x) graphed below, sketch the gradient function. y b y c y f(x)
−3
0
2
x
0
1
4
x
0
f(x)
x
5
f (x)
d
e
y Gradient = 0
f(x)
0
x
f
y
y
Gradient = 0
f(x) 0
x −3
0
x f(x)
g
f(x)
y Gradient = 0 2 0
x
6 MC a The figure at right has a positive gradient where: A −1 < x < 2 B x < −1 only C x > 2 only D x < −1 and x > 2 E x > 0 b The figure above has a negative gradient where: A x > −1 B x < 2 C −1 < x < 2 D x < −1 and x > 2 E x < 0 330
Maths Quest 12 Mathematical Methods CAS for the TINspire
y f(x)
−1
0
2
x
c The graph of the gradient function for the figure on the previous page is: b c a y y y f '(x)
f '(x) −1 0
2
x
d
2x
−1 0
f '(x)
e
y
y
−1 0
2
x
f '(x)
f '(x) −1 0
2
x
−1 0
2
x
7 For the functions graphed below, state the domain (where applicable) where the gradient: i is equal to zero ii is positive iii is negative iv does not exist. a
b
y
f(x)
1
d
x
y
0
e
g(x)
2
x
−1 0
f(x)
f
g(x) y 0
0
x
4
y
1 0
i
y
3
y
−1 0
2
f(x)
x 0
j
x f(x)
g(x)
Gradient = 0 x
2
x
y
−2 0
h
f(x)
g(x)
x Gradient = 0
−4
g
y 5
6
0
−1
c
y
x
y 4 2 −2 0
4
x f(x)
8 Sketch the gradient function for each function in question 7.
Chapter 7 Differentiation
331
9 WE 4 For each function f (x) graphed below, state the domain of the gradient function f ′(x). (Do not sketch the graph of f ′(x).) b c a y y y f(x)
f(x) 2 0
0
8 x
2
2
x
−4
−3
d
e
y
x
0 f(x)
f
y
y
f(x)
f(x) −1
0
x
2
0
x
0
x
3
f(x)
g
h
y
i
y
y
f(x)
f(x)
f(x) −2
j
x
0
0
2
x
0
x
5
y
f(x) −5
7B
0
x
4
Limits and differentiation from ﬁrst principles The limit of a function is the value that the function approaches as x approaches a given value. If the function is continuous at the point in question then the limit exists and can be found by direct substitution. For example, the limit of f (x) = 3x + 1 as x approaches 1 is denoted as lim (3 x + 1)
eBook plus eLesson
eles0093 Limits and differentiation from first principles
y
f(x)
4
Continuous
x →1
In this example f (x) is a continuous function at x = 1. (In fact, it is continuous for all values of x.) Therefore, the limit is found by direct substitution, that is, lim (3 x + 1) = 3(1) + 1 = 4
x →1
332
Maths Quest 12 Mathematical Methods CAS for the TINspire
1 0
1
x
If the function is discontinuous at the point in question, then the limit exists if the function is approaching the same value from the left as it is from the right. For example: For example: y
Discontinuous 3
y
Discontinuous at x = −2
f(x)
f(x) 5 3
–2 0
2
x
0
x
The limit exists at x = 2, as the function is approaching 3 from the left and from the right. Therefore: lim f ( x ) = 3 x→2
The limit does not exist at x = −2, as the function is approaching 3 from the left and 5 from the right. That is, if the lefthand limit is not equal to the righthand limit, then the limit does not exist.
WORKED EXAMPLE 5
Evaluate the following limits. a lim ( x 2 − 3 x ) x→ 5
eBook plus
x2 + 5 x + 6 x→ 0 x+3
Tutorial
b lim
int0053
THINK a
b
WRITE
1
Decide whether f (x) is continuous at x = 5. If so, substitute x = 5 into f (x).
2
Evaluate.
1 2
Worked example 5
a
lim ( x 2 − 3 x ) = 52 − 3(5)
x→5
= 10
Decide whether f (x) is continuous at x = 0. If so, substitute x = 0 into f (x).
b
lim
+ 5 x + 6 0 2 + 5(0) + 6 = x+3 0+3 6 = 3 =2
x2
x→0
Evaluate.
If direct substitution makes the denominator zero, the limit of a rational expression can be evaluated by first simplifying the expressions and then using direct substitution. WORKED EXAMPLE 6
Evaluate
lim
x→ − 3
x2 + 5 x + 6 . x+3
THINK
WRITE
x 2 + 5x + 6 ( x + 3) ( x + 2) = lim 3 x → 3 x+3 x+3
1
If the limit can’t be found because the denominator becomes zero, factorise the numerator.
2
Simplify by cancelling.
− = lim ( x + 2), x ≠ 3
3
Substitute x = −3.
= −3 + 2
4
Evaluate.
= −1
lim
x→
−
−
x → −3
Chapter 7
Differentiation
333
Differentiation using ﬁrst principles The gradient function is the rule for the instantaneous rate of change of a given function at any point. It also gives the gradient of the tangent drawn at any point of the given function. Consider the chord (straight line) PQ to the curve below. f(x) Q
y f(x + h)
f(x + h) − f(x) P
f(x)
h
0
x+h
x
x
rise run f ( x + h) − f ( x ) = h As Q moves along the curve towards P, the value of h gets smaller and smaller. Or as Q gets as close as possible to P, h → 0, and PQ becomes a tangent at P. The gradient of the curve at a point P is the gradient of the tangent at that point. f ( x + h) − f ( x ) That is, gradient at point P is f ′( x ) = lim , h ≠ 0. h→0 h Finding the gradient this way is known as differentiation from first principles. Differentiating f (x) gives f ′(x) or f ′(x) is the derivative of f (x). dy dy Differentiating y gives , or is the derivative of y with respect to x. dx dx The gradient of PQ =
WORKED EXAMPLE 7
Find the gradient of the chord PQ drawn to the curve f (x) = x2 + 2 in the diagram. y Q [1 + h, f(1 + h)]
P [1, f(1)] x
0 THINK
Find the gradient and simplify. rise Gr ient = Grad . run
WRITE
f ( x + h) − f ( x ) ,h≠0 h f (1 + h) − f (1) = h Gr ient = Grad
(1 + h)2 + 2 − (12 + 2) h 3 + 2h + h 2 − 3 = h =
334
Maths Quest 12 Mathematical Methods CAS for the TINspire
2h + h 2 h h(2 + h) = h =2+h The gradient is 2 + h. =
WORKED EXAMPLE 8
Find the gradient of f (x) = x2 − 1 at the point where x = 2 by: a sketching a graph and finding the gradient of the tangent at x = 2 b differentiating using first principles. THINK a
1 2
b
WRITE
Sketch the graph of f (x) over a domain which includes the given value of x. Construct a tangent at the given point on the curve. (It is difficult to be accurate.)
3
Find the gradient of this tangent by evaluating rise . run
1
Find f ′(x) using first principles.
a
y 8 7 (3, 6) 6 5 4 3 2 11 1 0 12345 −2 −3 −4 −5 (0, −5) 3
x
Gradient of tangent at x = 2 is 11 approximately = 3 23 . 3 f ( x + h) − f ( x ) b f ′( x ) = lim ,h≠0 h→0 h ( x + h)2 − 1 − ( x 2 − 1) h→0 h
= lim
x 2 + 2 xxhh + h 2 − 1 − x 2 + 1 h→0 h
= lim
2 xh + h 2 h→0 h h(2 x + h) = lim h→0 h = lim (2 x + h) = lim
h→0
= 2x 2
Evaluate f ′(2) to find the gradient at x = 2.
f ′(2) = 2(2) = 4
Note: Answer a (3 23 ) is very close to answer b (4).
Chapter 7
Differentiation
335
WORKED EXAMPLE 9
Use first principles to differentiate g(x) = x2 − x. THINK
WRITE
1
Find g(x + h) and simplify.
2
Find g ′(x) using first principles.
g(x + h) = (x + h)2 − (x + h) = x2 + 2xh + h2 − x − h g( x + h) − g( x ) g ′( x ) = lim , h ≠ 0, h→0 h x 2 + 2 xxhh + h 2 − x − h − ( x 2 − x ) h→0 h 2 2 x + 2 xxhh + h − x − h − x 2 + x = lim h→0 h 2 2 xh + h − h = lim h→0 h h(2 x + h − 1) = lim h→0 h = lim (2 x + h − 1) = lim
h→0
= 2x − 1 REMEMBER
1. The limit of a function is the value that y approaches as x approaches a given value. 2. If a function is continuous at a point then a limit exists at that point and can be found by direct substitution. 3. If a function is discontinuous at a point then the limit exists only if the function is approaching the same value from both left and right. If there is a break in the curve and the point lies within the break then the limit does not exist. 4. The limit of expressions with a denominator can be found. (a) If direct substitution makes the denominator equal to zero then factorise the numerator, cancel and then use direct substitution. (b) If direct substitution does not make the denominator equal to zero then use direct substitution. f ( x + h) − f ( x ) 5. The gradient of a chord or secant is found using . h f ( x + h) − f ( x ) , h ≠ 0. 6. The gradient at a point P on a curve is f ′( x ) = lim h→0 h 7. The gradient at a point P on a curve is the gradient of the tangent to the curve at that point. EXERCISE
7B
Limits and differentiation from ﬁrst principles 1 WE 5 Evaluate the following limits. a lim ( x + 4) b lim (2 p − 3) x→2
d lim ( x 2 − 5)
e
g
h lim
x→3
336
p→ −2
lim (10 − x + x 2 − x 3 )
x→ 2 −
c
lim ( x 2 + 4 x − 3)
f
x 2 + 5x + 6 x→0 x+2
i
x → −1
Maths Quest 12 Mathematical Methods CAS for the TINspire
lim (8 − 3h)
h→0
lim ( x 3 − 5 x + 2)
x→3
x2 − 2x − 3 x →1 x−3 lim
2 WE 6 By first simplifying the rational expression, evaluate the following limits. 2x2 + 2x x 2 + 3x 3x 2 − 3x a lim c lim b lim x→ 1 x +1 x→0 x →1 x − 1 x 2 −4 2 + 3x + 2 x x x 2 − 5x − 6 d lim e lim f lim x→2 x − 2 x→ 1 x→6 x +1 x−6 h3 − 8 x 3 + 27 x2 + 4x − 5 g lim h lim i lim h→2 h − 2 x→ 3 x + 3 x→ 5 x+5 −
−
−
3 Evaluate the following. a lim (3 x − 4) x→3
x3 + 4 x →1 x + 2 x 2 + 3x g lim x→3 x − 1 d lim
b
−
x2 − 9 3 x+3
lim
x→ −
c
h3 − 64 h→4 h − 4 x3 + 1 h lim x→ 1 x −1 e lim
f
−
i
x2 + x − 6 x→2 x−2 lim
lim ( x 3 + x 2 − 6)
x → −2
x 2 + 7 x + 12 4 x+4
lim
x→
−
Questions 4 and 5 refer to the following diagram. Consider the chord PQ drawn to the curve f (x) as shown. 4 WE 7 Find the gradient of the chord PQ drawn to the curve f (x) = x2 + 1 in the diagram at right.
y f(x) = x2+1 Q (2+h, f(2+h))
5 MC The gradient of the tangent at P in the diagram is: a h b 4+h c 4h e 4 d 4 + h2
P (2, f(2))
1 0
x
2
6 a Find the gradient of the chord PQ to the function f (x) = x(x + 2) if the xcoordinates of P and Q are −1 and −1 + h respectively. b Hence, find the gradient of the function at P.
7 If the gradient of the chord joining two points on the curve f (x) = x2 + 2x + 3 is find the gradient of the curve at the point where x = 1.
f (1 + h) − f (1) , h
8 WE 8 Find the gradient of g (x) = 4 − x2 at the point where x = 2 by: a sketching a graph and finding the gradient of the tangent at x = 2 b differentiating using first principles. 9 Find the gradient of h (x) = 2x2 − 6x at x = −1 by using: a a sketch graph to find the gradient of the tangent at x = −1 b differentiation from first principles. 10 MC The gradient of a function f (x) at the point where x = 3 is: f ( x + h) − f ( x ) f (3 + h) − f ( x ) f (3 + h) − f (3) a lim b lim c lim h→0 h→0 h→0 h h h f (3 + h) − f (3) f ( x + h) − f ( x ) d e h h 11 WE9 Use first principles to differentiate f (x) if: a f (x) = 3x + 5 b f (x) = x2 − 3 d f (x) = (x − 4)(x + 2) e f (x) = 8 − 3x2 dy 12 Use first principles to find if: dx a y = 9 − 4x b y = x2 + 3x 3 d y = x − 4x e y = 5x − 2x3
c f
f (x) = x2 + 6x f (x) = x3 + 2
c f
y = 3x2 + 8x − 5 y = −x2 − 2x
Chapter 7
Differentiation
337
The derivative of xn
7C
Instead of using the procedure of differentiating from first principles, rules can be applied to find derivatives. These rules can be derived from first principles and have been looked at in detail in Maths Quest 11 Mathematical Methods CAS. If f (x) = axn then f ′ (x) = naxn − 1, where a and n are constants and n is rational. If f (x) = c then f ′ (x) = 0, where c is a constant. (This is because c = x0 and, using the rule, the derivative of x0 is 0 × x −1 or 0.) dy For example, if y = x7 then = 7 x 6 . If f (x) = 5x4 then f ′(x) = 20x3. dx If f (x) = g(x) + h(x) then f ′(x) = g′(x) + h′(x), that is, differentiate each term of a function separately. If f (x) = a × g(x), where a is a constant, then f ′(x) = a × g′(x). WORKED EXAMPLE 10
Differentiate y = x 4 −
3 2 x + 7. 2
THINK
WRITE
y = x 4 − 32 x 2 + 7
1
Write the equation.
2
Differentiate each of the three terms separately.
dy = 4 x 4 − 1 − 32 (2) x 2 − 1 + 0 dx = 4x3 − 3x
WORKED EXAMPLE 11
Find the derivative of: 1 1 a f ( x) = + x x
b f ( x) =
x+ x . x2
THINK a
WRITE
1
Write the equation.
2
Rewrite
3
Differentiate each term.
a
1 1 and using negative indices. x x
f ( x) =
f ( x ) = x −1 + x
b
Write the function in the form originally given.
1
Write the equation.
−1 −1 −1 − 1 − 1 2 x 2 −3 2 x − −x 2 −
= b
f ( x) =
2
−
1 1 − x 2 2 x3
x+ x x2 1
2
338
Rewrite x using indices.
Maths Quest 12 Mathematical Methods CAS for the TINspire
−1 2
f ′( x ) = − x =
4
1 1 + x x
x + x2 = x2
1
3
Separate the function into two terms expressed in index form.
x x2 = 2+ 2 x x
4
Simplify each term.
=x 1+x
5
Differentiate each term.
−
f ′( x ) = −1x = −x
6
′( x ) . Simplify f ′(
= =
−
−
−3 2
−1 − 1
−2
−
1 − x2
−
3 − 32 − 1 x 2
3 − 25 x 2 3 5
2x 2 3
1 − x2 2 x5
WORKED EXAMPLE 12
If f ( x ) = x 3 − 2 x 2 + a f '(x)
8 , use a CAS calculator to find: x b f '(2).
THINK 1
2
WRITE
On a Calculator page, press: • MENU b • 4:Calculus 4 • 1:Derivative 1 Complete the entry lines as: 8 d 3 x − 2x2 + x dx d 3 8 x − 2x2 + x = 2 x dx Press ENTER · after each entry. Alternatively, press / r to obtain the expression template and choose the derivative template. Write the answers using the correct notation for the derivative.
a f '( x ) = 3 x 2 − 4 x − b f '(2) = 2
8 x2
REMEMBER
1. Rules for differentiating axn (a) If f (x) = axn then f '(x) = naxn − 1, where a and n are constants. (b) If f (x) = c then f '(x) = 0, where c is a constant. (c) If f (x) = ag(x), where a is a constant then f '(x) = ag'(x). 2. If f (x) = g(x) + h(x) then f '(x) = g'(x) + h'( x). Differentiate each term of a function separately. 3. To evaluate f (a) where a is a constant, replace every x in the f (x) equation with the value of a. For example, if f (x) = 2x2 − 3x + 1, f (2) = 2 × 22 − 3 × 2 + 1 = 3.
Chapter 7
Differentiation
339
EXERCISE
7C
The derivative of xn 1
Find the derivative of each of the following. a y = x6 b y = 3x2 c y = 5x4 f y = −5x g y = 12 x 3 j y = 8x5 Differentiate each of the following.
e y = −4x3 i y = 10 2 WE 10
eBook plus Digital doc
SkillSHEET 7.3 Index laws
d y = x20 x4 h y= 3
a f (x) = 4x3 + 5x
b g(x) = −5x2 + 6x + 1
c h( x ) = 9 +
d h(x) = 4 − 3x + 6x2 + x3
e g(x) = 7x11 + 6x5 − 8
f
g f (x) = −6x + 3x2 − 4x3
h g( x ) = 7 x 2 − 4 x + 23
x3 5
2x 5 x3 + + 10 5 3 i h (x) = (x + 4)(x − 1) f ( x) =
j f (x) = (x2 + 2x)(3x − 6) 3 WE 11 a
Find the derivative of each of the following. 1
2 x3
5
d 4x 4 g x
c x3
b 3 x
−1 2
j
3 4x
m
1 3
x − 2x2
e
f
x+3 x 2 k 5x 2
2
+ x3
h
i l
1 + x2 x x 2 + x3 x 2 − + 3x 2 x
1
x3
− 4x +
− x 3
n
x + x4 x
4 WE 12 If f (x) = 2x5 − 10x + 5 find: a f ′(x) b f ′(2). 5 MC If f (x) = x2 − 6x then f ′(4) is equal to: a 8 b −12 c 12
d 2
e
−16
3
6 MC The value of f ′(9) if f ( x ) = x 2 + x 2 − 10 x is: b 18 d 8 e 0 c 12 12 1 7 Find g′ (−2) if g( x ) = 2 + 3 x − 8. x 5 8 Find the gradient of the curve y = 4 at the point where a x = 2 and b x = 0. x 1 9 Find the gradient of f ( x ) = 2 x 3 − x 2 + at the point where x equals: x a 1 b 4 c 9. a
1 2
10 If g (x) =
3
x + 4 x , find:
a g′ (x)
b g′(1)
c g′ (8)
d g′ (−8).
11 Show that the derivative of y = k, where k is a constant, is zero. 12 For each of the following: i expand the brackets ii differentiate the expanded expression iii factorise. a (x + 1)2 b (x + 1)3 c (2x + 1)2 d (2x + 1)3 e (3x + 1)2 f (3x + 1)3 13 Using the results of question 12 give the derivative of (ax + b)n in factorised form. (a, b, n are constants.)
340
Maths Quest 12 Mathematical Methods CAS for the TINspire
The chain rule
7D
A function which can be expressed as a composition of two simpler functions is called a composite function. For example, y = (x + 3)2 can be expressed as y = u2 where u = x + 3. That is, to obtain y from x, the first function to be performed is to add 3 to x (u = x + 3), then this function has to be ‘squared’ ( y = u2). Composite functions can be differentiated using the chain rule. For example, using the previous function, y = (x + 3)2: Let u = x + 3, so y = u2. dy du = 2u. Then = 1 and du dx dy dy dy du = × . This is known as the chain rule. It is known as the But we require and dx dx du dx chain rule because u provides the ‘link’ between y and x. dy Now = 2u × 1 dx = 2(x + 3) × 1 (replacing u with x + 3) = 2(x + 3) The chain rule is used when it is necessary to differentiate a ‘function of a function’ as above. WORKED EXAMPLE 13
eBook plus
If y = (3x − is expressed as y = dy dy ddu u dy dy a b and hence c . du du dx dx dx dx 2)3
un,
find:
Tutorial
int0554 Worked example 13
THINK a
b
c
WRITE
1
Write the equation.
a y = (3x − 2)3
2
Express y as a function of u.
3
Differentiate y with respect to u.
1
Express u as a function of x.
2
Differentiate u with respect to x.
Let y = u3 where u = 3x − 2 dy = 3u 2 du b u = 3x − 2 du =3 dx
1
Find
dy using the chain rule. dx
c
dy dy du = × dx du dx = 3u2 × 3 = 9u2
2
= 9(3x − 2)2
Replace u as a function of x.
WORKED EXAMPLE 14
If f ( x ) =
1 2 x2
− 3x
, find f ′ (x). Check your answer using a CAS calculator.
THINK 1
Write the equation.
WRITE
f ( x) =
1 2x2
− 3x
Chapter 7
Differentiation
341
2
Express f (x) in index form.
3
Express y as a function of u.
4
Differentiate y with respect to u.
5
Express u as a function of x.
6
Differentiate u with respect to x.
7
Find f ′ (x) using the chain rule.
8
Replace u as a function of x and simplify.
y = (2 x 2 − 3 x ) −1
Let y = u 2 where u = 2x2 − 3x − dy − 1 32 = 2u du u = 2x2 − 3x du = 4x − 3 dx −3 dy − = f ′( x ) = 12 u 2 × (4 x − 3) dx = = =
9
−1 2
−1 (4 x − 3) (2 x 2 2 − (4 x − 3)
− 3x)
−3 2
3
2(2 x 2 − 3 x ) 2 3 − 4x 2 ( 2 x 2 − 3 x )3
On a Calculator page, press: • MENU b • 4:Calculus 4 • 1:Derivative 1 Complete the entry line as: d 1 2 dx 2 x − 3 x Press ENTER ·.
10
Note the answer from the CAS is not expressed in the same format obtained previously.
∴ f '( x) =
− (4 x − 3)
2 x (2 x − 3) x (2 x − 3)
A quicker way to apply the chain rule when a function can be expressed in index form is as follows. If f (x) = [g(x)]n then f ′(x) = n[g(x)]n − 1 × g ′(x). That is, differentiate the bracket and then what is inside the bracket; ‘outside then inside’. WORKED EXAMPLE 15
Find the derivative of f (x) = (x2 − 2 x)3. THINK
342
WRITE
1
Write the equation.
f (x) = (x2 − 2x)3
2
Let g(x) equal what is inside the bracket.
g(x) = x2 − 2x
3
Find g ′(x).
g′(x) = 2x − 2
Maths Quest 12 Mathematical Methods CAS for the TINspire
4
Use the rule f ′ (x) = n[g(x)]n − 1 × g′(x) to differentiate f (x).
f ′ (x) = 3(x2 − 2x)3 − 1 × (2x − 2) = 3(x2 − 2x)2 × [2(x − 1)] = 6(x − 1)(x2 − 2x)2
5
Simplify f ′(x) as far as possible.
= 6(x − 1)[x(x − 2)x(x − 2)] = 6x2(x − 1)(x − 2)2
REMEMBER
1. A composite function is a function composed of two (or more) functions. 2. Composite functions can be differentiated using the chain rule, dy dy du = × . dx du dx 3. A short way of applying the chain rule is: If f (x) = [g(x)]n then f ′(x) = n[g(x)]n − 1 × g′(x). EXERCISE
7D
The chain rule 1 If each of the following functions are expressed in the form y = un, state i u and ii n. 1 a y = (5x − 4)3 b y = 3x + 1 c y= (2 x + 3)4 4 1 − d y = 7 − 4x e y = (5x + 3) 6 f y = (4 − 3 x ) 3 2 MC If y = (x + 3)5 is expressed as y = u5 then: a u = (x + 3)5 b u=x+3 d u=3 e u = x5
c u=x
3 WE 13 If each of the following composite functions are expressed as y = un, find: du dy dy i ii and hence iii . dx du dx 1 b y = (7 − x)3 c y= a y = (3x + 2)2 2x − 5 3 1 d y= e y = 5x + 2 f y= 4 (4 − 2 x ) 3x − 2 −2
g y = 3(2x2 + 5x)5
h y = (4x − 3x2)
i
−4
1 y = x+ x
6
j y = 4(5 − 6x)
For questions 4, 5 and 6 below, y = x 2 − 3 x + 2 is expressed as y = un. 4 MC a
dy is equal to: du
1 u 2
b u
−1 2
c
1 2 u
d
3 1 2 u 2
d
1 2
e
1 1 2 u 2
du 5 MC is equal to: dx a 2x − 3
b x2 − 3x + 2
c x2 − 3x
3
x2 − 2 x + 1
Chapter 7
e x−3
Differentiation
343
6 MC Using the chain rule, dy is equal to: dx 2x − 3 2x − 3 b a 2 x 2 − 3x + 2 u d 7
1 (2 x 2
− 3)
e
1 (2 x 2
c
x 2 − 3x + 2 2 u
1 − 3)( x 2 − 3 x + 2) 2
Use the chain rule to find the derivative of the following. a y = (8x + 3)4
b y = (2x − 5)3
c f (x) = (4 − 3x)5
d y = 3x 2 − 4
e
1
−2
f g(x) = (2x3 + x)
f ( x) = ( x 2 − 4 x) 3
1 g g( x ) = x − x
6
−1
h y = (x2 − 3x) 1
8 WE 14 If f ( x ) =
4x + 7
, find f ′(x).
9 Use the chain rule to find the derivative of the following. (Hint: Simplify first using index notation and the laws of indices.) a y=
6x − 5 6x − 5
b f ( x) =
( x 2 + 2)2 x2 + 2
10 WE 15 Find the derivative of: a f (x) = (x2 + 5x)8 b y = (x3 − 2x)2 c
1
f ( x ) = ( x 3 + 2 x 2 − 7) 5
3
d y = (2 x 4 − 3 x 2 + 1) 2
11 If f (x) = (2x − 1)6, find f ′(3). −
12 If g (x) = (x2 − 3x) 2, find g′(−2). 13 If f ( x ) = x 2 − 2 x + 1, find: a f (3) c f ′ (3) eBook plus Digital doc
WorkSHEET 7.1
b f ′(x) d f ′(x) when x = 2.
14 Find the gradient of the function h( x ) = 3 x 2 + 2 x at the point where x = 2. 3 15 Find the value of f ′ (−1) if f ( x ) = . 5 − 4x 16 If f (x) = (2x − 1)5 and f ′ (x) = 10 (2x − 1)n, find the value of n. 17 If f (x) = (3x + 2)7 and f ′ (x) = a (3x + 2)6 × 3, find the value of a. 18 If f (x) = (5x − 3)10 and f ′ (x) = 10m (5x − 3)9, find the value of m.
7E
The derivative of e x If f (x) = ex then using first principles f ( x + h) − f ( x ) f ′( x ) = lim ,h≠0 h→0 h
344
ex + h − ex h→0 h
= lim
= lim
e x eh − e x h→0 h
Maths Quest 12 Mathematical Methods CAS for the TINspire
e x (e h − 1) h→0 h h e −1 = e x lim h→0 h h e −1 Note that lim can be deduced by using a calculator and substituting values of h close to h→0 h zero.
= lim
h
eh − 1 h
0.01
1.0050
0.0001
1.000 05
0.000 001
1.000 000
eh − 1 = 1. h→0 h Therefore, f ′ (x) = ex × 1 = ex That is, lim
If f (x) = ex then f ′ (x) = ex.
WORKED EXAMPLE 16
Differentiate y = e
−5x
.
THINK
WRITE
1
Write the equation.
2
Express u as a function of x and find
3 4 5
du . dx dy Express y as a function of u and find . du dy Find using the chain rule. dx Replace u as a function of x.
−5x
y=e
Let u = −5x so
du − = 5 dx
dy = eu du
y = eu so
dy − u = 5e du −5x
= −5e
This example shows that if f (x) = ekx then f ' (x) = kekx. WORKED EXAMPLE 17
Find the derivative of y = e2x + 1. THINK 1
Write the equation.
2
Express u as a function of x and find
3 4
5
du . dx dy Express y as a function of u and find . du dy Find using the chain rule. dx Replace u as a function of x.
WRITE
y = e2x + 1 Let u = 2x + 1 so y = eu so
du =2 dx
dy = eu du
dy = eu × 2 dx = 2eu = 2e2x + 1
Chapter 7
Differentiation
345
WORKED EXAMPLE 18
a f (x) = ex(ex − 2)
Differentiate:
b f ( x) =
e2 x − 2 e− x . ex
THINK a
b
WRITE
Write the equation.
2
Expand.
3
Differentiate.
4
Factorise in order to leave the answer in the form it was given.
1
Write the equation.
2
Write each term in the numerator over each term in the denominator. Divide the numerator of each term by its denominator using the laws of indices.
3
f (x) = ex(ex − 2)
a
1
= e2x − 2ex f ′ (x) = 2e2x − 2ex = 2ex(ex − 1) b
f ( x) =
e 2 x − 2e − x ex
=
e 2 x 2e − x − x ex e −x − x
= e2x − x − 2e − = ex − 2e 2x −
4
Differentiate each term.
5
Write your answer in the form it was given.
f ′ (x) = ex + 4e 2x 4 = ex + 2x e
WORKED EXAMPLE 19 3 − x
Find the derivative of y = ex
.
THINK
WRITE
1
Write the equation.
2
Express u as a function of x and find
3 4
5
3
y = ex
− x
du . dx dy Express y as a function of u and find . du dy Find using the chain rule. dx
y = eu so
Replace u as a function of x.
Let u = x3 − x so
du = 3x 2 − 1 dx
dy = eu du
dy = eu × (3 x 2 − 1) dx = (3x2 − 1)eu 3 − x
= (3x2 − 1)ex
This example shows that if f (x) = eg(x) then f ′ (x) = g′ (x) e g (x). REMEMBER
1. 2. 3. 4.
346
If f (x) = ex, f ′ (x) = ex. If f (x) = ekx, f ′ (x) = ke kx. If f (x) = aekx + c, f ′ (x) = ake kx + c. If f (x) = ae g (x), f ′ (x) = g ′(x) × ae g(x).
Maths Quest 12 Mathematical Methods CAS for the TINspire
EXERCISE
7E
The derivative of ex 1 WE 16
Differentiate each of the following. 1x
x
−
a y = e10x b y = e3 c y = e4 −5x −2x − f y = 4e g y = 6e h y = 5e0.2x 2 WE 17 Find the derivative of each of the following. a y = e6x − 2 b y = e8 − 6x 7 − 2x d y = 4e e y = −3e8x + 1 6 − 9x g y = 10e h y = −5e3x + 4 j
x +1
y = 2e 2
k y = 3e
c y = 2e5x + 3 f y = −2e6 − 5x − i y = 6e 7x
2− x 3
l
3 MC The derivative of y = e3x + 2 is equal to: 3x + 2 a 3e b (3x + 2)e3x + 2 c 3e3x d 3xe3x + 2 4 WE 18 Differentiate each of the following: a f (x) = 2(ex + 1) b f (x) = 3e2x(ex + 1) −x
d f (x) = (ex + 2)(e + 3)
e
f ( x) =
3e3 x
e2
4e5x
2
2x2
d f (x) = e2 − 5x
e f (x) = e6 − 3x + x
e
+ 2x) − 2e − x
−2x
−3
2
f g(x) = ex + 3x − 2
2
i y = e(2x + 1)
−2
4
f ( x) =
4e7 x
c y = ex − 2x
h y = −5e1 − 2x − 3x
j f (x) = e(4 − x)
f
−4x
2
b y = ex − 3x + 1 2
e 3xe3x
−e
2
a y = ex + 3x g h(x) = 3e4x − 7x
x +5
y = − 4e 4
c f (x) = 5(e
− e 6x
+ ex
g f (x) = + h f (x) = + 5 WE 19 Find the derivative of each of the following: ex
e y = 2e3x
d y=e x − i y = −2e 11x
k g(x) = e(x + 2)
3
3
l
y=e
3x + 4
2
1 3
n h(x) = e(x2 + 3x) m f ( x ) = e( x +1) 6 MC The derivative of 6ex3 − 5x is equal to: a 3x2 − 5 b 6(3x2 − 5)ex3 − 5x 3 − 5x 3 x d 6(x − 5x)e e 6(3x2 − 5)e3x2 − 5 9 − 4x 7 If f (x) = 5e , find the exact value of f ′ (2).
c (3x2 − 5)ex2 − 5x
8 If g (x) = 2ex2 − 3x + 2, give the exact value of g′(0). 9 Find the exact value of h′(−1) if h(x) = −5ex3 + 2x.
7F
The derivative of loge (x) −
The inverse of the function f (x) = ex is f 1(x) = loge (x). If y = loge (x) then ey = x, as shown in chapter 4, Exponential and logarithmic equations. Let x = ey dx = ey dy dy 1 But = dx dx and ey = x dy dy 1 Therefore, = . dx e y 1 = x
Chapter 7
Differentiation
347
That is, if f (x) = loge (x) then f'( x ) =
1 . x
WORKED EXAMPLE 20
Differentiate y = loge (7x). THINK 1 2 3 4
WRITE
Write the equation. du Express u as a function of x and find . dx dy Express y as a function of u and find . du dy Find . dx
y = loge (7x) du u = 7x, so =7 dx dy 1 y = loge (u), so = du u dy 1 = ×7 dx 7 x 1 = x
If f (x) = loge (kx), where k is a constant then f ′( x ) =
1 . x
WORKED EXAMPLE 21
Find the derivative of y = 2 loge (3x − 4). THINK 1
WRITE
2
Write the equation. Express u as a function of x.
3
Differentiate u with respect to x.
4
Express y as a function of u.
5
Differentiate y with respect to u.
6
Find
7
Replace u with 3x − 4.
dy using the chain rule. dx
y = 2 loge (3x − 4) Let u = 3x − 4. du =3 dx y = 2 loge (u) dy 1 = 2× du u 2 = u dy 2 = ×3 dx u 6 = u 6 = 3x − 4
WORKED EXAMPLE 22
Differentiate y = loge (x2 + 4x − 1). THINK
348
WRITE
1
Write the equation.
y = loge (x2 + 4x − 1)
2
Let u equal the section in brackets.
Let u = x2 + 4x − 1
Maths Quest 12 Mathematical Methods CAS for the TINspire
3
Differentiate u with respect to x.
4
Express y as a function of u.
5
Differentiate y with respect to u.
6
Find
7
Replace u with what is in the brackets.
du = 2x + 4 dx y = loge (u) dy 1 = du u dy 1 = × (2 x + 4) dx u 2x + 4 = 2 x + 4x − 1
dy using the chain rule. dx
This example shows that if f (x) = loge [g(x)] then f ′( x ) =
g ′( x ) . g( x )
WORKED EXAMPLE 23
Differentiate y = loge (x2 + 5x − 2)2. THINK
WRITE
1
Write the equation.
y = loge (x2 + 5x − 2)2
2
Simplify the right side by using log laws. du Express u as a function of x and find . dx dy Express y as a function of u and find . du dy Use the chain rule to find . dx
y = 2 loge (x2 + 5x − 2) du u = x2 + 5x − 2, so = 2x + 5 dx dy 2 y = 2 loge (u), so = du u dy dy du dy 2(2 x + 5) = × so = dx du dx dx u 2(2 x + 5) = 2 x + 5x − 2
3 4 5
REMEMBER
1 1. If f (x) = loge (x), then f ′( x ) = . x 1 2. If f (x) = loge (kx), then f ′( x ) = . x g ′( x ) . 3. If f (x) = loge [g(x)], then f ′( x ) = g( x ) EXERCISE
7F
The derivative of loge (x) 1 If y = loge (4x) is expressed as y = loge (u), then find: du a u b dx c
dy du
d
dy using the chain rule. dx Chapter 7
Differentiation
349
2 WE 20
Differentiate each of the following.
a y = loge (10x)
b y = loge (5x)
c y = loge (−x)
d y = loge (−6x)
e y = 3 loge (4x) x h y = log e 3
f y = −6 loge (9x) x i y = 4 log e 5
x g y = log e 2 j
− 2x y = − 5 log e 3
3 MC The derivative of loge (8x) is:
8 1 d x x 4 MC To differentiate y = loge (3x + 7) using the chain rule: a ‘u’ would be used to represent d loge (3x) a 3x + 7 b 3x c loge (x) b
a 8
b
1 8
x
c
e loge (8)
e x
dy du and are respectively du dx
1 1 1 1 and 3x c 3 and b and 3x + 7 d and 3 3x u u u dy c Hence is equal to dx 3 1 1 d c a 3 b 3x + 7 3x + 7 x 5 WE 21 Find the derivative of each of the following. a y = loge (2x + 5) b y = loge (6x + 1) c y = loge (3x − 4) a
e
d y = loge (8x − 1)
e y = loge (3 − 5x)
f y = loge (2 − x)
g y = loge (4 − 7x)
h y = 6 loge (5x + 2)
i y = 8 loge (4x − 2)
j y = −4 loge (12x + 5)
k y = −7 loge (8 − 9x)
6 WE 22
3 x
Differentiate the following.
a y = loge (3x4)
b y = loge (x2 + 3)
d y = loge (x2 − 2x3 + x4) g y = log e (5 x +
1 2) 3
j f (x) = loge (3x − 2)4
e y = log e
(
h f ( x ) = log log e
2x + 1
c y = loge (x3 + 2x2 − 7x)
)
1 (2 − 3 x ) 5 (2 −2
k f (x) = loge (5x + 8)
(
3 − 4x
a y = loge (x2 + 1)2
b y = loge (3 − x2)2
c y = loge (x2 − 2x + 3)3
d y = loge (x2 + 4x + 4)3
y = log e
i
1 f ( x ) = log e x + 3
l
2 f ( x ) = log e 4 + 3 x
8 MC Using the chain rule the derivative of f (x) = loge (x2 − 5x + 2) would be: − 1 5 c 2x − 5 a 2 b 2 x − 5x + 2 x − 5x + 2 1 2x − 5 d e 2 x (2 x − 5) x − 5x + 2 9 Find the gradient of the function f (x) = 6 loge (4 − 3x) when x = −1. Maths Quest 12 Mathematical Methods CAS for the TINspire
)
f
7 WE 23 Differentiate each of the following.
350
e 1 and 3
10 If g(x) = 3 loge (3x + 5) find the value of g′(0). 11 Find the exact value of f ′(2) if f (x) = 3x2 + 4 loge (x2 + x). 12 If y = eloge (x), find: dy a dx b the exact gradient when i x = 1 result?
ii x = 2
iii x = 4 iv x = 10. Can you explain this
13 If f (x) = eloge (x2), find: a f ' (x) b the exact value of i f ' (1) ii f ' (5) iii f ' (−2).
7G
The derivatives of sin (x), cos (x) and tan (x) The derivatives of sin (x), cos (x) and tan (x) can be found by differentiation from first principles but are beyond the requirements of this course. The derivatives of sin (x) and cos (x) can be shown by drawing the graphs of the gradient functions. The domain of each of these functions is R, but we will use only part of that domain. Consider the graph of f (x) = sin (x), domain [0, 2π] shown below. y π (— , 1) 2
0
π
2π
x
π ( 3— , −1) 2
π 3π , 2 2 π 3π f '(x) < 0 when < x < 2 2 3π π f '(x) > 0 when 0 < x < and < x < 2π 2 2 By sketching the graph of the gradient function, we can see that it is y = cos (x). f '(x) = 0 when x =
y 1 0 −1
π– 2
π
3π π — 2
2π x
Similarly, by sketching the graph of the gradient function of y = cos (x), we can see that the derivative of y = cos (x) is y = −sin (x). The derivative of tan (x) can be found using the quotient rule y (which appears later in this chapter). If f (x) = sin (x) then f ′ (x) = cos (x). y = cos (x) 1 If f (x) = cos (x) then f ′ (x) = −sin (x). y = −sin (x) 1 0 — π 3π 2π x π — If f (x) = tan (x) then f '( x ) = , which can be 2 2 2 ( x) –1 2 cos s written as sec (x).
Chapter 7
Differentiation
351
WORKED EXAMPLE 24
Find the derivative of y = sin (5x). THINK
WRITE
y = sin (5x)
1
Write the equation.
2
Express u as a function of x and find
du . dx
Let u = 5x so
du =5 dx
3
Express y as a function of u and find
dy . du
y = sin (u) so
dy = cos (u) du
4
Find
5
Replace u with 5x.
6
It is important to note what happens when a CAS calculator is used to answer this question. On a calculator page, complete the entry line as: d (sin (5x)) dx Press ENTER ·. Note: The calculator gives an incorrect answer as it is set in Degree mode.
dy using the chain rule. dx
dy = 5cos (u) dx = 5 cos (5x)
In differentiating circular functions, the calculator needs to be set in Radian mode as the angles are given in radians. Write the answer, using the correct notation for the derivative.
∴
dy = 5 cos (5 x ) dx
This example shows that if f (x) = sin (ax), then f ′ (x) = a cos (ax). Similarly, if f (x) = cos (ax), then f ′ (x) = −a sin (ax). 352
Maths Quest 12 Mathematical Methods CAS for the TINspire
WORKED EXAMPLE 25
Find the derivative of y = tan (3x). THINK
WRITE
1
Write the equation.
2
Express u as a function of x and find
3 4
y = tan (3x)
du . dx dy Express y as a function of u and find . du dy Find using the chain rule. dx
du =3 dx dy y = tan (u) so = secc 2 (u) du dy 3 = 3 secc 2 (u) or dx coss2 (u) 3 = 3 sec 2 (3 x ) or 2 cos (3 x ) Let u = 3x so
This example shows that if f (x) = tan (ax), then f ' ( x ) =
a or a sec2 (ax). coss 2 ( ax )
WORKED EXAMPLE 26
eBook plus
Differentiate y = cos (x2 + 2x − 3).
Tutorial
THINK 1
Write the equation.
2
Express u as a function of x and find
3 4 5
int0555
WRITE
Worked example 26
y = cos (x2 + 2x − 3)
du . dx dy Express y as a function of u and find . du dy Find using the chain rule. dx Replace u with the part in brackets in the rule and simplify.
Let u = x2 + 2x − 3 so y = cos (u) so
du = 2x + 2 dx
dy − = sin(u) du
dy − = sin u × (2 x + 2) dx = −2(x + 1) sin (x2 + 2x − 3)
This example shows that the chain rule can be applied as follows. If f (x) = sin [g(x)] then f ′(x) = g′(x) cos [g(x)]. If f (x) = cos [g(x)] then f ′(x) = −g′(x) sin [g(x)]. g'( x ) If f (x) = tan [g(x)] then f '( x ) = cos 2 [ g( x ))]] = g′(x) sec2 [ g(x)] REMEMBER
1. If f (x) = sin (x) 2. If f (x) = cos (x) 3. If f (x) = tan (x) 4. If f (x) = sin (ax) 5. If f (x) = cos (ax)
then f ′(x) = cos (x). then f ′(x) = −sin (x). 1 then f ′( x ) = = secc 2 ( x ). cos2 ( x ) then f ′(x) = a cos (ax). then f ′(x) = −a sin (ax).
Chapter 7
Differentiation
353
a
6. If f (x) = tan (ax)
then f ′( x ) =
7. If f (x) = sin [g(x)] 8. If f (x) = cos [g(x)]
then f ′(x) = g′ (x) cos [g(x)]. then f ′(x) = −g′ (x) sin [g(x)]. g ′( x ) = g ′( x ) secc 2 [ g( x ))]]. then f ′( x ) = cos2 [ g( x)] x
9. If f (x) = tan [g(x)]
= a sec 2 (aax ).
cos2 (ax )
EXERCISE
7G
The derivatives of sin (x), cos (x) and tan (x) 1 WE 24
Find the derivative of each of the following.
a y = sin (8x)
b y = sin (−6x)
c y = sin (x)
x d y = sin 3
− x e y = sin 2
f
2
2x y = sin 3
Differentiate each of the following. a y = cos (3x)
b y = cos (−2x)
x c y = cos 3
π x d y = cos 4
− x e y = cos 8
f
3 WE 25
2x y = cos 5
Differentiate each of the following.
a y = tan (2x)
b y = tan (−4x)
x ta c y = tan 5
− 3x d y = tan ta 4
4 MC a The derivative of sin (6x) is: a 6 cos (6x)
b 6 cos (x)
d −6 cos (6x)
e
1 6
cos (6 x )
b The derivative of cos (4x) is: a 4 sin (4x) b 4 sin (x) c The derivative of sin (−4x) is: a 4 cos (−4x) b 4 cos (−x) d The derivative of cos (−8x) is: a 8 cos (−8x) b 8 sin (−8x) e The derivative of tan (7x) is: x ta a 7 tan 7
b
1 (7 x )
coss2
c 6 sin (x)
c −4 sin (x)
d 4 cos (4x)
e
−4
sin (4x)
c −4 cos (4x)
d −4 cos (−4x)
e
−4
sin (4x)
c −8 sin (−8x)
d −8 sin (−x)
e 8 sin (−x)
c 7 sec2 (7x)
d sec2 (7x)
e
5 WE26 If y = sin (4x + 3) is expressed as y = sin (u), find: a
dy du
b
du dx
c
dy using the chain rule. dx
6 If y = cos (3x + 1) is expressed as y = cos (u), find: a
354
dy du
b
du dx
Maths Quest 12 Mathematical Methods CAS for the TINspire
c
dy using the chain rule. dx
1 7
sec ( x )
7
Differentiate each of the following. a y = sin (2x + 3) b y = sin (6 − 7x)
c y = sin (5x − 4)
3x + 2 d y = sin 4
f
e y = 5π sin (2π x)
8 Differentiate each of the following. a y = cos (3x − 2) b y = cos (4x + 7) 2x + 3 d y = cos 3
3x y = − 4 sin 8
c y = cos (6 − 5x) f y = −6 cos (−2x)
e y = 4π cos (10π x)
9 Differentiate each of the following. a y = tan (2x + 1) b y = tan (8 − x) d y = tan 2(x + 1) e y = −3 tan (−x)
c y = tan (5x − 2)
10 Find the derivative of each of the following. a cos (x2 − 4x + 3)
b sin (10 − 5x + x2)
c sin (ex)
d cos (x2 + 7x)
e tan (4x − x2)
f tan (x2 + 3x)
g cos [loge (x)]
h sin (e4x)
i
1 cos x
j sin [loge (2x − 1)]
k cos (2e3x)
l 3 cos [loge (10x)]
m 4 tan (x3 + 2x2)
n
o
− 2 cos
x 4
− 8 tan ta
− 3x 5
p cos (x2 + 2x) + sin (3x − 9)
11 If f (x) = 3 sin (x2 + x), find f ′ (1) (answer correct to 3 decimal places). 12 Find the gradient of the curve g (x) = 2 cos (x3 − 3x) at the point where x = 0. 13 For each of the following functions find: i f ′ (x) and π ii the exact value of f ′ . 6 a f (x) = esin (x)
7H
b f (x) = ecos (x)
c f (x) = loge [sin (x)]
d f (x) = loge [cos (x)]
The product rule Any function which is a product of two simpler functions, for example, f (x) = (x + 2)(x − 5) or f (x) = (x2 − 5x + 6) sin (3x + 5) can be differentiated using the product rule of differentiation. Although the first example can be expanded and then differentiated, the second example cannot and therefore can be differentiated only using the product rule.
Product rule
dy dy dv dv ddu u = u× +v× . ddxx dx dx dx dx Or if f (x) = u(x) × v(x) then f ′(x) = u(x) × v′(x) + v(x) × u′(x). If y = uv then
Chapter 7
Differentiation
355
WORKED EXAMPLE 27
If y = (3x − 1)(x2 + 4x + 3) is expressed as y = uv, find: du du dv dv dy dy dy dy dv dv du du =u +v . a u and v b and c using dx dx dx dx dx dx dx dx dx dx dx dx THINK a
b
c
WRITE a y = (3x − 1)(x2 + 4x + 3)
1
Write the equation.
2
Identify u and v, two functions of x which are multiplied together.
1
Differentiate u with respect to x.
2
Differentiate v with respect to x.
1
Apply the product rule to find
Let u = 3x − 1
and
v = x2 + 4x + 3.
du =3 dx dv = 2x + 4 dx c dy = u × dv + v × du dx dx dx
b
dy . dx
= (3x − 1)(2x + 4) + (x2 + 4x + 3) × 3 2
= 6x2 + 10x − 4 + 3x2 + 12x + 9 = 9x2 + 22x + 5
Expand and simplify where possible.
WORKED EXAMPLE 28
Find the derivative of y = loge (4x) × sin (3x − 2). THINK
WRITE
1
Write the equation.
y = loge (4x) × sin (3x − 2)
2
Identify u and v. du dv Find and . dx dx dy Find using the product rule. dx
Let u = loge (4x) and let v = sin (3x − 2). du 1 dv = = 3 cos (3 x − 2) dx x dx dy 1 = log e (4 x ) × 3 cos (3 x − 2) + sin (3 x − 2) × dx x 1 = 3 log e (4 x ) × cos (3 x − 2) + sin (3 x − 2) x
3 4 5
Simplify wherever possible.
REMEMBER
dy dv du =u× +v× . dx dx dx 2. If f (x) = u(x) × v(x) then f ′ (x) = u(x) × v′(x) + v(x) × u′(x). 1. If y = u × v then
EXERCISE
7H
The product rule 1 WE27 If y = (x + 3)(2x2 − 5x) is expressed as y = u × v, find: du dv dy dy dv du a u and v b and c using the product rule, =u× +v × . dx dx dx dx dx dx
356
Maths Quest 12 Mathematical Methods CAS for the TINspire
2 WE28 a y=
Find the derivative of: 4x3
× loge (6x)
b g (x) = (3x − 2) loge (2x).
3 MC The derivative of f (x) = a f ′ (x) = 2x cos (2x) c f ′ (x) = 2x sin (2x) + x2 cos (2x) e f ′ (x) = 2x sin (x) + 2x2 cos (x)
x2
sin (2x) is: b f ′ (x) = 4x cos (2x) d f ′ (x) = 2x sin (2x) + 2x2 cos (2x)
4 Use the product rule to differentiate each of the following. a y = x cos (x) b y = 3x sin (x) c y = (5x − 2) ex d y = e3x (2 − 11x) e y = x5 cos (3x + 1) f y = 2x3 loge (7x) −2x g y = e loge (2x − 5) h y = 8 tan (5x) loge (5x) 4x i y = 5 cos (2x) sin (x) j y = sinn cos ( x ) 3 − 4x − 3 k f (x) = e loge (6x) l f (x) = 4e 5x sin (2 − x) − 1 n f ( x) = x e 3 x cos (6 x ) m f ( x) = x −
−
o f (x) = 2x 3 sin (2x + 3) − q f (x) = (x2 + e3x)(4 − e 3x)
p f (x) = e 2x loge (3x2 + 5) r f (x) = (x2 − 6)(2 + 3x − x2)
5 If f (x) = (2x + 1) loge (x + 3), find the exact value of f ′(1). 6 Find g′(0) if g(x) = 5e2x cos (4x). eBook plus
7 Find the value of f ′(−2) if f (x) = (x2 + 2) sin (4 − 3x) (answer correct to 3 decimal places).
Digital doc
8 If g (x) = (6x + x2) ex − 3, find the exact value of g′(2).
WorkSHEET 7.2
9 Find the exact value of f ' (π) if f (x) = (3 − x) tan (2x).
7I
The quotient rule The quotient rule is used to differentiate functions which are rational expressions (that is, one function divided by another). For example, e3 x + 8 x2 − 6x + 3 or f ( x ) = f ( x) = coss ((6 − x ) 5x + 2
Quotient rule
du du dv dv −u× u dy v × dx dy dx dx . dx If y = then = v ddxx v2 u( x ) v ( x )u ′( x ) − u( x )v ′( x ) Or if f ( x ) = then f ′( x ) = . v( x) [v ( x)] x 2 WORKED EXAMPLE 29
3− x u is expressed as y = , find: v x2 + 4 x du du dv dv a u and v b and dx dx dx dx
If y =
c
dy dy . ddxx
Chapter 7
Differentiation
357
THINK a
b
c
WRITE
Write the equation.
2
Identify u and v.
1
Differentiate u with respect to x.
2
Differentiate v with respect to x.
1
Apply the quotient rule to obtain
2
3− x x2 + 4x Let u = 3 − x and v = x2 + 4x. du − b = 1 dx dv = 2x + 4 dx du dv v× −u × dy dx dx c = dx v2 ( x 2 + 4 x ) × −1 − (3 − x )( )(2 x + 4) = 2 2 ( x + 4 x) a y=
1
dy . dx
dy where possible, factorising the dx final answer where appropriate. Simplify
= =
− x2
− 4 x − (12 + 2 x − 2 x 2 ) ( x 2 + 4 x )2
− x2
− 4 x − 12 − 2 x + 2 x 2 ( x 2 + 4 x )2
=
x 2 − 6 x − 12 ( x 2 + 4 x )2
=
x 2 − 6 x − 12 x 2 ( x + 4) 2
WORKED EXAMPLE 30
Find the derivative of f ( x ) =
eBook plus 3x
2e . cos ( 2 x − 3)
THINK
Tutorial
int0556
WRITE
2e 3 x coss ((2 (2x 2xx − 3)
1
Write the equation.
f ( x) =
2
Identify u(x) and v(x).
Let u(x) = 2e3x. Let v(x) = cos (2x − 3).
3
Differentiate u(x) and v(x) with respect to x.
4
Apply the quotient rule to obtain f ′(x).
u′(x) = 6e3x v′(x) = −2 sin (2x − 3) v ( x )u ′( x ) − u( x )v ′( x ) f ′( x ) = [v ( x)] x 2 =
5
358
Simplify where possible.
Worked example 30
coss ((22x − 3) × 6e3 x − 2e3 x [− 2 ssin in (2 x − 3))]] [cos cos (2 (2 x − 3))]]2
6e3 x cos (2x − 3) + 4 e3 x sin (2 x − 3) [cos cos ((22xx − 3))]]2 3 x 2e [3 cos (2 x − 3) + 2 sin (2 x − 3)] = cos2 (2 x − 3)
=
Maths Quest 12 Mathematical Methods CAS for the TINspire
REMEMBER
du dv v× −u × u dy dx dx . 1. If y = then = 2 v dx v u( x ) v ( x )u ′( x ) − u( x )v ′( x ) 2. If f ( x ) = then f ′( x ) = . v( x) [v ( x)] x 2
EXERCISE
7I
The quotient rule 1 WE 29 If y =
x+3 u is expressed as y = , find: x+7 v
a u and v du dv b and dx dx du dv v× −u × dy dy dx dx . c using the quotient rule, = dx dx v2 x2 + 2x u( x ) 2 If f ( x ) = is expressed as f ( x ) = , find: 5−x v( x) a u(x) and v(x) b u′ (x) and v′ (x) c f ′(x) using the quotient rule. sin ( x ) 3 f (x) = tan (x) can be written as f ( x ) = . If u(x) = sin (x) and v(x) = cos (x), use the cos ( x ) 1 quotient rule to show that the derivative of tan (x) is . coss2 ( x ) 4 WE 30 Find the derivative of each of the following. a d g j
x2
2x − 4x
4x − 7 10 − x sin (2 x ) cos (2 x ) x2
4x + 3x − 2
− 3x
b
x2 + 7x + 6 3x + 2
c
e
e2 x x
f
h
log e ( x + 1) x2 + 2
i
k
2x3 + 7x e5x
l
m
e 3x + 8
n
4 log e (8x 8x) 2 x − 2x
o
p
2 cos (3 − 2 x ) x2
q
3e 2 − 7 x x+3
r
5 MC If h( x ) = 9x2 − 8 x2 − 3x 2 − 8 d x2 a
8 − 3x 2 then h′(x) equals: x 8 − 9x 9 2 b x2 − 3x 2 + 8 e x
c
cos ( x ) ex 3x 2 logg e (4 x ) e3 x + 2 cos (2 x ) x2 − 5 x sin ( x ) x 2 x e 2x − 3
− 3x 2
+8
x2
Chapter 7
Differentiation
359
sinn (4 x ) is: 4x + 1 4(4 x + 1) cos ( x ) − 4 sin (4 x ) a f ′( x ) = (4 x + 1)2 4(4 x + 1) cos (4x ) − 4 sin (4 x ) c f ′( x ) = 4x + 1 4 ssin (4x ) − 4(4 x + 1) cos (4x ) e f ′( x ) = (4 x + 1)2 cos (33 x − 2) 7 MC If g( x ) = then g′(x) is equal to: ex 3e x sin (3 x − 2) − e x cos (3 x − 2) a e2 x e x cos (3 x − 2) + 3e x cos (3 x − 2) c e2 x − 3e x sin (3 x − 2) − e x cos (3 x − 2) e e2 x 6 MC The derivative of f ( x ) =
8 If y =
cos (2 x ) dy find when x = 0. dx e3 x
9 Find the gradient of the function f ( x ) = 10 Find the exact value of g′(5) if g( x ) =
7J
(4 x + 1) cos (4 x ) − 4 ssin in (4 x ) (4 x + 1)2 4(4 x + 1) cos (4 x ) − 4 sin (4 x ) d f ′( x ) = (4 x + 1)2 b f ′( x ) =
−ex
sin (3 x − 2) − e x cos (3 x − 2) e2 x − 3e x sin (3 x − 2) − e x cos ((33x − 2) d ex b
2 x − 3x 2 at the point where x = −1. log e (3 x + 4)
4 log e (2 x ) . 3x
eBook plus Interactivity
Mixed problems on differentiation
int0252 Mixed problems on differentiation
Problems on differentiation may involve any combination of chain, product and quotient rules. WORKED EXAMPLE 31
For each of the following decide which rule of differentiation, that is, chain, product or quotient rule, would be useful to find the derivative. logg e (x ( ) 2 a b (x2 − 5x)6 c (x2 + 2x − 3) cos (2x) d ex + 3x sin ( x ) THINK a
b
c
360
WRITE
1
Write the equation.
2
It is of the form
1
u , that is, a rational function. v Write the equation.
2
It is a composite function of the form u6, where u = x2 − 5x.
1
Write the equation.
2
It is of the form u × v, that is, the product of two functions.
Maths Quest 12 Mathematical Methods CAS for the TINspire
a y=
log e ( x ) sin ( x )
Quotient rule b y = (x2 − 5x)6
Chain rule c y = (x2 + 2x − 3) cos (2x)
Product rule
d
d y = ex2 + 3x
1
Write the equation.
2
It is a composite function of the form eu, where u = x2 + 3x.
Chain rule
WORKED EXAMPLE 32
Find the derivative of y = 2x3 cos (x2 + x). THINK
WRITE
1
Write the equation.
y = 2x3 cos (x2 + x)
2
Decide which rule to use and identify u and v to apply the rule.
Use the product rule. u = 2x3 and v = cos (x2 + x).
3
Differentiate u with respect to x.
du = 6x2 dx
4
v is a composite function, so differentiate v with respect to x using the chain rule.
5
Apply the product rule to find
v = cos (x2 + x) Let w = x2 + x. dw = 2x + 1 dx v = cos (w) dv − = sin ( w) dw dv dv dw = × dx dw dx dv − So = ( ssin in ( w)))) (2 x + 1) dx = −(2x + 1) sin (x2 + x) dy dv du =u× +v× dx dx dx
dy . dx
= 2x3[−(2x + 1)] sin (x2 + x) + cos (x2 + x) × 6x2 6
= −2x3(2x + 1) sin (x2 + x) + 6x2 cos (x2 + x) = 2x2[3 cos (x2 + x) − x(2x + 1) sin (x2 + x)]
Simplify where possible.
Derivatives involving the absolute value function x if x ≥ 0 Consider the absolute value function, f ( x ) = x = − . x if x < 0 As discussed earlier, the graph of f (x) = x is continuous for all x and has a cusp (sharp point) at x = 0.
y
1 if x > 0 ′( x ) = − The derivative of f (x) = x, f ′( . x 1 if x < 0 In order for the derivative to be defined at x = 0 for a function f (x), the limit as x approaches 0 from the left, written as lim f '( x ), and the limit as x approaches 0 from the right, written as x →0−
lim f '( x ), must be equal.
x →0+
Chapter 7
Differentiation
361
y
This is not the case, as lim f ' ( x ) = − 1 and lim f ' ( x ) = 1. Thus, x → 0−
x →0+
1 the function f '(x) = x is not differentiable at x = 0 and the graph of y = f '(x) has open circles at x = 0 as shown right. x −1 The chain rule is used to differentiate the composite function f (x) = h(g(x)), to give f '(x) = g′(x) × h′(g(x)). 1 if g( x ) > 0 Hence, for f (x) = h(g(x)), where h (x) = x, then f '( x ) = g'( x ) × − . 1 if g( x ) < 0 Note the derivative is a hybrid function and the domain is obtained by examining the graph of the function y = g(x). WORKED EXAMPLE 33
eBook plus
= x2
For the function: f (x) − 4x: a find the derivative b sketch the graphs of y = f (x) and y = f ′(x) on the same set of axes. THINK a
1
2
int0557 Worked example 33
WRITE
As f (x) = x2 − 4x is a composite function, apply the chain rule to find the derivative, f ′(x) where g(x) = x2 − 4x and h(x) = x.
To determine the domain of the derivative, consider the graph of the function g(x) = x2 − 4x. From the graph, x2 − 4x > 0 if x < 0 or x > 4 and x2 − 4x < 0 if 0 < x < 4.
f (x) = h(g(x)) f ′(x) = g ′(x) × h′(g(x)) 1 f '( x) = 2 x − 4 × − 1 2 x − 4 if f '( x) = − 2 x + 4 if
x 2 − 4 x > 0. x2 − 4x < 0
1 2 3 4 5x
2 x − 4 if x < 0 or x > 4 ∴ f ' ( x ) = − 2 x + 4 if 0 < x < 4
3
Write the derivative with the correct domain.
4
To differentiate f (x) = x2 − 4x with a CAS calculator, on a Calculator page, press: • MENU b • 4:Calculus 4 • 1:Derivative 1 Alternatively press /r to obtain the expression template and choose the derivative template. d Complete the entry line as: x2 − 4x dx Press ENTER ·.
(
if x 2 − 4 x > 0 if x 2 − 4 x < 0
y 4 3 2 1 0 −1 −1 −2 −3 −4
)
Note: The answer from the CAS is not expressed as a hybrid function with the correct domain and hence cannot be used.
362
Tutorial
x.sign
(x − 4) + sign (x). x − 4 is not an acceptable answer.
Maths Quest 12 Mathematical Methods CAS for the TINspire
b
1
Sketch the graph of f (x) = x2 − 4x.
2
Sketch the graph of the derivative, 2 x − 4 if x < 0 or x > 4 f '( x) = − 2 x + 4 if 0 < x < 4 For the graph of y = f ′(x), put open circles at x = 0 and x = 4 as the derivative is not defined at these points.
3
y
f'(x) = 2x − 4, x > 4
10 8 6 4 2 f(x) =x − 4x 2 x −3 −2 −1 0 1 2 3 4 5 6 7 −2 −4 f '(x) = −2x + 4, 0 < x < 4 −6 −8 f '(x) = 2x − 4, x < 0
To graph f (x) = x2 − 4x and y = f ′(x) with a CAS calculator, on a Graphs page, complete the function entry lines as: f1(x) = x2 − 4x d f 2(x) = x2 − 4x dx and press ENTER · after each entry. Note: The open circles at x = 0 and x = 4 are not displayed.
(
)
WORKED EXAMPLE 34
For the function f (x) = sin (x) for x ∈[0, 2π]: a find the derivative b sketch the graphs of y = f (x) and y = f ′(x) on the same set of axes. THINK a
1
WRITE
As f (x) = sin (x) is a composite function, apply the chain rule to find the derivative, f ′(x) where g(x) = sin (x) and h(x) = x.
f (x) = h(g(x)) f ′(x) = g ′(x) × h′(g(x)) sinn ( x ) > 0 1 if si f ' ( x ) = cos cos ( x ) × − sin ( x ) < 0 1 if sin cos ( x ) if sin ( x ) > 0 f '( x) = − in ( x ) < 0 cos ( x ) if ssin
2
3
To determine the domain of the derivative, consider the graph of the function g(x) = sin (x) for x ∈ [0, 2π]. From the graph, sin (x) > 0 if 0 < x < π and sin (x) < 0 if π < x < 2π. Write the derivative with the correct domain.
y 1 0.5 0 −0.5 −1
π 2
π
3π 2
2π
x
cos ( x ) if 0 < x < π ∴ f '( x) = − cos ( x ) if π < x < 2π
Chapter 7
Differentiation
363
b
1
Sketch the graph of f (x) = sin(x) for x ∈ [0, 2π].
y 1
2
Sketch the graph of the derivative, cos ( x ) if 0 < x < π f '( x) = − cos ( x ) if π < x < 2π For the graph of y = f ′(x), put open circles at x = 0, x = π and x = 2π as the derivative is not defined at these points.
f(x) =sin (x)
0.5 0
π 2
−0.5 −1
π
2π
3π 2
x
f '(x) = cos (x), 0 < x < π f '(x) = −cos (x), π < x < 2π
REMEMBER
1. Chain rule: dy dy du (a) = × dx du dx (b) A short way to apply the chain rule is: If f (x) = [g(x)]n then f ′(x) = n[g(x)n − 1 × g′(x). 2. Product rule: dy dv du (a) If y = uv then =u× +v× . dx dx dx (b) If f (x) = u(x) × v(x) then f ′(x) = u(x) × v′(x) + v(x) × u′(x). 3. Quotient rule: du dv v× −u × u dy dx dx. (a) If y = then = v dx v2 u( x ) v ( x ) u ′( x ) − u( x ) v ′( x ) (b) If f ( x ) = then f ′( x ) = . v( x) [v ( x)] x 2 EXERCISE
7J
Mixed problems on differentiation 1 WE 31 For each function given below, state which rule of differentiation would be used to find the derivative, that is, chain (C), product (P) or quotient (Q). 3x + 7 a f (x) = loge (8x) b f (x) = 3x sin (x) c g( x ) = 4x2 d h( x ) =
4x cos ( x )
e g(x) = e5x sin (x) −
g h(x) = cos (x2 − 4x)
h f (x) = e x loge (5x)
j f (x) = sin2 (x)
k h( x ) =
m g(x) = ecos (x)
x−2 ex n f (x) = tan (x)
f
g( x ) =
x2 + 9x − 8 log e ( x )
i g(x) = loge [sin (x)] l
f ( x ) = log e ( x )
2 Using the appropriate rule find the derivative of each function in question 1. 3 WE 32 Find the derivative of each of the following. (Note that more than one rule will need to be applied in some cases.) x−2 − a y = e 5x cos (4x − 7) b y= 3x + 1 c y = loge (x + 1)3
364
Maths Quest 12 Mathematical Methods CAS for the TINspire
d y = cos (x2 − 6x)
e f (x) = ex cos (2x) 1 sin ( x )
g f ( x) =
j f (x) = (x − 1)(x2 + 5x + 3)
k g(x) = ex(x2 + 3) −4x
l
cos (4x − 3)
s y=
p f ( x) =
cos (2 x ) sin (2 x )
log e
g( x ) =
(2 x + 3)5 x3 − 5
n y = cos2 (3x)
o y = loge [cos (3x)] q f ( x) =
sin (2 x ) cos (2 x )
h y = loge [sin (3x)]
i y = 4e3x2 − 5x + 2
m f (x) = e
f ( x) =
f
sin ( x 4 ) x2
r f (x) = [loge (5x − 1)]4
( x)
t
x
x + 3 y = sin x − 2 3
log e ( x 2 ) v g( x ) = x2 − x f (x) = 3 cos2 (x) + e 7x − x3
u f (x) = 3x5 cos (2x + 1) w y = ex sin (x) y f (x) = 3 sin (6x) + loge
(5x2)
−
x 2 4e
z h (x) = cos3 (x)
4 WE 33 For the following functions: a f (x) = x2 − 1 b f (x) = x2 + 2x i find the derivative ii sketch the graphs of y = f (x) and y = f ′(x) on the same set of axes. 5 WE 34 For the following functions: a f (x) = sin (2x) for x ∈ [0, π] b f (x) = cos (x) for x ∈ [0, 2π] i find the derivative ii sketch the graphs of y = f (x) and y = f ′(x) on the same set of axes.
Chapter 7
Differentiation
365
SUMMARY Review — gradient and rates of change
• The gradient of a function exists wherever the graph of the function is smooth and continuous. • If the gradient of a function, f (x), is zero at x = a, then the graph of its gradient function, f ′(x), will have an xintercept at x = a. • When the gradient of a function is positive, the graph of the gradient function is above the xaxis and when the gradient of a function is negative, the graph of the gradient function is below the xaxis. • A polynomial function has a gradient function which is also a polynomial function, but its degree is reduced by one. Limits and differentiation from first principles
• The gradient of a chord (secant) or the average rate of change is given by: f ( x + h) − f ( x ) . h • A limit is the value that y approaches as x approaches a given value. • A limit exists if the function is approaching the same value from both left and right. • The gradient of the tangent to a curve at a point P is the gradient of the curve at P and is f ( x + h) − f ( x ) given by lim . h→0 h dy • For a function y = f (x), its derivative is expressed as either or f ′ (x). dx The derivative of xn
• • • •
If f (x) = axn then f ′(x) = naxn − 1, where a and n are constants. If f (x) = c then f ′(x) = 0, where c is a constant. If f (x) = ag(x) where a is a constant then f ′(x) = ag′(x). If f (x) = g(x) + h(x) then f ′(x) = g ′(x) + h ′(x). Differentiate each term separately. The chain rule
• The chain rule of differentiation is: dy dy du = × dx du dx The derivative of ex
• • • •
If f (x) = ex then f ′(x) = ex. If f (x) = ekx then f ′(x) = kekx. If f (x) = aekx + c then f ′(x) = akekx + c. If f (x) = aeg(x) then f ′(x) = g ′(x) × ae g(x). The derivative of loge (x )
1 . x 1 • If f (x) = loge (kx) then f ′(x) = . x g ′( x ) • If f (x) = loge [g(x)] then f ′(x) = . g( x )
• If f (x) = loge (x) then f ′(x) =
The derivatives of sin (x ), cos (x ) and tan (x )
• If f (x) = sin (x) then f ′(x) = cos (x). • If f (x) = cos (x) then f ′(x) = −sin (x).
366
Maths Quest 12 Mathematical Methods CAS for the TINspire
1 = sec2 (x). cos2 ( x ) If f (x) = sin (ax) then f ′(x) = a cos (ax). If f (x) = cos (ax) then f ′(x) = −a sin (ax). a If f (x) = tan (ax) then f ′(x) = = a sec2 (ax). 2 cos (ax ) If f (x) = sin [g(x)] then f ′(x) = g ′(x) cos [g(x)]. If f (x) = cos [g(x)] then f ′(x) = −g ′(x) sin [g(x)]. g ′( x ) If f (x) = tan [g(x)] then f ′(x) = = g ′(x) sec2 [g(x)]. cos2 [ g( x )]
• If f (x) = tan (x) then f ′(x) = • • • • • •
The product rule
• The product rule of differentiation states: dy dv du 1. if y = u × v, then =u× +v× dx dx dx 2. if f (x) = u(x) × v (x) then f ′(x) = u(x) × v′(x) + v(x) × u′(x). The quotient rule
• The quotient rule of differentiation states: du dv v× −u × u dy dx dx . 1. if y = then = v dx v2 u( x ) 2. if f ( x ) = then v( x) v ( x )u ′( x ) − u( x )v ′( x ) f ′( x ) = [v ( x)] x 2 • Summary of derivatives: f (x)
f ′ (x)
c
0
axn
naxn − 1
[g(x)]n
ng′(x)[g(x)]n − 1
ex
ex
ekx
kekx
eg(x)
g′(x)eg(x)
loge (x)
1 x 1 x g ′( x ) g( x ) a cos (ax) −a sin (ax) a cos2 (ax )
loge (kx) loge [g(x)] sin (ax) cos (ax) tan (ax)
(= a sec2 (ax))
Chapter 7
Differentiation
367
CHAPTER REVIEW y
SHORT ANSWER
1 The graph of a cubic function is shown below. y
−4
−1 0
1 2 3
3
4
5
Sketch the graph of its gradient function. h 3 + 2h 2 + 4 h Find lim . h→0 h a Find the derivative of f (x) = x3 + 2x using first principles. b Hence find the gradient at the point where x = 1. x3 a Find the gradient function if g( x ) = − 4x . 3 b Find the gradient of g (x) when x = 3. 3x 4 x3 If h( x ) = + − 3 x, find: 2 4 a h′ (x) b i h′(−1) ii h′(2). 3 (4 x + 1) 2 ,
6 If y = find the gradient when a x = 2 and b x = 1. 7 Find the derivative of x 2 + 4 . 8 Differentiate f (x) = e2x − 1. −x2
9 a Find f ′ (x) if f (x) = e . b What is the value of x when f ′ (x) = 0? dy if y = loge (2x3 − 4). 10 Find dx 11 The tangent to the curve f (x) = loge (ax − 1) when x = 2, has a gradient of 1. Find the value of a.
a On the same set of axes, sketch the graph of the derivative function. b Write down the domain of the derivative function. EXAM TIP
a Common errors were drawing the cusp point at x = 0 as a closed circle as well as the point at x = 3. b In this question students often gave responses that were inconsistent with the graph they had drawn in a. Union and intersection were sometimes confused as was the use of round, square or curly brackets. [Assessment report 1 2007]
[© Vcaa 2007]
19 Find the derivative of y = f (x) where f (x) = x2 − 2x . 20 Find the derivative of y = f (x) where f (x) = sin (x) − π π for x ∈ , . 2 2 MULTIPLE CHOICE
1 The graph of f (x) is shown below. y
0
13 Find f '(x) if f (x) = tan (5x). 14 Find f '(x) if f (x) = tan (2x2 − 3). dy 15 If y = 3x2 loge (6x), find . dx cos ( x 2 ) 16 Find f '(x) if f ( x ) = . x2 17 Differentiate esin (2x).
f(x)
4
12 Find f '(x) if f (x) = 3 sin (2x) and hence find f '(2π).
3
The graph of its gradient function is: y b a
18 The diagram (top, right) shows the graph of a function with domain R.
368
x
3
x
f(x)
2
0
−3
Maths Quest 12 Mathematical Methods CAS for the TINspire
0
3
x
x
y 0
3
x
c
y
0
e
y
d
3
c
x
0
x
3
y 4
e
4
2 4x − 9 −6x
9 If y = 5e
then
−
0
a −30e 6x − c 5e 6x − 1 − e 5e 7x
x
2 For the function g(x) graphed below, the gradient function g′(x) is defined over the domain: y a R b R \{1} c R \{4} d R \{1, 4} x e [1, 4] 0 1 4 x2 − 2x − 8 is: x→4 x−4 a undefined b 0 c 1
d 4
− 5x is: x +1 a 22 b −22 c 6 d undefined 5 The derivative of f (x) = 4x3 − x2 + 3x is: a 12x2 − 2x + 3 b 4x2 − 2x + 3 c 12x2 − 2x d 12x2 − x + 3 e 4x2 − 2x 1 6 The derivative of g( x ) = 2 − 2 x is: x −1 −2 2 2 a − b 3 − x x x x −2 −2 1 2 c d − − x x3 x x −2 1 − 3 e x3 x2 4 lim−
e 6
x3
7 The derivative of (2x + 5)6 is: a 6(2x + 5)5 b 12x(2x + 5)5 5 d 12(2x + 5) e 12(2x + 5)4 1 8 The derivative of is: 4x − 9 b
e
−6
c 6x(2x + 5)5
−2
(4 x − 9)3
dy is equal to: dx − b −6e 6x − d −30e 6x − 1 dy is: dx
a 4e4x + 6 c e4x + 7 e 4e3x + 7
b e4x + 6 d 4e4x + 7
11 The derivative of loge (3x − 2) is: 1 1 a b 3x − 2 3x 1 3(3 x − 2)
e
c
1 x
3 3x − 2
12 The derivative of 2 loge (x2 + x) is: 2(2 x + 1) c 2x + 1 x x2 + x 4x e 2 x +x dy 13 If y = cos (8x) then is: dx a 8 sin (8x) b sin (8x) 2(2 x + 1) x2 + x 2x d 2 x +x a
x→ 3
a 2 4x − 9
10 If y = e4x + 7 then
d
3 The value of lim
d 4 4x − 9
3
(4 x − 9) 2
b
c −8 sin (8x) e
−sin
d −8x sin (8x)
(8x)
14 If y = 2 sin (2x + 3) then a 4x cos (2x + 3) c 4 cos (2x + 3) e 2 cos (2x + 3)
dy is equal to: dx b −4 cos (2x + 3) d 4 cos (2x)
15 If f (x) = tan (6 − 5x), then f ′(x) is equal to: −5 5 a b 2 (6 − 5x 2 cos s 5 ) coss (6 − 5 x ) 2 2 c sec (6 − 5x) d 5 cos (6 − 5x) e −5 cos2 (6 − 5x) 16 If f (x) = x2 e2x then f ′ (x) is equal to: a 2xe2x + 2x2e2x c 4xe2x e 2xe2x + x2e2x
b 2xe2x d 2xe2x − 2x2e2x
Chapter 7
Differentiation
369
17 If g (x) = 2x loge (3x) then g′(x) must be: 2
a 2 log e (3x 3 )+ 3
b 2 loge (3x) + 2
c 2 loge (3x) + 6x
3 )− 3 d 2 log e (3x
e 2 loge (3x) + 6x loge (3x) 2x + 1 18 The derivative of is: x−2 4x − 5 a ( x − 2)2 4x − 3 c ( x − 2)2 −5 e ( x − 2)2 19 The derivative of
2
b
−3
( x − 2)2
d 4x − 5
e4 x is: x2
( x − 2)e 4 x x3 2(2 x − 1)e 4 x c x3 a
b
2(1 − 2 x ) x3
d
x 2 e 4 x − 2e 4 x x4
2e 4 x x3 20 If g(x) = (x2 + 3x − 7)5 then g′(x) is equal to: a 5(x2 + 3x − 7)4 b (2x + 3)(x2 + 3x − 7)4 4 c 5(2x + 3) d 5(2x + 3)(x2 + 3x − 7)4 2 4 e (x + 3x − 7) e
21 The derivative of sin (x) cos (x) is: a 2 sin (x) cos (x) b sin2 (x) − cos2 (x) 2 2 c sin (x) + cos (x) d cos2 (x) − sin2 (x) − 2 2 e sin (x) − cos (x) 22 Which one of the following is not true about the function f: R → R, f (x) = 2x + 4? a The graph of f is continuous everywhere. b The graph of f ′ is continuous everywhere. c f (x) ≥ 0 for all values of x. d f ′(x) = 2 for all x > 0 e f ′(x) = −2 for all x < −2 [© Vcaa 2007] 23 If y = cos (x), the rate of change of y with respect to x at x = k, π < k < 3π , is: 2 2 a −sin (k) b sin (k) c −cos (k) d −sin (1) e sin (1) [© Vcaa 2005] 24 Let f: R → R be a differentiable function. For all real values of x, the derivative of f (e2x) with respect to x will be equal to: a 2e2x f ′(x) b e2x f ′(x) c 2e2x f ′(e2x) d 2f ′(e2x) e f ′(e2x) [© Vcaa 2005]
EXTENDED RESPONSE 1
1 A section of a rollercoaster ride follows part of the curve with the equation y = 200 ( x 3 + 30 x 2 ) as shown below. a For what values of x (domain) is the gradient: y i zero? ii positive? iii negative? b Sketch the gradient function. c Use the graph of the gradient function to find the value of x where the 0 −28 −20 gradient is steepest over the domain [−25, 10]. dy d Find . dx e Find the gradient where x equals: i −25 ii −10 iii 10. f Does this verify your answer to part c? Brieﬂy explain. g What is the highest point reached by the rollercoaster? (Give your answer in metres.) 2 Consider the functions f ( x ) = 2 x and g(x) = x2 + 1. a State the domain and range for each function.
370
Maths Quest 12 Mathematical Methods CAS for the TINspire
12
x
b Find the composite functions: i f (g(x)) ii g(f (x)). c State the domain and range for f (g(x)) and g(f (x)). d Find: d i ( f ( g( g( x )))) dx d ii ( g( f ( x )))) dx e Evaluate: i f ′(g(2)) ii g ′(f (2)) 4 − x 2 , x ≤ 2 f ( x ) = 3 Consider the function 2 x − 4, 2 < x < 5 . x − 1, x ≥ 5 a Sketch the graph of f (x). b For what values of x is f (x) discontinuous? c For what values of x is f (x) not differentiable? d Find f ′(x). e Sketch the graph of f ′(x).
eBook plus Digital doc
Test Yourself Chapter 7
Chapter 7
Differentiation
371
eBook plus
ACTIVITIES
chapter opener Digital doc
• 10 Quick Questions: Warm up with ten quick questions on differentiation. (page 324) 7a
Review — gradient and rates of change
Tutorial
• WE 3 int0552: Watch how to sketch a gradient function. (page 327) Digital docs
• SkillSHEET 7.1: Practise identifying positive negative and zero gradients. (page 328) • SkillSHEET 7.2: Practise sketching the gradient function given the original function. (page 328) 7b
Limits and differentiation from first principles
eLesson eles0093
• Limits and differentiation from first principles. Watch an eLesson on related rates of change (page 332) Tutorial
• WE 5 int0553: Watch how to evaluate limits. (page 333) 7c
The derivative of x n
Digital doc
• SkillSHEET 7.3: Practise using index laws. (page 340) 7d
The chain rule
Tutorial
• WE 13 int0554: Watch a worked example on using the chain rule. (page 341) Digital doc
• WorkSHEET 7.1: Sketch gradient functions, identify where the derivative exists, evaluate limits, apply first principles and differentiation rules to determine derivatives. (page 344) 7G
The derivatives of sin (x), cos (x) and tan (x)
7H
Digital doc
• WorkSHEET 7.2: Differentiation of mixed expression involving the product, quotient and chain rules. (page 357) 7I
The quotient rule
Tutorial
• WE 30 int0556: Watch a worked example on how to use the quotient rule. (page 358) 7J
Mixed problems on differentiation
Interactivity int0252
• Differentiation: Consolidate your understanding of differentiation. (page 360) Tutorial
• WE 33 int0557: Watch a worked example on how derivatives of an absolute value function. (page 362) chapter review Digital doc
• Test Yourself: Take the endofchapter test to test your progress. (page 371) To access eBookPLUS activities, log on to www.jacplus.com.au
Tutorial
• WE 26 int0555: Watch a worked example on using the chain rule to differentiate trigonometric functions. (page 353)
372
The product rule
Maths Quest 12 Mathematical Methods CAS for the TINspire
8
8A Equations of tangents and normals 8B Sketching curves 8c Maximum and minimum problems when the function is known 8D Maximum and minimum problems when the function is unknown 8e Rates of change 8F Related rates 8G Linear approximation
Applications of differentiation AreAS oF STudy
• Application of differentiation to curve sketching and identification of: – key features of curves – intervals over which a function is constant, stationary, strictly increasing or strictly decreasing – the maximum rate of increase or decrease in a given application context (consideration of second derivative is not required) – tangents and normals to curves
• Identification of: – local maximum/minimum values over an interval and application to solving problems – interval end point maximum and minimum values • Average and instantaneous rates of change, including formulation of expressions and solution and interpretation of problems involving rates of change and simple cases of related rates of change • The relationship f ( x + h) ≈ f ( x ) + hf hf '( x ) for a small value of h and its geometric interpretation eBook plus
8A
Digital doc
equations of tangents and normals As we have seen, a tangent to a curve is a straight line that touches the curve at a given point and whose gradient represents the gradient of the curve at that point. A normal to a curve is a straight line passing through the point where the tangent touches the curve and is perpendicular (at right angles) to the tangent at that point.
10 Quick Questions
y
0
f(x)
Tangent
Point of tangency Normal x
− 1 If the gradient of the tangent to a curve is m, then the gradient of the normal is (as the m − product of the gradients of two perpendicular lines equals 1).
The equation of a straight line passing through the point (x1, y1) and having a gradient of m is: y − y1 = m(x − x1) The gradient of the tangent at x = a is f ′(a). − 1 Therefore the gradient of the normal is . f ′(a)
Chapter 8
Applications of differentiation
373
y
The equation of the tangent at x = a is y − f (a) = f ′(a) (x − a) and the equation of the normal is −1 y − f (a) = ( x − a). f ′(a)
f(x)
f(a) 0
(a, f(a)) a
Worked exAmple 1
Find the equation of the tangent to y = x3 − 2x + 3 at the point (1, 2). Think
WriTe
1
Write the equation.
2
Find
dy . dx
dy where x = 1 to find the gradient of dx the tangent where x = 1.
y = x3 − 2x + 3 dy = 3x2 − 2 dx
Evaluate
At x = 1, dy =3−2 dx =1 So gradient of tangent is 1.
4
Substitute (1, 2) for (x1, y1) and m = 1 into the rule for the equation of a straight line: y − y1 = m(x − x1).
Equation of tangent at the point (1, 2) is y − 2 = 1(x − 1) y−2=x−1
5
Rearrange the rule to a simple form.
3
y=x+1
Worked exAmple 2
Find the equation of: a the tangent b the normal to the curve with the equation y = 3 loge (2x) at x = 1 c the tangent and the normal to the curve y = 3 loge (2x) at x = 1 using a CAS calculator. Think
Write the equation.
y = 3 loge (2x)
2
Evaluate y when x = 1. dy Find . dx
At x = 1, y = 3 loge (2) dy 3(2) = dx 2x 3 = x dy 3 a At x = 1, = dx 1 =3 So gradient of tangent is 3.
a
374
Tutorial
int0558 Worked example 2
WriTe
1
3
eBook plus
dy when x = 1 to obtain the dx gradient of the tangent at x = 1.
1
Evaluate
2
Determine the equation of the tangent at (1, 3 loge (2)).
Equation of tangent is y − 3 loge (2) = 3(x − 1) = 3x − 3 y = 3x − 3 + 3 loge (2)
maths Quest 12 mathematical methods CAS for the Tinspire
x
b
1
Evaluate the gradient of the normal which is
c
−1
dy dx
−1 . 3
b Gradient of normal is
.
2
Determine the equation of the normal at (1, 3 loge (2)).
1
On a Calculator page, press: • MENU b • 4:Calculus 4 Select either 9:Tangent Line 9 or A:Normal Line A. Complete the entry lines as: tangentLine(3ln (2x),x,1) normalLine(3ln (2x),x,1) Press ENTER · after each entry.
2
Note the answers from the CAS are not written as equations.
Equation of normal is 1 y − 3 loge (2) = 3 (x − 1) 3y − 9 loge (2) = −1(x − 1) = −x + 1 x + 3y = 1 + 9 loge (2) −
c
∴ y = 3 x + 3 (log e (2 − 1)) is the equation of the tangent. −x 9 log e (2 + 1) ∴y= is the equation + 3 3 of the normal.
rememBer − 1 1. If m is the gradient of the tangent, then is the gradient of the normal. That is, − m m1 × m2 = 1. 2. The equation of a straight line passing through the point (x1, y1) with gradient m is y − y1 = m(x − x1)
3. Parallel lines have the same gradient. exerCiSe
8A
equations of tangents and normals 1 We1 Find the equation of the tangent to the curve y = x2 + x at the point (2, 6). 2 3
Find the equations of the tangent to the curve y = x2 + 5x − 6 at the points where it crosses the xaxis.
eBook plus Digital doc
Spreadsheet 129 Tangent and normal
Find the equation of the normal to the curve y = 3x2 − 5x + 4 at the point where x = 1.
Chapter 8
Applications of differentiation
375
4
Find the equation of the normal to the curve y = 12 x2 + 3x − 7 at the point where it crosses the yaxis.
5 We2 For each of the following functions, find the equation of: i the tangent ii the normal at the given value of x. a y = x2 + 1, x = 1 b y = x3 − 6x, x = −2 1 c y = , x = 2 d y = (x − 1)(x2 + 2), x = −1 x e g i k
y = x , x = 4 y = x(x + 2)(x − 1), x = −1 y = 2x3 + x2 − 6x + 2, x = 1 y = e3x + 2, x = −1
y = 2 x + 3, x = 3 y = x3 − 3x2 + 4x, x = 0 y = e2x, x = 0 y = loge (x), x = 2 π n y = sin (2x), x = 3 π p y = sin 2 x + , x = 0 4 f h j l
m y = loge (2x + 3), x = 0 x o y = 3 cos , x = π 2
6 mC If y = (2x + 3)4 then at the point (−1, 1) a the equation of the normal is: A y + 4x − 3 = 0 B 8y + x − 7 = 0 D y + 2x + 8 = 0 e 2y − x = 0
c y − 4x + 5 = 0
b the value of x where the gradient of the tangent is parallel to the xaxis is: A
−2 3
B
1 3
c
−3 2
D
−1 3
e
2 3
7 Find the equation of the tangent to f (x) = x2 + 4x + 1 which is parallel to the line y = 2x + 3. x2 + 1 at x = 0. x2 − 1 π 9 Find the equation of the normal to y = x sin (x) at x = . 2 10 Find the equation of the normal to y = loge (x + 2) which is parallel to the line with equation y + 3x − 5 = 0. 8 Find the equation of the tangent to y =
11 Find the equations of the tangent and normal for each of the following curves. a f (x) = x2 + 1 at x = a b f (x) = x at x = a
eBook plus Digital doc
WorkSHEET 8.1
c f (x) = e x at x = 2a. 2
12 Find the equation of the tangent to the curve f (x) = e4x that is perpendicular to the line x + 8y = 16. 13 The graph of y = x has a normal with equation y = −8x + b, where b is a real constant. Find the value of b.
8B
Sketching curves When the graphs of polynomial functions are being sketched, four main characteristics should be featured: 1. the basic shape (whenever possible) 2. the yintercept 3. the xintercept(s) 4. the stationary points.
376
maths Quest 12 mathematical methods CAS for the Tinspire
Stationary points A stationary point is a point on a graph where the function momentarily stops rising or falling; that is, it is a point where the gradient is zero. y
y
or x Gradient = 0 where function stops rising momentarily, then continues to rise again after this point
0 0
x Function stops falling and rises after this point
The stationary point (or turning point) of a quadratic function can be found by completing a perfect square in the form y = (x + h)2 + k. In this case the stationary point is (−h, k). For cubics, quartics or higherdegree polynomials there is no similar procedure. Differentiation enables stationary points to be found for any polynomial function where the rule is known. The gradient function of a function f (x) is f ′(x). Stationary points occur wherever the gradient is zero. f (x) has stationary points when f ′(x) = 0 or y has stationary points when
dy = 0. dx
The solution of f ′(x) = 0 gives the xvalue or values where stationary points occur. If f ′(a) = 0, a stationary point occurs when x = a and y = f (a). So the coordinate of the stationary point is (a, f (a)).
Types of stationary points There are four types of stationary points. 1. A local minimum turning point at x = a. If x < a, then f ′(x) < 0 (immediately to the left of x = a, the gradient is negative). If x = a, then f ′(x) = 0 (at x = a the gradient is zero). If x > a, then f ′(x) > 0 (immediately to the right of x = a, the gradient is positive).
y
f(x)
f '(x) > 0 f '(x) < 0 0
2. A local maximum turning point at x = a. If x < a, then f ′(x) > 0. If x = a, then f ′(x) = 0. If x > a, then f ′(x) < 0. The two cases (1 and 2) can be called ‘turning points’ because the gradients each side of the stationary point are opposite in sign (that is, the graph turns). The term ‘local turning point at x = a’ implies ‘in the vicinity of x = a’, as polynomials can have more than one stationary point.
f '(x) = 0 x
a
y 0
a
x
f(x)
Chapter 8 Applications of differentiation
377
y
3. A positive stationary point of inflection at x = a. If x < a, then f ′(x) > 0. If x = a, then f ′(x) = 0. If x > a, then f ′(x) > 0. That is, the gradient is positive either side of the stationary point.
Gradient = 0
0
x
a
f'(x)
4. A negative stationary point of inflection at x = a. If x < a, then f ′(x) < 0. If x = a, then f ′(x) = 0. If x > a, then f ′(x) < 0. In cases 3 and 4 above the word ‘stationary’ implies that the gradient is zero. Not all points of inflection are stationary points. y
y
f(x)
Gradient = 0
x
0
a
y
or 0
x
0
x
Gradient ≠ 0
When determining the nature of stationary points it is helpful to complete a ‘gradient table’, which shows the sign of the gradient either side of any stationary points. This is known as the ﬁrst derivative test. Gradient tables are demonstrated in the examples that follow.
Worked exAmple 3
eBook plus
a Find the stationary points and their nature for the function b c d e
f (x) = x3 + 5x2 − 8x − 12. Show that the curve passes through (−1, 0). Find the coordinates of all other intercepts. Hence, sketch the graph of f (x). Use a CAS calculator to find the stationary points.
Think a
378
Tutorial
int0559 Worked example 3
WriTe/drAW a f (x) = x3 + 5x2 − 8x − 12
1
Write the rule for f (x).
2
Differentiate f (x) to find f ′(x).
f ′(x) = 3x2 + 10x − 8
3
Find all values of x where f ′(x) = 0.
For stationary points 3x2 + 10x − 8 = 0 (3x − 2)(x + 4) = 0 x = 23 or x = −4
maths Quest 12 mathematical methods CAS for the Tinspire
4
2
2
2
2
2
When x = 3 , f ( 3 ) = ( 3 )3 + 5( 3 )2 − 8( 3 ) − 12
Find the value of f (x) for each value of x where f ′(x) = 0.
22
= −14 27 so ( 23 , −14 22 ) is one stationary point. 27 When x = −4, f (−4) = (−4)3 + 5(−4)2 − 8(−4) − 12 = 36 so (4, 36) is another stationary point.
5
6
b
c
d
Complete a gradient table to determine the nature of the stationary points.
To find the xintercepts, factorise f (x) by long division, or by another appropriate method, knowing that (x + 1) is a factor of f (x).
x
−5
−4
0
2 3
1
f ′(x)
+
0
−
0
+
Slope
/
—
\
—
/
Therefore (−4, 36) is a local maximum stationary point and ( 23 , −14 22 ) is a local minimum stationary 27 point.
State each stationary point and its nature.
Show that f (−1) = 0.
1
Gradient table:
b f (−1) = (−1)3 + 5(−1)2 − 8(−1) − 12
= −1 + 5 + 8 − 12 =0 Therefore f (x) passes through (−1, 0).
c As f (−1) = 0 then (x + 1) is a factor of f (x)
and f (x) = (x + 1)(x2 + 4x − 12) = (x + 1)(x + 6)(x − 2)
2
Solve f (x) = 0.
xintercepts: (x + 1)(x + 6)(x − 2) = 0 x = −1 or −6 or 2
3
State the coordinates of the xintercepts.
The xintercepts are (−1, 0), (−6, 0) and (2, 0).
4
Evaluate f (0) to determine the yintercept.
5
State the coordinate of the yintercept.
The yintercept is (0, −12).
Sketch the graph of f (x) showing all intercepts and stationary points.
f (0) = (0)3 + 5(0)2 − 8(0) − 12 = −12
d (−4, 36)
f(x)
y
(−6, 0) (−1, 0)
0
(2, 0) x
(0, −12) —) ( 2–3 , −14 22 27
Chapter 8 Applications of differentiation
379
e
1
To find the stationary points with a CAS calculator, on a Calculator page: • define the function f (x) • find f ′(x) • solve f ′(x) = 0 • find the yvalues of the stationary points. To do this, complete the entry lines as: Define f (x) = x3 + 5x2 − 8x − 12
e
d ( f ( x)) dx solve(3x2 + 10x − 8 = 0,x) 2 f − 4, 3 Press ENTER · after each entry. 2
∴ The stationary points are (−4, 36) and 23 ,
Write down the coordinates of the stationary points.
− 400 27
.
Worked Example 4
Sketch the graph of g(x) = x2(4 − x2), clearly indicating all stationary points and intercepts. Think
Write/draw
1
Write the rule for g(x).
g(x) = x2(4 − x2)
2
Expand g(x) to make it easier to differentiate.
g(x) = 4x2 − x4
3
Differentiate g(x).
g′(x) = 8x − 4x3
4
Solve g′(x) = 0.
For stationary points, g′(x) = 0 8x − 4x3 = 0 4x(2 − x2) = 0 x = 0 or x2 = 2 x = 0 or x = − 2 or 2
5
Find g(x) for each value of x where g′(x) = 0.
When x = 0, g(0) = 0 When x = − 2 , g(− 2 ) = 4(− 2 )2 − (− 2 )4 =4 When x = 2, g( 2) = 4( 2)2 − ( 2)4 =4 Therefore the stationary points are (− 2 , 4), (0, 0) and ( 2, 4).
6
380
Complete a gradient table to determine the nature of the stationary points.
Gradient table: x
−2
g′(x)
+
Slope
/
Maths Quest 12 Mathematical Methods CAS for the TINspire
−
−1
0
1
0
−
0
+
0
−
—
\
—
/
—
\
2
2
2
7
State the stationary points and their nature.
Therefore (− 2 , 4) is a local maximum stationary point. (0, 0) is a local minimum stationary point. ( 2, 4) is a local maximum stationary point.
8
Solve g(x) = 0 to determine the xintercepts.
xintercepts: When g(x) = 0, x2(4 − x2) = 0 x2 = 0 or x2 = 4 x = 0 or x = −2 or 2 The xintercepts are (−2, 0), (0, 0) and (2, 0).
9
Find g(0) to determine the yintercept.
yintercept: When x = 0, g(0) = 02(4 − 02) =0 The yintercept is (0, 0).
10
Sketch the graph of g(x).
(− 2, 4)
y
( 2, 4)
(−2, 0) (0, 0) 0
(2, 0)
x g(x)
Worked Example 5
If f (x) = x3 + 4x2 − 3x − 7: a sketch the graph of f ′(x) b state the values of x where f (x) is i increasing and i i decreasing. Think a
Write/draw a f (x) = x3 + 4x2 − 3x − 7
1
Write the rule for f (x).
2
Differentiate f (x) to find f ′(x).
f ′(x) = 3x2 + 8x − 3
3
Solve f ′(x) = 0 to find the xintercepts of f ′(x).
xintercepts: When f ′(x) = 0, 3x2 + 8x − 3 = 0 (3x − 1)(x + 3) = 0 x = 13 or −3 The xintercepts of f ′(x) are ( 13 , 0) and (−3, 0).
4
Evaluate f ′(0) to find the yintercept of f ′(x).
5
Sketch the graph of f ′(x) (an upright parabola).
yintercept: When x = 0, f ′(0) = −3 so the yintercept of f ′(x) is (0, −3). y
−3
0
f '(x)
1– 3
x
−3
Chapter 8 Applications of differentiation
381
b
i
f (x) increases where f ′(x) > 0. By inspecting the graph of f ′(x) deduce where f ′(x) is positive (that is, above the xaxis).
b
i f ′(x) > 0 where x < −3 and x >
so f (x) is increasing where x < −3 and x > 13 .
i i f ′(x) < 0 where −3 < x <
ii f (x) decreases where f ′(x) < 0.
By inspecting the graph of f ′(x), deduce where f ′(x) is negative (that is, below the xaxis).
1 3
1 3
so f (x) is decreasing where −3 < x < 13 .
Worked Example 6
Consider the function f ( x ) = ( x + 3)( x − a)2 where a is a positive real constant. a Find f ′(x). b Find the coordinates of the stationary points. c Determine the nature of the stationary points. d Find the value of a in exact form if the straight line with equation y = 5x + 15 intersects y = f (x) at the maximum turning point. Think a
b
382
Write
1
Store the function to f (x). On a Calculator page, press: • MENU b • 1:Actions 1 • 1:Define 1 Complete the entry line as: Define f (x) = (x + 3)(x − a)2 Press ENTER ·.
2
On the same Calculator page, select the derivative template: Complete the entry line as: d ( f ( x)) dx Press ENTER ·.
3
Write the derivative with the correct notation.
1
Solve f ′(x) = 0 to find the xvalues of each stationary point. On the same Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve((x + 3)(x − a)2 = 0, x) Press ENTER ·.
a
∴ f ′(x) = (x − a)(3x − a + 6) b
Maths Quest 12 Mathematical Methods CAS for the TINspire
2
a − 6 4(a + 3)3 ∴ and (a, 0) are the , 27 3
Substitute each value of x into f (x) to find the ycoordinates of the stationary points. On the same Calculator page, complete the entry line as:
coordinates of the stationary points.
a − 6 f , a . 3 Press ENTER ·. Write the stationary points in coordinate form. c
1
To determine the nature of the stationary points, let a = 1 (as a is given as a positive real constant).
c 1 − 6 4(1 + 3)3
3 ,
− 5 256 ∴ , and (1, 0) are the stationary points. 3 27
Substitute a = 1 into the answer found previously to obtain the stationary points. 2
3
4
d
Complete a gradient table to determine the nature of the stationary points.
State the stationary points and their nature.
The maximum turning point lies on the straight line y = 5x + 15. Substitute the coordinates of the turning point into y = 5x + 15 and solve for a.
2
On a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as:
x
−2
f ′(x)
+
Slope
/
−5
0
1
2
0
−
0
+
—
\
—
/
3
(1, 0) is a local minimum turning point. − 5 256 3 , 27 is a local maximum turning point. ∴ (a, 0) is a local minimum turning point and
State the stationary points and their nature in terms of a.
1
and (1, 0)
27
a − 6 4(a + 3)3 is a local maximum turning point. 3 , 27 d
4 ( a + 3)3 5(a − 6) solve = + 15, a 27 3 Press ENTER ·. 3
Select the appropriate value of a given that it is positive.
∴a=
3
(
5−2
)
2
Chapter 8 Applications of differentiation
383
REMEMBER
1. A stationary point (SP) occurs when f ′(x) = 0. 2. There are four types of stationary points: (a) local maximum, where the gradient is positive on the left of the SP and negative on the right (b) local minimum, where the gradient is negative on the left of the SP and positive on the right (c) positive point of inflection, where the gradient is positive on both sides of the SP (d) negative point of inflection, where the gradient is negative on both sides of the SP. Exercise
8B
Sketching curves 1 WE3 Find the stationary points and their nature for each of the following functions. a y = 8 − x2 b f (x) = x3 − 3x c g(x) = 2x2 − 8x 2 3 3 4 d f (x) = 4x − 2x − x e g(x) = 4x − 3x f y = x2(x + 3) 2 3 g y = 5 − 6x + x h f (x) = x + 8 i y = −x2 − x + 6 4 3 2 2 j y = 3x − 8x + 6x + 5 k g(x) = x(x − 27) l y = x3 + 4x2 − 3x − 2 3 3 m h(x) = 12 − x n g(x) = x (x − 4) 2 WE3 Sketch the graph of each function in question 1 , clearly indicating all stationary points. 3 a WE3 Find the stationary points of the function f (x) = x3 − 2x2 − 7x − 4 and state their nature. b Show that the graph passes through (4, 0). c Give the coordinates of all other intercepts and hence sketch the graph of f (x). 4 a Find the stationary points, and their nature, for the curve y = x3 − x2 − 16x + 16. b Show that the graph passes through (1, 0) and give the coordinates of all other intercepts. c Sketch the graph. 5 a WE3 Find the stationary points of the function g(x) = x4 − 4x2 and state their nature. b Find the coordinates of all the intercepts. c Sketch the graph of g(x). 6 a b c d
If y = x4 − 6x2 + 8x − 3, find each stationary point and its nature. Show that the point (1, 0) lies on the curve. Find all other intercepts. Sketch the graph.
7 a If y = x4 + x3 − 5x2 − 6, find each stationary point and its nature. b Find the yintercept. c Sketch the graph without finding the xintercepts. 8 WE4 Sketch the graph of each of the following functions, clearly indicating all stationary points and intercepts. a f (x) = x4 − x2 b f (x) = x3 − 3x2 3 4 c g(x) = x + 3x d g(x) = x3 − 4x2 + 4x 3 2 e h(x) = x − 4x − 11x + 30 f h(x) = x(x + 3)(x − 5) g f (x) = x4 − 2x2 + 1 h f (x) = x(x2 + 1) 3 2 i g(x) = x + 9x + 24x + 20 j h(x) = (x2 − 1)3 9 MC If f ′(x) < 0 where x > 2 and f ′(x) > 0 where x < 2, then at x = 2, f (x) has a: A local minimum B local maximum c point of inflection D discontinuous point e gradient of 2 384
Maths Quest 12 Mathematical Methods CAS for the TINspire
10 mC The function f (x) = x3 + x2 − 8x − 3 has stationary points when x is equal to: A −2 and
4 3
c −2 only
B 3 only
D 3 and
4 3
e 0 and 2
11 mC The graph of y = + has: A a local maximum where x = 0 B a local minimum where x = 0 x4
x3
−3 4
c a local minimum where x =
−3 4
D a local maximum where x =
−4 3
e a local maximum where x =
12 mC A quadratic function has a turning point (2, 1) and a yintercept of (0, 9). The equation must be: A y = (x − 2)2 + 9 B y = (x − 1)2 + 8 c y = (x − 2)2 + 1 2 2 D y = 2(x − 2) + 9 e y = 2(x − 2) + 1 13 The graphs of f ′(x) are shown below. Find all values of x for which f (x) has stationary points and state their nature. y y y a b c
f '(x)
f '(x)
x
0
−3
−2
0
1
−2
x
4
0
3
x
f'(x) y
d
−5
0
2
y
e
f '(x)
x
−3
f
0
x
2
y f '(x)
0 1
5
x
f'(x)
14 Show that f (x) = x2 − 4x + 3 is decreasing for x < 2 and increasing for x > 2. 15 We5 For each of the following functions i sketch f ′(x) and, hence, state the values of x where f (x) is ii increasing and iii decreasing. a f (x) = 13 x3 + 2x2 + 2 b g(x) = x3 + 2x2 − 7x − 5 c h(x) = x4 + 4x3 + 4x2 16 If y = f (x) has the following properties then sketch its graph. f ′(x) = 0 if x = −2 and x = 3 f ′(x) < 0 if −2 < x < 3 f ′(x) > 0 for all other x −
17 If f (x) = x3 + ax2 + bx has stationary points at x = 2 and x = 43 , find: a the value of a and b b the nature of the stationary points. eBook plus Digital doc
SkillSHEET 8.1 Review of derivatives other than polynomials
18 If f (x) = x4 + ax2 + b has a stationary point (1, 4), find: a a and b b the other stationary points c the nature of each stationary point. 19 We6 Consider the function f ( x ) = (b − x )( x + 2)2 where b is a positive real constant. a Find f ′(x). b Find the coordinates of the stationary points. c State the nature of the stationary points. d Find the equation of the tangent at x = 2. e If the maximum turning point occurs at x = 4, find the value of b.
Chapter 8
Applications of differentiation
385
20 WE6 Consider the function f ( x ) = ( x + b)2 (2 x − 1) where b is a positive real constant. a Find f ′(x). b Find the coordinates of the stationary points. c Find the equation of the tangent to the curve when x = 2. d Find the value of b in exact form if the straight line with equation y = 4x − 2 intersects y = f (x) at the minimum turning point.
8c
Maximum and minimum problems when the function is known In many practical situations the maximum local or minimum value of a quantity is desired. y maximum maximum For example it is important for manufacturers or business operators to minimise the costs involved in running their businesses. Equally, it is just as important to maximise their profits. local A graph is always useful and helps us to find minimum x 0 approximately where a maximum or minimum minimum occurs. As we have seen, local maximum and/or minimum stationary points occur where the derivative is zero. dP If P = f (x) then a local maximum and/or minimum may exist where = 0. dx To decide whether a solution is a maximum or a minimum the first derivative test must be applied (by setting up a gradient table). A graph of the function could be sketched if it is not difficult, to ensure that the maximum or minimum value is applicable. y f(x)
0
1 2 3 4 5
x
The function f (x) has a limited domain of [1, 5]. The maximum value occurs at the point where x = 4. The minimum value, however, is not at x = 2 but at the end point x = 5. This example shows that the derivative test on its own is not always reliable for finding maximum or minimum values and a graph is sometimes necessary.
Solving maximum and minimum problems When finding the maximum/minimum value of f (x) the following steps are taken. 1. Find f ′(x) (to obtain the gradient function). 2. Solve for x where f ′(x) = 0 (to find the values of x where the maximum or minimum occur). 3. Apply the first derivative test as a check. Also, sketch the graph of f (x). 4. Check end points if domain is restricted. 5. Substitute the appropriate value of x into f (x) to obtain the maximum or minimum.
386
Maths Quest 12 Mathematical Methods CAS for the TINspire
Worked exAmple 7
The population of a colony of birds at any time, t months, after observation began can be modelled by the function: −t
P(t) = 400 te 5 + 600, where P is the number of birds. Find: a the initial population b when the largest number of birds is reached c the maximum number of birds. Think a
b
WriTe
1
Write the rule for P(t).
2
The initial value occurs when t = 0 and is P(0).
1
Find P′(t) using the product rule.
a P(t) = 400te
P(0) = + 600 = 0 + 600 = 600 So the initial population of birds is 600. b P′(t) = 400te
= Solve P′(t) = 0.
+ 400t
( )
−t −t 400 e 5 − 80te 5 −t 80 e 5 (5 − t)
−t 5
−1 5
e
−t 5
+0
(5 − t) = 0
t = 5 (as 80 e Check that it is a maximum using the first derivative test.
−t 5
For maximum and/or minimum, P′(t) = 0 80 e
3
+ 600
400(0)e0
= 2
−t 5
−t 5
cannot equal 0)
Gradient table: t
0
5
10
P′(t)
+
0
−
Slope
/
—
\
Therefore the maximum population occurs after 5 months. c
Evaluate P(5) to find the maximum number of birds.
c P(5) = 400(5)e
−5 5
+ 600 − = 2000e 1 + 600 = 735.8 + 600 = 1335.8 Therefore the maximum number of birds is approximately 1335 (as it does not reach 1336).
Worked exAmple 8
eBook plus
The displacement of a particle moving in a straight line from the origin at any time, t, is given by x(t) = 13 t3 − 4t2 + 12 t + 1, 0 ≤ t ≤ 7. Find the maximum and minimum displacement.
Chapter 8
Tutorial
int0560 Worked example 8
Applications of differentiation
387
Think
Write
t3 − 4t2 + 12t + 1 3
1
Write the rule.
x(t) =
2
Find x′(t).
x′(t) = t2 − 8t + 12
3
Solve x′(t) = 0.
For maximum and/or minimum, t2 − 8t + 12 = 0 (t − 2)(t − 6) = 0 t = 2 or t = 6
4
Test for maximum and minimum using the first derivative test.
Gradient table: x
0
2
4
6
8
x′(t)
+
0
−
0
+
Slope
/
—
\
—
/
When t = 2 a local maximum occurs and when t = 6 a local minimum occurs. 5
Evaluate x(2) to find the local maximum.
x(2) = 13 (2)3 − 4(2)2 + 12(2) + 1
= 83 − 16 + 24 + 1
= 11 3
2 2
so a local maximum occurs at (2, 11 3 ). 6
Evaluate x(6) to find the local minimum.
x(6) = 13 (6)3 − 4(6)2 + 12(6) + 1 = 72 − 144 + 72 + 1 =1 so a local minimum occurs at (6, 1).
7
8
Sketch the graph of x(t) over the domain [0, 7] to test that the maximum and minimum have been found.
Evaluate x(0) and x(7) to see if the end points give a smaller or larger value.
This graph indicates that we have to find the value of x at the end points in case they produce larger or smaller values than the stationary points do.
x
0
x(t)
1 2 3 4 5 6 7 t
x(0) = 1 which is equal to the local minimum above x(7) = 13 (7)3 − 4(7)2 + 12(7) + 1
1
= 114 3 − 196 + 84 + 1 1
= 3 3 which is between the local maximum and minimum above. Therefore the maximum displacement is 2 11 3 units from the origin and the minimum displacement is 1 unit from the origin.
388
Maths Quest 12 Mathematical Methods CAS for the TINspire
REMEMBER
1. To find a maximum and/or minimum, let f ′(x) = 0. 2. If the function has a restricted domain, f ′(x) may not be useful for finding maximum and/or minimum values. 3. You can prove your maximum and/or minimum value using a slope or gradient diagram. Exercise
8c
Maximum and minimum problems when the function is known 1
For each of the following, write the expression for the derivative indicated. a C = 5x − 2x2 + 10,
dC dx
dV dh dM e M = k loge 3k, dk 1
2
c V = 4 h4 − 3 h3,
b P = n3 + 2n2 − 5 n, d h = t3 + 2e−t, f L = t2 +
dP dn
dh dt
4 dL − 2 cos (πt), t dt
Questions 2 and 3 refer to the following information. The function Q = n3 − 3n2 + 5, 0 ≤ n ≤ 5, is graphed at right.
Q
2 MC The minimum value occurs where n equals: A −2 B 0 c 1 d 2 e 5 3 MC The maximum value occurs where n is equal to: A 0 b 2 c 5 d 1 e −1
0
2
5
n
4 WE7 The profit, $P per week, of a small manufacturing company is related to the number of workers, n, by: P = − 12 n3 + 96n + 600 Find: a the number of workers needed for maximum profit per week b he maximum profit per week. 5
The cost, $C, of producing xmetre lengths of a certain tow rope is: C = 15 x2 − 8x + 100 per rope, x > 0 What is the length of the cheapest tow rope that can be produced?
6 WE8 The number of people, P, visiting a certain beach on a particular day in January depends on the number of hours, x, that the temperature is below 30 °C according to the rule P = x3 − 12x2 + 21x + 105 where x ≥ 0. Find the value of x for the maximum and minimum number of people who visit the beach. 7 The number of rabbits, N, which feed on a particular farmland on any night can be modelled − as N = 13 x3 + 5x2 + 75x + 500 where x is the average overnight temperature in °C and x ≥ −5. Find: a the temperature for the minimum and maximum number of rabbits b the minimum and maximum number of rabbits that will be found on the farmland.
Chapter 8 Applications of differentiation
389
8 The velocity, v cm/s, of an engine’s piston t seconds after the engine is started is approximated by v = 0.8 sin (2t). a Find the minimum velocity and the time it first occurs. b Find the maximum velocity and the time it first occurs. 9 The length of a snake, L cm, at time t weeks after it is born is modelled as: πt L = 12 + 6t + 2 sin , 0 ≤ t ≤ 20 4 Find: a the length at i birth and ii 20 weeks b R, the rate of growth, at any time, t c the maximum and minimum growth rate. 10 The population of cheetahs, P, in a national park in Africa since 1 January 1986 can be −t
modelled as N = 100te 12 + 500 where t is the number of years. a When does this model predict that the maximum population will be reached? b What is the maximum population of cheetahs that will be reached? c How many cheetahs will there be on 1 January in i 2010 and ii 2070? 11 In the same park as in question 10 the number of elephants, N, since 1 January 1986 is − t
modelled as N = 100 + 4t + 400 e 10 . Find: a the minimum number of elephants predicted b when this minimum will occur. 12 The profit, $P, per item that a store makes by selling n items of a certain type each day is P = 40 n + 25 − 200 − 2 n. a Find the number of items that need to be sold to maximise the profit on each item. b What is the maximum profit per item? c Hence, find the total profit per day by selling this number of items. 13 The weight (in kg) of a bodybuilder t months after starting a training program is W = 5t − 20 loge (t + 1) + 90, 0 ≤ t ≤ 15 a Find the weight of the bodybuilder at the start of the program. b Find the minimum weight obtained and how many months it takes to reach it.
8d
Maximum and minimum problems when the function is unknown When solving maximum and/or minimum problems when the function is not given directly, a rule for the function must be obtained from the given information. This rule should have the quantity being maximised or minimised in terms of one variable only. Sometimes a diagram will assist in establishing the rule. Then solve the problem using differentiation. These steps should be followed. 1. Draw a diagram if appropriate. 2. Identify the quantity to be maximised or minimised. 3. Express this quantity in terms of one variable only. 4. Solve f ′(x) = 0. 5. Verify it is a maximum or minimum using the first derivative test. 6. Sketch a graph to confirm the maximum or minimum found. 7. Answer the question that has been asked.
390
Maths Quest 12 Mathematical Methods CAS for the TINspire
The following formulas may be useful in problem solving: Area of circle Curved surface area of a cylinder Surface area of a sphere
A = π r2 S = 2π rh S = 4π r2
Volume of a sphere
V = 3 πr3
Volume of a cylinder
V = π r2h
Volume of a cone
V = 13 π r 2 h
4
1
V = 3 Ah , where A is the area of the base
Volume of a right pyramid
d = ( x2 − x1 )2 + ( y2 − y1 )2
Distance between two points
Worked Example 9
The sum of two positive numbers is 10. Find the numbers if the sum of their squares is a minimum. Think
Write
1
Let one number be x and the other number be y. Form an equation.
Let x = 1st number and y = 2nd number. x + y = 10
2
Express y in terms of x.
So y = 10 − x
3
Write an expression for S(x), the sum of the squares of x and y, in terms of x only.
S(x) = x2 + y2 = x2 + (10 − x)2
4
Simplify the equation.
5
Find S′(x).
S′(x) = 4x − 20 Function has a stationary point when S′(x) = 0
6
Find x for minimum S by solving S′(x) = 0.
4x − 20 = 0 4x = 20 x=5
7
Verify that it is a minimum by the first derivative test.
Gradient table:
= x2 + 100 − 20x + x2 = 2x2 − 20x + 100
x
4
5
6
P′(t)
−
0
+
Slope
\
—
/
So x = 5 gives a minimum for S. 8
Find y.
When x = 5, y = 10 − 5 =5
9
State the two numbers.
Therefore, the two numbers which give a minimum of their squares are both 5.
Note: The actual sum was not asked for in this example.
Chapter 8 Applications of differentiation
391
Worked exAmple 10
eBook plus
A cuboid container with a base length twice its width is to be made with 48 m2 of metal. 8 2x a Show that the height is given by the expression h = − , x 3 where x is the width of the base. b Express the volume, V, in terms of x. c Find the maximum volume. Think a
1
Tutorial
int0561 Worked example 10
WriTe
Draw a diagram of a cuboid.
a
h
2x x
2
Let x be the width of the base and hence express length in terms of x.
Let x = width and h = height so length = 2x
3
Calculate the total surface area (TSA) of the cuboid in terms of x and h only.
TSA = 2[2x(x) + 2x(h) + x(h)] = 2(2x2 + 3xh) = 4x2 + 6xh
4
Express h in terms of x.
As TSA = 48 m2 4x2 + 6xh = 48 6xh = 48 − 4x2 h=
48 − 4 x 2 6x
=
48 4 x 2 − 6x 6x
h= b
1
Find the volume, V, in terms of x and h.
2
Express the volume in terms of x by substituting for h.
b V = x(2x)h
8 2x V(x) = 2x2 − x 3 = 16x −
c
1
2
Solve V′(x) = 0.
Verify that x = 2 gives a maximum.
8 2x − x 3
4 x3 3
c V′(x) = 16 − 4x2 = 0 for maximum or minimum
4x2 = 16 x2 = 4 x = 2 or −2 (reject −2 as width cannot be negative)
Gradient table: x
1
2
3
V′(x)
+
0
−
Slope
/
—
\
The maximum volume is achieved when x = 2 m. 392
maths Quest 12 mathematical methods CAS for the Tinspire
3
Substitute x = 2 into the rule for V(x) to obtain the maximum volume.
V(2) = 16(2) −
4(2)3 3
32 3
= 32 −
= 32 − 10 23
= 21 13 1
Therefore the maximum volume is 21 3 m3.
Worked Example 11 y
Find the minimum distance from the straight line with equation y = x − 4 to the point (1, 1).
1
P (1, 1)
0 1 −4 Think
y=x−4 x
4 Q (x, y)
Write
1
The minimum distance between a straight line and a point is a perpendicular line from a point on the straight line, Q (x, y) to the point P (1, 1).
Let Q be the point on the line with coordinates (x, y).
2
From the given rule for the straight line, find y in terms of x.
As Q is on the line y = x − 4 then Q is (x, x − 4).
3
Find the distance, d(x) between P and Q in terms of x only using the formula for the distance between 2 points.
d(x) = ( x2 − x1 )2 + ( y2 − y1 )2
4
Differentiate d(x) using the chain rule (or differentiate the expression inside the square root only, as the smallest value of this expression will give the minimum distance when the square root is taken).
= ( x − 1)2 + ( x − 4 − 1)2
= ( x − 1)2 + ( x − 5)2
= x 2 − 2 x + 1 + x 2 − 10 x + 25
= 2 x 2 − 12 x + 26
1 d (2 x 2 − 12 x + 26) 2 dx
= 12 × (4x − 12) × =
1 1
(2 x 2 − 12 x + 26) 2
4 x − 12 2 2 x 2 − 12 x + 26
OR d (2 x 2 − 12 x + 26) dx = 4x − 12
Chapter 8 Applications of differentiation
393
5
Solve for x where the derivative equals zero.
For maximum or minimum, 4 x − 12 =0 2 2 x 2 − 12 x + 26 4x − 12 = 0 4x = 12 x=3
6
Verify that x = 3 gives a minimum.
Gradient table: x
2
3
4
Derivative (4x − 12)
−
0
+
Slope
\
−
/
So x = 3 gives the minimum distance. 7
Evaluate d(3) to obtain the minimum distance. The exact answer is 8 and 2.828 is an approximate answer.
d(3) = 2(3)2 − 12(3) + 26
d = 18 − 36 + 26
d= 8=2 2
or
≈ 2.828
Therefore the minimum distance is 2 2 units or approximately 2.828 units.
REMEMBER
1. Express the quantity being maximised or minimised in terms of one variable only. 2. Find the derivative and let it equal zero to find the maximum or minimum. 3. Test for maximum and/or minimum using the first derivative test (gradient table). 4. Use exact answers where appropriate. If using an approximate answer show by use of the ≈ symbol. 5. Ensure you have answered the question asked. Exercise
8d
Maximum and minimum problems when the function is unknown 1 WE9 The sum of two positive numbers is 10. Find the numbers if their product is a maximum. 2 The sum of two positive numbers is 8. Find the numbers if the sum of the cube of one and the square of the other is a minimum. 3 A rectangular frame is to be made from a piece of wire 120 cm long. a If the width is x cm, show that the length is 60 − x. b Find an expression for the area of the rectangle in terms of x. c Hence, find the dimensions for maximum area. d Find the maximum area. 4 Find the area of the largest rectangular paddock that can be enclosed with 400 metres of fencing.
394
Maths Quest 12 Mathematical Methods CAS for the TINspire
5 Cylindrical, cardboard postal tubes are made with the restriction that the sum of the length and the circular circumference are 120 cm. What should the dimensions be for maximum volume?
Length
Circumference
6 The frame of a container in the shape of a cuboid is shown at right. If it is to be made with a total length of 18 metres of steel edging, find: a the value of L in terms of x 2x b the expression for the volume in terms of x only c the length of each edge for maximum volume d the maximum volume.
L
x
7 WE10 A cuboid with a square base is to be made with 200 cm2 of material. 50 x − , where x is the side length of the base. x 2 b Express the volume, V, in terms of x. c Find the maximum volume (to the nearest unit).
a Show that the height, h =
8 A window frame is in the shape of a semicircle joined to a rectangle. Find the maximum area of a window using 300 cm of framework. 9 A rectangular sheet of cardboard, 50 cm by 40 cm, is to have square corners cut out so it can be folded into a rectangular tray. Find the maximum volume possible for such a tray.
10 A bushwalker can walk at 5 km/h through clear land and 3 km/h through bushland. If she has to get from point A to point B following a route indicated at right, find the value of x so that the route is covered in a minimum time. (Note: time =
50 cm 40 cm
B Clear
Bush 3 km
distance ) speed
x
A
11 The cost of running a train at a constant speed of v km/h is C = 50 +
2 km
v2 dollars/hour. 1000
a Find the time taken for an 800 km journey in terms of v. b Hence, find an expression for the cost of an 800 km journey. c Find the most economical speed for this journey. 12 Find the side length of the largest cube which can fit inside a sphere of diameter 24 cm.
24 cm
x
Chapter 8 Applications of differentiation
395
13 A cylinder of cheese is to be removed from a spherical piece of cheese of radius 8 cm. What is the maximum volume of the cylinder of cheese? (Express the answer to the nearest unit.) 8 cm
y
14 We11 Find the minimum distance from the line y = 2x + 3 to the point (1, 0).
y = 2x + 3 3
Minimum distance (1, 0)
−2
0 y
eBook plus Digital doc
WorkSHEET 8.2
x
1 y = x2
15 Find the minimum distance from the parabola y = to the point (5, 0). (Express the answer to the nearest hundredth.) x2
(5, 0) 5 x
0
8e
rates of change
eBook plus
P (x1, f (x1)) and Q (x2, f (x2)) are two points on the function with rule y = f (x) as shown in the diagram.
Interactivity
int0253 Rates of change
y y = f(x) Q (x2, f(x2))
P (x1, f(x1)) 0
x1
x2
x
The average rate of change of y with respect to x over the interval x ∈ [ x1, x2] is equal to the gradient of the straight line (or chord) PQ. Average rate of change =
change in f ( x ) change in x
f ( x2 ) − f ( x1 ) x2 − x1 The instantaneous rate of change is the rate of change at a specific point. dy The instantaneous rate of change of y with respect to x is given by the derivative . dx dy If > 0, then y is increasing as x increases (gradient is positive). dx dy If < 0, then y is decreasing as x increases (gradient is negative). dx =
396
maths Quest 12 mathematical methods CAS for the Tinspire
Note: Rates of change are often calculated with respect to time, but not always. If you are required to find the rate of change with respect to some quantity other than time then the quantity must be stated. If this quantity is not stated then the rate of change is taken as being with respect to time. Worked exAmple 12 a Find the rate of change of the surface area of a melting ice cube with respect to its side length (x). b What is the rate of change when x = 2 cm? (Assume that the ice cube remains in the shape of
a cube.)
Think a
b
WriTe
1
Express the surface area, S, of the cube in terms of its side length, x.
2
Find
3
Place a negative sign in front of the rate as the ice cube is melting (that is, the rate is decreasing). dS Substitute x = 2 into . dx
1
2
dS . dx
State the solution.
a S = 6x2 (total surface area of a cube)
dS = 12x dx But
dS − = 12x because the surface area is decreasing. dx
b When
x=2 dS − = 12(2) dx = −24
Therefore, the rate of change of the surface area when x = 2 cm is −24 cm2/cm (it is decreasing at a rate of 24 cm2/cm).
Worked exAmple 13
eBook plus
The number of mosquitoes, N, around a dam on a certain night can be modelled by the equation N = 100 loge (2t + 1) + 5t + 1000
Tutorial
int0562 Worked example 13
where t equals hours after sunset. Find: the initial number of mosquitoes the average rate of change in the first 4 hours the rate of change at any time, t the rate of change when t = 4 hours.
a b c d
Think a
WriTe
1
Write the rule.
2
Find N when t = 0.
a N = 100 loge (2t + 1) + 5t + 1000
When t = 0, N = 100 loge (2 × 0 + 1) + 5 × 0 + 1000 = 100 loge 1 + 1000 = 100 × 0 + 1000 = 1000 The initial number of mosquitoes is 1000.
Chapter 8
Applications of differentiation
397
b
1
2
Find N when t = 4.
b When t = 4,
N = 100 loge (2 × 4 + 1) + 5 × 4 + 1000 = 100 loge (9) + 20 + 1000 = 1020 + 100 loge (9) = 1020 + 219.7 = 1239.7 that is, 1239 mosquitoes (as the number has not reached 1240).
Calculate the average rate between t = 0 and t = 4 using
The average rate of change in the first 4 hours 1239 − 1000 = 4−0
f ( x2 ) − f ( x1 ) . x2 − x1
239 4 = 59.75 mosquitoes per hour. =
c Differentiate N with respect to t, to find
dN . dt
c
dN 100 × 2 = +5 dt 2t + 1 =
d Find
dN when t = 4. dt
200 +5 2t + 1
d When t = 4 the rate of change is:
dN 200 = +5 dt 2 × 4 × 1 =
200 9
+5
2
= 22 9 + 5 2
= 27 9 mosquitoes per hour.
rememBer
1. The average rate of change is the gradient of a straight line between two points. f ( x2 ) − f ( x1 ) . x2 − x1 3. The instantaneous rate of change at any point is given by the value of the derivative at that point. 2. The average rate of change =
exerCiSe
8e
rates of change In the following exercise, use a CAS calculator to assist with any graphing. 1 Express the following in simplest mathematical notation. a The rate of change of volume, V, with respect to radius, r. b The rate of change of surface area, S, with respect to height, h. c The rate of change of area, A, with respect to time, t.
398
maths Quest 12 mathematical methods CAS for the Tinspire
eBook plus Digital doc
SkillSHEET 8.2 Review of rates of change
d The rate of change of cost, C, with respect to distance, x. e The rate of change of intensity, I, with respect to pressure p. f The rate of change of velocity. 2 WE12 a Find the rate of change of the area, A = π r2, of an increasing circular oil spill with respect to the radius, r. b What is the rate of change when r = 10 metres? 3 a Find the rate of change of the volume, V = 43 π r3, of a deflating spherical balloon with respect to the radius, r. b Hence, find the rate when r = 5 cm. 4 A sugar cube dissolves in a cup of tea. Find the rate of decrease of its surface area, S, when the side length, x, is 0.5 cm. 5 The height of a projectile t seconds after being fired is h = 10 + 20t − 5t2 metres. a Find the rate of change of height after i 1 second and ii 3 seconds. b Explain the difference between these two answers. 6 The volume of water (in litres) which has flowed through a swimming pool filter t minutes t4 1 (30t3 − ) where 0 ≤ t ≤ 90. after starting it is V = 100 4 a At what rate is water flowing through the filter at any time, t? dV over the domain [0, 90]. b Sketch a graph of dt c When is the rate of flow greatest? 7 A particle moves in a straight line so that its displacement from a point, O, at any time, t, is x = 3t 2 + 4 . Find: a the velocity as a function of time b the acceleration as a function of time c the velocity and acceleration when t = 2. 8 If a particle is moving in a straight line so that its displacement from the origin at any time, t, is x = t3 − 12t2 + 36t, find: a the velocity b the time and displacement when the velocity is zero c the acceleration when the velocity is zero. 9 WE13 The number of people with the flu virus, N, in a particular town t days after a vaccine is introduced is N = 3000 − 500 loge (8t + 1). a How many people are infected in the town before the vaccine is introduced? b Find the average rate of change over the first 5 days. c Find the rate of change of the number of people in the town infected with flu. d Find the rate of change after 5 days. 10 MC If V = t3 − 2t2 + 8 t , where V is in m3 and t is in hours a the rate of change of V with respect to t when t = 4 is: B 32 m3/h c 28 m3/h d 34 m3/h e −20 m3/h A 48 m3/h b the average rate of change from t = 1 to t = 4 is: d −10 13 m3/h e 0 m3/h A 13 23 m3/h B 10 13 m3/h c 3 m3/h 11 MC Temperature, T °C, is related to height, h metres, by T = h2 + 4eh. a The rate of change, °C/m, at h = 0 is: A 0 b 4 c 4e d −4e e 1 Chapter 8 Applications of differentiation
399
b The rate of change, °C/m, at h = 4 is: A 16 + 4e4 b 16 + e4 D 8 + e4 e 8 + 16e4
c 8 + 4e4
c The average rate of change between h = 0 and h = 4 in °C/m is: A 1 + e4 b e4 c 12 + 4e4 D 4 + e4 e 3 + e4 12 The height of water (in metres) at the entrance to a bay t hours after high tide is: πt H = 10 + 2 cos 12
Find: a the rate of change of H at any time, t b the rate of change of H: i 6 hours ii 15 hours iii 20 hours after high tide c the minimum and maximum value of H and the time when they first occur after the initial high tide. 13 The value of an antique ornament t years after being purchased is A = 2000 e 0.1 t dollars. a Find the value of the antique i when purchased and ii 3 years after purchase. b Hence, find the average rate of change of value in the 3 years since purchased. c Find the rate of change of value 3 years after purchase. 20 14 The amount of chlorine in a jug of water t hours after it was filled from a tap is C = , t +1 where C is in millilitres. Find the rate of decrease of chlorine 9 hours after being poured. 15 The graph below shows the price of the shares of a particular company over a period of time. Price (¢) 180 170 160 150 140 130 120 0
2
4
6
8
10 12 14 16 t (months)
Use the graph to estimate the answers to the following questions. a Estimate the share price when t = 5. b State the time interval over which the share price was decreasing. c Find the average rate of change of the share prices over the entire period shown on the graph. d Estimate the rate of change of the share price at t = 11. e What was the maximum share price over the period shown? f When was the share price at its lowest value? 400
Maths Quest 12 Mathematical Methods CAS for the TINspire
16 The graph below shows how the velocity of a car varies over a period of time. v (m/s) 14 12 10 8 6 4 2 0
2
4
6
8
10 12 14 16
t (s)
Use the graph to estimate the answers to the following questions. a Estimate the velocity of the car at t = 12. b Estimate the acceleration of the car at t = 5. c State the time intervals over which the car is accelerating. d What was the average rate of change of velocity of the car over the interval [4, 8]?
8F
eBook plus
related rates
eLesson
eles0094
When two variables are both functions of the third variable, we may need to Related rates use related rates to solve the problem. For example, we may need to dx dy dx dx dy find when given . In such instances we would need to use the chain rule: = × . dt dt dt dy dt When solving problems involving related rates, the following steps may be helpful: 1. Draw a diagram where appropriate. Sometimes two or more diagrams may be necessary. 2. Identify the variables. 3. Identify which rate is given and which rate is required. 4. Use the chain rule to connect the required and given rates. 5. Find an equation, or relationship, that connects the variables if not given. 6. Differentiate the equation. 7. Substitute into the chain rule and simplify. 8. Answer the question noting correct units. Tools for finding relationships: • similar triangles • Pythagoras’ theorem • rightangled triangle trigonometry • sine and cosine rules. Worked exAmple 14
eBook plus
cm3
An ice cube is melting at a constant rate of 2 per hour. At what rate is the side length of the cube changing when the side length is 1.2 cm? Think 1
We are given the rate of change of volume with respect to time. Write it using appropriate notation. Since the volume is decreasing, the rate is negative.
Tutorial
int0563 Worked example 14
WriTe
dV − = 2 dt
Chapter 8
Applications of differentiation
401
dx when x = 1.2 dt
2
State the rate of change that needs to be found.
Need to find
3
Use the chain rule to connect the rate that we need to find with the one that is given.
dx dx dV = × dt dV dt
4
We do not know dx . To find it, we need the dV rule that connects x and V, that is, the rule that connects the side length of the cube with its volume. So write the formula for the volume of the cube.
5
Differentiate.
6
Reciprocate both sides to obtain dx . dV
7
8
9
Substitute 1 2 for dx and (−2) for dv into the dV dt 3x chain rule and simplify.
Vcube = x3, where x is the length of the side.
dV = 3x2 dx dx 1 = 2 dV 3 x dx 1 = 2 × (−2) dt 3 x
=
−2
3x 2
We now have the formula that allows us to find the rate of change of the side length for any value of x. To find the rate of change when length is 1.2 cm, simply substitute 1.2 for x into this formula and evaluate.
So when x = 1.2, −2 dx = dt 3 (1.2 )2
State the answer, adding appropriate units.
When the side length of the cube is 1.2 cm, its
=
−
25 54
− length is changing at a rate of 25 cm/h. 54
Worked Example 15
An empty inverted right circular cone has a radius of 4 cm and a height of 20 cm. Water is being poured in at a constant rate of 0.1 cm3/second. Find the rate at which the depth of the water is increasing at the instant the depth is 6 cm. Give your answer correct to 2 decimal places. Think 1
Draw a diagram showing half the crosssection at any depth, h cm.
Write 4 cm
r cm
20 cm h cm
402
Maths Quest 12 Mathematical Methods CAS for the TINspire
2
Identify the variables.
3
Identify the given rate. Identify the required rate.
4
Use the chain rule to connect the rate that we need to find with the one that is given.
5
Find a rule that connects volume, V with the depth of water, h. Use similar triangles to connect r and h.
Depth of the water, h cm Radius, r cm, of surface of the water Time, t seconds Volume, V cm3 dV = 0.1 cm3/second dt dh Find when h = 10 cm. dt dh dh dV = × dt dV dt 1 V = π r 2h 3 r 4 = h 20 h ∴ r= 5 2
Rewrite the formula with the two variables V and h.
1 h V = π h 3 5
π 3 h 75 dV 3π 2 = h dh 75 dV π 2 = h dh 25 dh 25 = dV π h 2 dh dh dV = × dt dV dt dh 25 = × 0.1 dt π h 2 dh 2.5 = dt π h 2 dh 2.5 = dt π × 62
∴V= 6
7
8
Differentiate V with respect to h.
dh by taking reciprocals of both sides. dV Substitute into the Chain Rule and simplify. Obtain
Substitute h = 6 cm to find the required rate of change. State the answer with appropriate units.
dh = 0.02 cm/second (correct to dt 2 decimal places) ∴ The depth of water is increasing at a rate of 0.02 cm/second when the depth is 6 cm.
REMEMBER
dx dx dy 1. To solve a problem involving related rates, use the chain rule, for example: = × . dt dy dt 2. Show appropriate units.
Chapter 8 Applications of differentiation
403
Exercise
8F
Related rates 1 WE 14 A gym fitness ball is being inflated and its volume is increasing at a constant rate of 5 cm3 per second. At what rate is the radius of the ball changing when the radius is 34 cm? 2 A particle moves along a path that can be described using the Cartesian equation y = 3x3 − 2x + 1. If
dy − dx = 3 when x = 5, find at that moment. dt dt
3 The surface area of a cube is decreasing at a constant rate of 9 cm2/s. Find the rate at which the sides of the cube are decreasing when the sides are 1.5 cm long. 4 A spherical balloon is being deflated and its radius, r cm, is decreasing at a constant rate of 5 cm/min. At what rate is its volume, V cm3, decreasing when the radius of the balloon is 4 cm? 5 A bowl is being filled with water at a rate of 12 cm3/s. The volume, V cm3, of water in the 5
bowl is given by V = 8h 2 where the depth of water in the bowl is h cm. Find the rate at which the depth of water in the bowl is increasing when the depth is 9 cm. 6 The radius of a circular puddle of water is increasing at a rate of 2.5 cm/s. Find the exact rate at which the area is increasing at the instant the radius is 12 cm. 7 MC A cylinder with radius of 4 m is being filled with water. If the rate of change of the depth 1 of the water is m/s, the rate of change of volume, in m3/s, in the cylinder is: 8π 1 1 1 A B 2 c 2 d e 2π 2 2π 108π 8 WE15 An inverted right circular cone is filled with liquid. The cone has a radius of 3 m and height of 7 m. The liquid flows from the apex of the cone at a constant rate of 0.6 m3/min. Find the rate at which the depth of the liquid is dropping, correct to 2 decimal places, when the depth of the liquid is 2 m. 9 The upper end of an 8 m ladder rests against a vertical wall with the lower end on the horizontal ground. The lower end of the ladder on the ground slips away from the wall at a rate of 6 m/s. Find the rate at which the upper end of the ladder is moving the instant the ladder is 4 m from the wall. 10 A stainless steel cylindrical tank of radius 6 m is being filled with milk at a constant rate of 1.5 m3/min. At what rate is the level of milk rising? Give your answer in terms of π. 11 A sand timer consists of two cones joined at the apex. Each cone has height h, radius r and an angle at the apex of 60°. a Express the radius of the top cone in terms of its height. Give your answer in exact form. b Write the volume of the top cone as a function of its height. c When the timer is turned over, the sand starts pouring from the top cone into the bottom 1 one at a constant rate of 32 cm3/s. Find the rate of change of the depth of the sand in the top cone when the depth is 0.8 cm.
8G
Linear approximation It is useful to be able to find how much a small change in the independent variable affects the dependent variable. For example, how will a small increase in the radius of a circle affect the area of a circle?
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Maths Quest 12 Mathematical Methods CAS for the TINspire
This concept can be illustrated graphically. y
y = f(x)
B δy
δy = f(x + h) − f(x)
A
0
x
δx =h
Consider the function y = f (x) as shown with the points A (x, f (x)) and B (x + h, f (x + h)). The gradient of the chord AB =
f ( x + h) − f ( x ) . If h is very small, the gradient of the chord is h
very close to the gradient of the tangent at the point A (x, f (x)). f '( x ) ≈
f ( x + h) − f ( x ) h
h f '( x ) ≈ f ( x + h) − f ( x ) f ( x + h) ≈ f ( x ) + h f '( x ) The formula f ( x + h) ≈ f ( x ) + h f '( x ) is called the linear approximation formula. Alternatively, if a small change in y, δ y, corresponds to a small change in x, δ x, then ∴ δ y ≈ δx
dy δ y . ≈ dx δ x
dy where δ y ≈ f ( x + h) − f ( x ). dx
If δ y is the small change in a quantity, y, the percentage change (or error) in y is given by h f '( x ) δy × 100% . × 100% or f ( x) y
Worked Example 16
Use the approximation formula, f (x + h) ≈ f (x) + hf '(x), with f (x) = x2 and x = 1 to find an approximate value of (1.01)2. Think
Write
f (x) = x2, f '(x) = 2x, x = 1 and h = 0.01
1
Write f (x), f '(x), x and h.
2
Find x + h and hence f (x + h).
x + h = 1.01 and f (x + h) = (x + h)2 = 1.012
3
Write the formula and hence find the value of 1.012.
f (x + h) ≈ f (x) + hf '(x) 1.012 ≈ 12 + 0.01 × 2 1.012 ≈ 1.02
Chapter 8 Applications of differentiation
405
Worked Example 17
A circular metal disc is being cooled. If the radius is decreased from 10 cm to 9.8 cm, find correct to 2 decimal places: a the approximate change in the area b the percentage change in the area. Think a
b
Write
1
Write the relationship between area and radius.
A(r) = πr2
2
Write A′(r), r and h.
A′(r) = 2πr, r = 10 cm, h = −0.2
3
Write the approximation formula and substitute.
A(r + h) ≈ A(r ) + hA'(r ) A(9.8) ≈ A(10) − 0.2 A'(10)
4
State the answer with correct units.
1
Write the change in area as fraction of the original area, expressing as a percentage.
2
State the answer.
Change in A ≈ A(9.8) − A(10) ≈ −0.2 × 20π Area decreases by 12.57 cm2 correct to 2 decimal places.
δ A − 0.2 × 20 × π = × 100% A π × 10 2 = −4% ∴ Area is decreased by approximately 4%
REMEMBER
1. Where h is a small change in x then f ( x + h) ≈ f ( x ) + hf '( x ). 2. Alternatively f ( x + h) − f ( x ) ≈ hf '( x ) dy or δ y ≈ δ x × . dx Exercise
8G
Linear approximation 1 WE 16 Use the approximation formula, f (x + h) ≈ f '(x)h + f (x), with each of the following. a For f (x) = x3 and x = 1, find 1.013. b For f (x) = x2 and x = 1, find 0.9992. c For f (x) = 5x2 and x = 1, find 5 × 0.992. d For f (x) = x and x = 1, find 1.001 2 If a spherical balloon has a radius of 5 cm, find the increase in volume of the balloon when the radius expands by 0.02 cm. 3 A circular disc has a radius of 10 cm. Find the percentage increase in its area if the radius is increased by 2%. 4 WE 17 An isosceles triangle has its equal sides of length 10 cm θ with an included angle of θ as shown in the diagram. 10 cm
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Maths Quest 12 Mathematical Methods CAS for the TINspire
10 cm
If θ changes from 60° to 61°, find, correct to 2 decimal places: a the approximate area of the triangle when θ = 61° b the approximate increase in the area, A, of the triangle. 5 MC If f (x) = x2 and x = 3, an approximate value of (2.9)2 is given by: A f (3) + 0.1 f '(3) B f (2.9) + 0.1 f '(2.9) C f (3) − f (2.9) D f (2.9) − 0.1 f '(2.9) E f (3) − 0.1 f '(3) 6 The volume, V cm3, of a gas is inversely proportional to the pressure, P atmospheres, of a gas, k that is, V = where k is a constant (that depends on the mass of the gas and the P temperature). If the pressure of the gas increases from 2 atmospheres to 2.25 atmospheres, find the approximate change in the volume of the gas in terms of k. Hence find the corresponding percentage change in the volume. 2 7 Find the approximate change in x as y decreases from 2 to 1.5 if f ( x ) = . x−4 8 The length of a rectangle is four times its width. If the width increases by 5%, find the corresponding percentage change in the perimeter and area of the rectangle.
Chapter 8 Applications of differentiation
407
Summary Stationary points
• A function f (x) has stationary points wherever its derivative f ′(x) = 0. • If f ′(a) = 0, then f (x) has a stationary point (a, f (a)) which is: 1. a local minimum turning point if: for x < a, f ′(x) < 0 and x > a, f ′(x) > 0. 2. a local maximum turning point if: for x < a, f ′(x) > 0 and x > a, f ′(x) < 0. 3. a positive stationary point of inflection if: for x < a, f ′(x) > 0 and x > a, f ′(x) > 0. 4. a negative stationary point of inflection if: for x < a, f ′(x) < 0 and x > a, f ′(x) < 0. Equations of tangents and normals
• The gradient of the tangent at x = a to the curve y = f (x) is f ′(a). −1 • The gradient of the normal at x = a to the curve y = f (x) is . f'(a) • The equation of a straight line with gradient m and passing through the point (x1, y1) is: y – y1 = m(x – x1) Maximum and minimum problems
• When solving maximum or minimum problems follow these steps. 1. Draw a diagram if appropriate. 2. Identify the quantity to be maximised or minimised (say, f (x)). 3. Express the quantity in terms of one variable only (say x). 4. Solve f ′(x) = 0. 5. Verify it is a maximum or minimum using the first derivative test. 6. Sketch a graph to confirm the maximum or minimum found. 7. Answer the question. Rates of change
• The instantaneous rate of change of y = f (x) at x = x1 is found by evaluating f ′(x1) (or finding x = x1). • The average rate of change of y = f (x) between x = x1 and x = x2 is: f ( x2 ) − f ( x1 ) . x2 − x1
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Maths Quest 12 Mathematical Methods CAS for the TINspire
dy where dx
Related rates
• To solve a problem involving related rates, use the chain rule, for example
dx dx dy = × . dt dy dt
Linear approximation
• f ( x + h) ≈ f ( x ) + hf '( x ) dy • δ y ≈ δ x dx • Percentage change: δ y × 100% y
Chapter 8 Applications of differentiation
409
chapter review Short answer
1 Determine the stationary points and their nature for the function f (x) = 2x3 + 3x2 − 36x + 5.
9 P is the point on the line 2x + y − 10 = 0 such that the length of OP, the line segment from the origin O to P, is a minimum. Find the coordinates of P and this minimum length.
2 Sketch the graph of y = x2(4 − x2), clearly indicating all stationary points and intercepts.8B
Exam tip Students failed to recognise this as a maximum/minimum question. A common, incorrect response was to find intercepts (0, 10) and (5, 0) on the line then use the distance between two points to
3 Find the equation of the tangent to the curve y = 6 − 3x + 2x2 − x3 at the point where x = 1. 4 Find the equation of the tangent to the curve y = 3 loge (x2) (x ≠ 0) which is parallel to the line y − 3x − 7 = 0.
find the distance to be 5 5. Other common, incorrect responses involved taking P as the midpoint of the line, guessing and checking solutions without justification, incorrectly differentiating the distance equation, or using an incorrect diagram.
5 Find the equation of the normals to the curve 1 y = x + at the point where the normals are x
[Assessment report 1 2007]
[© VCAA 2007]
parallel to the line 3y + 4x − 10 = 0.C 6 The number of bees, N, in a hive can be modelled by the function N = 2t(50 − t) + 180 where t is the number of days the hive has existed. What is the maximum number of bees in the hive?
10 A normal to the graph of y = x has the equation y = −4x + a, where a is a real constant. Find the value of a.
7 The area of a certain triangular shape is: A=
Exam tip Students found this question quite challenging, and a significant number seemed to have little or no idea of what was required. Many were able to find the correct derivative but did not know how to proceed, or equated the derivative to –4 1 instead of 4 . Students who got to this point without error often did not solve correctly for x or y or both. It was not uncommon to see two values found for x and y and therefore two values for a, 18 and –14.
2x2 ,x>6 3( x − 6)
where x is the length of the base of the triangle. a Find the value of x for minimum area. b Hence, find the minimum area. 8 A wine glass is being filled with wine at a rate of 8 cm3/s. The volume, V cm3, of wine in the glass when the depth of wine in the glass is x cm is given 3 4x 2.
by V = Find the rate at which the depth of wine in the glass is increasing when the depth is 4 cm. Exam tip Some students had difficulty deciding on how to label variables; x was sometimes interchangeable with d or h and little or no consistency was evident through the question. Most students were able to correctly differentiate V with respect to x but then either incorrectly used the chain rule or could not manage the arithmetic required. Common errors were to simply substitute 4 into V or to substitute into the derivative. [Assessment report 1 2007]
[© VCAA 2007]
410
[Assessment report 1 2006]
[© VCAA 2006]
Multiple choice
1 The graph of y = (x + 2)3 has: A 1 turning point B 2 turning points C 1 point of inflection D 2 points of inflection E 0 stationary points 2 The graph of x3 + 2x2 + x − 2 has: A 2 points of inflection B 1 point of inflection C 1 turning point and 1 point of inflection D 3 turning points E 2 turning points
Maths Quest 12 Mathematical Methods CAS for the TINspire
3 The graph of 13 x3 − 4x2 − 9x + 5 has a local maximum turning point at: 2
A (−1, 9 3 ) B (9, 20) C (9, 18) d (1, −10 23 ) E (−1, 7 13 ) 4 If the graph of g(x) has the following properties: i g'(x) = 0 if x = −3, 1 and 4 ii g'(x) < 0 if x < −3 and 1 < x < 4 iii g'(x) > 0 for all other x then the graph of g(x) could be: y
A
g(x) −3
x
0 1
4
01
4 g(x)
y
B
−3
C
x
y g(x) 0
−3
1
x
4
g(x) 01
−3
4
x
y
E
g(x) −3
01
4
8 The tangent to the curve y = 2 loge (3 − 2x) at the point where x = 1 intercepts with the yaxis at the point where y equals: A 1 B 3 C 0 D −2 E 4 9 The graph of f ′(x) y shown at right f '(x) indicates that the graph of f (x) has: x A a turning point at 0 –4 2 x = 2 and x = −4 B a turning point at x = 2 and point of inflection at x = −4 C a turning point at x = −4 and point of inflection at x = 2 D 3 stationary points E 2 points of inflection at x = −4 and x = 2 Questions 10 and 11 refer to the following information. The volume of liquid in a container varies with time according to the rule V = 1 + t2e−t, where t is in hours and V is in thousands of litres, and 0 ≤ t ≤ 40. 10 The minimum value of V occurs when t equals: A 1 hour B 2 hours C 0 hours d 3 hours E 40 hours
y
D
7 The equation of the normal to the curve y = 3ex + 2 at the point where x = −2 is: A 3y + x − 7 = 0 B 3y − x + 7 = 0 C y − 3x − 9 = 0 d y − 3x − 3 = 0 E y + 3x − 9 = 0
11 The maximum volume, to the nearest tens of litres, is: A 1450 B 1540 C 1390 d 1360 E 1630 Questions 12 to 16 follow from the isosceles triangle below which has a perimeter of 40 cm.
x
5 The equation of a line with a gradient of 2 and passing through the point (1, −3) is: A y = 2x + 5 B y = 2x C y = 2x − 1 d y = 2x − 5 E y = x + 1 6 The equation of the tangent to the curve y = x (x + 2)(x − 1) at the point where x = −1 is: A y − x + 1 = 0 B y − x + 3 = 0 C y + x + 3 = 0 d y + x − 1 = 0 E y − x − 1 = 0
y
x
x
8D
12 The value of y in terms of x is: A 40 − 2x B 20 − x C 40 − x d 20 − 2x E x2
Chapter 8 Applications of differentiation
411
13 The height of the triangle in terms of x is: A 400 − 40 x
B 20 − 40 x
C
400 − 40 x + 2 x 2
E
40x
D
400 − 40 x + x 2
14 The area in terms of x is: A ( 20 − x ) 400 − 40 x
B x 400 − 40 x + x 2
C 2 x 400 − 40 x + x 2
D x 400 − 40 x
E 2 x 400 − 40 x 15 The maximum area of the triangle is obtained if x equals: A 6 23 cm B 10 cm C 20 cm d 5 cm
e 10 25 cm
16 Therefore, the maximum area possible is: A 50 2 cm2 d 40 3 9
B 64 cm2 400 3 E 9
C 32 5 cm2
17 The average rate of change of the function f (x) = x4 − 3x3 + 5x between x = 1 and x = 3 is: A 15 B 6 C 12 d 4 12 E 16 18 An iceberg in the shape of a cube is slowly melting. The rate of change of the surface area of the iceberg in m3/m when the side length is 40 metres is: A 480 B 240 C 720 d 480 E 240 19 The radius of a sphere is increasing at a rate of 3 cm/min. When the radius is 6 cm, the rate of increase of the volume, in cm3/min, in the sphere is: A 432π B 48π C 144π d 108π E 16π [© VCAA 2006] 20 Using the approximation formula f ( x + h) ≈ f ( x ) + hf '( x ) , with f (x) = x3 and x = 2, an approximate value of (1.8)3 is given by: A f (2) + 0.2 f '(2) B f (1.8) + 0.2 f '(1.8) C f (2) − f (1.8) d f (2) − 0.2 f '(2) E f (1.8) − 0.2 f '(1.8) [© VCAA 2004]
Extended response
1 Consider the cone with a slant side of 10 cm shown at right. 10 cm
a Show that the height, h, is 100 − r 2 . h
b Show that the volume of the cone is V = 13 πr2 100 − r 2 . c Hence, find the maximum volume.
r 2 The number of bacteria present in a lasagne, t minutes after it is placed in a − microwave to heat up, can be modelled by N = 3000e 0.5t, where t > 0. a Find the initial number of bacteria. b Find the rate of change of bacteria after being in the microwave for 10 minutes. 3 The height above the ground of a person on a ferris wheel at any time, t, seconds after the ride has πt started is h = 5.4 − 4 cos metres. 15 a Find the initial height of the person above ground level. d (m) b Find the height after 5 seconds. 700 c Hence, find the average rate of change of height during the first 5 seconds. 600 d Find the rate of change of height at t = 5. 500 4 Lena is walking in the park. The graph at right shows her 400 displacement from the park entrance over a period of time. a Estimate Lena’s displacement from the park entrance at t = 20. 300 b Find Lena’s average velocity over the interval [20, 30]. 200 c What was Lena’s maximum displacement from her starting 100 point? d Estimate Lena’s velocity at t = 15. 0
412
Maths Quest 12 Mathematical Methods CAS for the TINspire
5
10 15 20 25 30 35 t(min)
5 A manufacturing company is required to produce cylindrical cans (for tuna) of volume 50 cm3. Tin used to produce the cans costs 40 cents per 100 cm2. a Find the area of tin required, A, in terms of the radius, r. b Find the radius of the can (to the nearest tenth) for minimum area. c Hence, find the minimum area (to the nearest tenth). d What is the cost of tin to produce 10 000 such cans? 6 A small manufacturing company needs to order new cardboard boxes for packaging their product. Each box is to be in the shape of a prism with a square end w and is to have a volume of 27 000 cm3. To hold each box together, tape is used all around the box as shown in the diagram at right. w l a Express the length of the box (l) in terms of its width (w). b Write the formula for the total area of cardboard (A) in terms of w. c Find the dimensions of the box which uses the least amount of cardboard. d Write the formula for the total length of the tape (L) in terms of w. (Ignore the width of the tape and any overlaps.) e Find the dimensions of the box that uses the least amount of tape. f The cardboard used to make the boxes costs 0.01 cents per square centimetre, and the tape can be purchased at $0.50 per metre. Write the formula for the total cost (in cents) of the package (that is, the cost of the cardboard and tape) in terms of w. g Find the dimensions of the box with the minimum total cost. h Find the minimum total cost to the nearest cent. 7 The crosssection of a pencil head when it is first placed in a sharpening device is shown at right. The gradient of the tangent at point A is 4. The equation of the pencil head is y = 4x2. If the x and yaxes are as indicated and all distances are in centimetres, find: a the coordinates of points A and B b the equation of the tangent to the curve y = 4x2 at point A c the coordinate of point C d the minimum distance from the pencil head to point C e the length of the pencil head if it starts at the point where the normal at point A meets the yaxis.
y
y = 4x2
Pencil head B
A
Tangent x
0 C
8 The population of rabbits on a particular island t weeks after a virus is introduced is modelled by − P = 1200e 0.1t, where P is the number of rabbits. Find: a the time taken for the population to halve (to the nearest week) b the rate of decrease of the population after: i 2 weeks and ii 10 weeks. After 15 weeks the virus has become ineffective and the population of rabbits starts to increase again according to the model P = P0 + 10(t − 15) loge (2t − 29) where t is the number of weeks since the virus was first introduced. Find: c the value of P0 d the population after 30 weeks e the rate of change of the population after i 20 weeks and ii 30 weeks f how many weeks the population takes to get back to its original number.
Chapter 8 Applications of differentiation
413
9 After washing the kitchen floor, Alex put his favourite mop flat against the wall and left it there. A few minutes later, the mop starts to slide down the wall. Let h be the height of the top end of the mop above the floor and let y be the horizontal distance of h the bottom end of the mop from the wall at any time, t. a If the mop is 1.2 m tall, express y in terms of x. b If the top end of the mop slides down with a constant speed of 5 cm per second, find the speed (in terms of h) with which the bottom end of the mop moves away y from the wall. c Find the speed with which the bottom end of the mop moves away from the wall when the top of the mop: i is 0.8 m from the floor ii has slid down by 20 cm. d Find the speed with which the bottom end of the mop moves away from the wall 6 seconds after the top of the mop started sliding down. 10 Consider the function f ( x ) = − ( x − 3)( x + 2a)2 where a is a positive real constant. a Find f ′(x). b Find the coordinates of the stationary points. c Determine the nature of the stationary points. d Find the equation of the tangent at x = −2. e Find the xintercept of the tangent. 11 Consider the function y = (x2 − a)2 where a ∈ R. dy a Find at x = 2. dx b Find the equation of the tangent to the curve at x = 2 in terms of a. c Hence find the xintercept of the tangent line in terms of a. d A straight line with equation y = −2x + 1 passes through the xintercept of the tangent line. Find the value of a. e Using the value of a found previously, what is the equation of the tangent line? eBook plus Digital doc
Test Yourself Chapter 8
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maths Quest 12 mathematical methods CAS for the Tinspire
eBook plus
ACTiviTieS
chapter opener Digital doc
• 10 Quick Questions: Warm up with ten quick questions on applications of differentiation. (page 373) 8A
equations of tangents and normals
Tutorial
• We 2 int0558: Watch a worked example on finding equations of tangents and normals. (page 374) Digital docs
• Spreadsheet 129: Investigate tangents and normals. (page 375) • WorkSHEET 8.1: Find stationary points and their nature and equations of normals and tangents. (page 376) 8B
Sketching curves
Tutorial
• We 3 int0559: Watch a worked example on sketching curves and finding stationary points. (page 378) Digital doc
• SkillSHEET 8.1: Practise differentiating expressions other than polynomials. (page 385) 8c
Maximum and minimum problems when the function is known
Tutorial
• We 8 int0560: Watch a worked example on applications of maximum and minimum problems. (page 387) 8D
Maximum and minimum problems when the function is unknown
Tutorial
• We 10 int0561: Watch a worked example on maximising volume. (page 392) Digital doc
• WorkSHEET 8.2: Apply differentiation skills to a variety of problems. (page 396) 8e
Tutorial
• We 13 int0562: Watch a worked example on application of rates of change and average rates of change. (page 397) Digital doc
• SkillSHEET 8.2: Practise identifying rates of change. (page 398) 8F
Related rates
eLesson eles0094
• Related rates: Learn about using differentiation for related rates. (page 401) Tutorial
• We 14 int0563: Watch a worked example on using related rates to determine an instantaneous rate of change. (page 401) 8G
Linear approximation
chapter review Digital doc
• Test Yourself: Take the endofchapter test to test your progress. (page 414) To access eBookPLUS activities, log on to www.jacplus.com.au
Rates of change
Interactivity int0253
• Rates of change: Consolidate your understanding of applying differentiation techniques by riding and rolling. (page 396)
Chapter 8
Applications of differentiation
415
9
9A 9B 9C 9D 9E 9F 9G 9H 9I 9J
Antidifferentiation Integration of ex, sin ((x) and cos ((x) Integration by recognition Approximating areas enclosed by functions The fundamental theorem of integral calculus Signed areas Further areas Areas between two curves Average value of a function Further applications of integration
Integration areaS of STudy
• Informal treatment of the fundamental theorem
• Antiderivatives of polynomial functions and of f (ax + b), where f is xn for n ∈ Q, ex, sin (x), cos (x) or linear combinations of these • Definition of the definite integral as the limiting b
a
n
∑ f ( xi* ) δ xi, δ x→0
value of a sum ∫ f ( x ) dx = lim
i =1
where the interval [a, b] is partitioned into n subintervals, with the ith subinterval of length δx δ i and containing xi*, and δx = max{δxi: i = 1, 2, … n} and evaluation of numerical approximations based on this definition • Informal approximation using areas under curves by left and right rectangles • Examples of the definite integral as a limiting value of a sum involving quantities such as area under a curve, distance travelled in a straight line and cumulative effects of growth such as inflation • Antidifferentiation by recognition that F ′( x ) = f ( x ) implies ∫ f ( x ) dx = F ( x ) + c
9a
b
of calculus, ∫ f ( x ) dx = F (b) − F (a) a
• Properties of antiderivatives and definite integrals: ( x ) ± bg( x )] dx = a ∫ f ( x ) ddxx ± b ∫ g( x ) dx ∫ [af (x b
∫a
b
∫a
a
∫a
c
c
f ( x ) dx +
∫b f ( x) dx = ∫a f ( x) dx
f ( x ) dx =
−
a
∫b
f ( x ) dx
f ( x ) dx = 0
• Application of integration to problems involving calculation of the area of a region under a curve and simple cases of areas between curves, such as distance travelled in a straight line; average value of a function; other situations modelled by the use of the definite integral as a limiting value of a sum over an interval; and finding a function from a known rate of change
antidifferentiation
eBoo k plus eBook Digital doc
10 Quick Questions
As we have seen, the process of differentiation enables us to find the gradient of a function. The reverse process, antidifferentiation (or integration), will find the function for a particular gradient. Integration has wider applications including calculation of areas, volumes, energy, probability and many more quantities in science and business. Note that d f ( x ) means differentiate f (x) with respect to x; that is, d f ( x ) = f ′( x ). dx dx
416
maths Quest 12 mathematical methods CaS for the TInspire
So f (x) is the antiderivative of f '(x), denoted as f ( x ) = ∫ f '( x ) dx where ∫ means antidifferentiate, or integrate, or find an indefinite integral and dx indicates that the integration of the function is with respect to x. Since
d (ax + c) = a, where a and c are constants dx
then
∫ a dx = ax + c
Since then
d ax n + 1 = ax n dx n + 1
∫ ax n dx =
ax n + 1 + c, n ≠ − 1 n +1
In the expression above, the term c is used to denote a constant. In the antiderivative of a function, there are an infinite number of possibilities for c. However, when we are finding an antiderivative, we set c to zero. That is, finding an antiderivative means ‘let c = 0’, or ‘do not add on the c’. For example, the antiderivative of 3x2 + 4x + 5 is x3 + 2x2 + 5x + c. An antiderivative of 2 3x + 4x + 5 is x3 + 2x2 + 5x.
Properties of integrals Since d is a linear operator, so too is ∫ ..Therefore, dx ∫ [ f ( x) ± g( x)] dx = ∫ f ( x) dx ± ∫ g( x) dx That is, each term can be integrated separately, and
∫ k f ( x) dx = k ∫ f ( x) dx That is, a ‘constant’ factor of the function can be taken to the front of the integral. So ∫ [af ( x) ± bg( x)]dx = a ∫ f ( x )dx ± b ∫ g( x )dx Worked Example 1
Antidifferentiate each of the following, expressing answers with positive indices. − 3 a 2x7 b 4x 3 c x Think a
b
c
Write
1
Integrate by rule; that is, add 1 to the index and divide by the new index.
2
Simplify.
1
Integrate by rule.
2
Simplify.
3
Express the answer with a positive index.
1
When a square root is involved, replace it with a fractional index.
a
∫ 2 x 7 dx =
2 x8 +c 8
x8 +c 4 − 4x 2 dx = − + c 2 =
b
∫ 4x
−3
−
c
∫
= −2x 2 + c − 2 = 2 x 3 3 dx = ∫ 1 dx x x2
Chapter 9 Integration
417
2
= 3∫ x
Bring the x to the numerator and change the sign of the index.
−1 2 dx
1
3x 2
+c
3
Integrate by rule.
=
4
Simplify.
= 6x 2 + c
5
Write the answer in the form it was given.
=6 x +c
1 2 1
Worked Example 2
Find the following indefinite integral. ∫(x.  1)(3x + 5) dx Think
Write
∫ ( x − 1)(3x + 5) dx = ∫ (3x 2 − 3x + 5x − 5) dx = ∫ (3 x 2 + 2 x − 5) dx
1
Expand the expression.
2
Collect like terms.
3
Integrate each term separately.
=
4
Simplify each term.
= x3 + x 2 − 5x + c
3x3 2 x 2 + − 5x + c 3 2
Integration of (ax + b)n where n ≠ − 1 By applying the chain rule for differentiation: d (ax + b) n + 1 = a( n + 1)(ax + b) n dx
so
∫ a(n + 1)(ax + b)n dx = (ax + b)n + 1 + c
or
a( n + 1) ∫ (ax + b) n dx = (ax + b) n + 1 + c
or
∫ (ax + b)n dx =
(ax + b) n + 1 +c a( n + 1)
Worked Example 3 n+1
Antidifferentiate 4(5x − 2)3 by using ( ax + b) n dx = ( ax + b) ∫ a( n + 1) Think
418
+ c.
Write
∫ 4(5x − 2)3 dx = 4 ∫ (5x − 2)3 dx
1
Express as an integral and take 4 out as a factor.
2
Apply the rule where a = 5 and n = 3.
=
3
Simplify the antiderivative by cancelling the fraction.
= 5(5x − 2)4 + c
Maths Quest 12 Mathematical Methods CAS for the TINspire
4(5 x − 2)4 +c 5(4) 1
Integration of x1 Since then
1 d log e ( x ) = dx x 1 ∫ x dx = loge ( x) + c, where x > 0
or
∫x
−1 dx
= log e x + c.
Worked Example 4
Antidifferentiate 4 . 7x Think 1
Write
4
4
= 2
1
∫ 7 x dx = ∫ 7 × x dx
Take 47 out as a factor.
4 7
1
∫ x dx
= 47 log e x + c
Integrate by rule.
 1
Integration of (ax + b)
This can be done by applying the chain rule for differentiation: d a log e (ax + b) = , where a and b are constants. dx ax + b Multiplying both sides by 1 gives a 1 d a 1 × log e (ax + b) = × a dx ax + b a 1 ax + b 1 1 d log e ax + b dx dx = × ax + b a dx =
So
∫
so
∫ ( ax + b)
−1
dx =
1 log e ax + b + c a
1 Note that the a in the fraction a is the derivative of the linear function ax + b. Worked Example 5
Antidifferentiate
5 . 2x + 3
Think
Write
Method 1: Technologyfree 1
Express as an integral and take 5 out as a factor.
2
Integrate by rule where a = 2.
5
1
∫ 2 x + 3 dx = 5 ∫ 2 x + 3 dx =
5 log e 2 x + 3 + c 2
Chapter 9 Integration
419
Method 2: Technologyenabled 1
On a Calculator page, press: • MENU b • 4:Calculus 4 • 3:Integral 3 Alternatively, select the indefinite integral template from the Maths expression template and fill in the required fields. Complete the entry line as: 5 dx ∫ 2 x + 3 Press ENTER ·.
2
Write the answer and put in the constant of integration, as the CAS calculator omits it. Also note that the CAS uses ln instead of loge.
5 5 ∴ ∫ dx = log e 2 x + 3 + c 2 2 x + 3
Worked Example 6
Find ∫
6x + 5 dx . x2
Think
Write
∫
6x + 5 6x 5 dx = ∫ 2 + 2 dx 2 x x x
1
Express as separate fractions.
2
Simplify each fraction.
= ∫ (6 x
3
Integrate each term separately by rule.
= 6 log e x +
4
Simplify leaving the answer with positive indices.
= 6 log e x − 5 x + c 5 = 6 log e x − + c x
−1
−
+ 5 x 2 ) dx 5x
−1
−1 −1
+c
Worked Example 7
Find the equation of the curve g(x) given that g′(x) = 3 x + 2 and the curve passes through (1, 2). Think
420
Write
1
Write the rule for g′(x).
g′(x) = 3 x + 2
2
Rewrite g′(x) in index form.
g′(x) = 3 x 2 + 2
3
Express g(x) in integral notation.
g(x) =
4
Antidifferentiate to obtain a general rule for g(x).
Maths Quest 12 Mathematical Methods CAS for the TINspire
1
1
∫ 3x 2 + 2 dx 3
= 3x 2 ÷ 3 + 2 x + c 2
5
Simplify.
3 2
= 3x × 2 + 2 x + c 3 1
3
g(x) = 2 x 2 + 2 x + c
6
Substitute coordinates of the given point into g(x).
7
Find the constant of antidifferentiation, c.
8
State the rule for g(x) in the form that it is given.
3
As g(1) = 2, 2(1) 2 + 2(1) + c = 2 2+2+c=2 so c = −2 3
g(x) = 2 x 2 + 2 x − 2
= 2 x3 + 2x − 2
Worked Example 8
If a curve has a stationary point (2, 3), and a gradient of 2x − k, where k is a constant, find: a the value of k b y when x = 1. Think a
1
2
Write
The gradient is dy so write the rule for the dx gradient.
a
Let dy = 0 (as stationary points occur when dx the derivative is zero) and substitute the value of x into this equation.
dy = 2x − k dx For stationary points, dy = 0, so 2x − k = 0 dx 2(2) − k = 0 as x = 2
b
3
Solve for k.
1
Rewrite the rule for the gradient function, using the value of k found in a above.
2
Integrate to obtain the general rule for y.
4 − k = 0 so k = 4 b
dy = 2x − 4 dx y = ∫ (2 x − 4)dx = x2 − 4x + c
3
Substitute the coordinates of the given point on the curve to find the value of c.
Since curve passes through (2, 3), 3 = 22 − 4(2) + c 3=4−8+c c=7
4
State the rule for y.
So y = x2 − 4x + 7
5
Substitute the given value of x and calculate y.
When x = 1, y = (1)2 − 4(1) + 7 =4
The relationship between the graph of an antiderivative function and the graph of the original function f (x) is the antiderivative of f ′(x) and is written as f ( x ) = ∫ f ′( x )dx. f ′(x) is the gradient function of the antiderivative function f (x).
Chapter 9 Integration
421
The graph of the antiderivative function f (x) can be derived from the graph of f ′(x) and results in a family of curves. For example if f ′(x) = 1 then the antiderivative function is f (x) = x + c, where c can take any real value.
Sketching the antiderivative function from the graph of the original function 1. The general shape of the graph of the antiderivative function can be determined from the graph of a polynomial function by increasing the degree by one. For example, if f ′(x) is a quadratic function, then f (x) is a cubic function. 2. The xintercepts of f ′(x) become the turning points on the graph of f (x). 3. When f ′(x) is above the xaxis, the gradient of f (x) is positive. 4. When f ′(x) is below the xaxis, the gradient of f (x) is negative. Worked Example 9
Sketch the graph of the antiderivative function from the graph of the gradient function f ′(x) shown. f ′(x)
(1, 0)
(−2, 0) 0
x
(0, −1) Think
Write
1
State the shape of the antiderivative function.
The antiderivative function will be a positive cubic function.
2
Find the xintercepts of the gradient function, f ′(x), and hence find the xcoordinates of the turning points.
There are xintercepts when x = −2 and when x = 1, so f (x) has turning points when x = −2 and x = 1.
3
Find when the given graph, f ′(x), is above the xaxis and hence find when f (x) has a positive gradient.
f ′(x) is above the xaxis when x < −2 and when x > 1, so f (x) has a positive gradient when x < −2 and when x > 1.
4
Find when the given graph, f ′(x), is below the xaxis and hence find when f (x) has a negative gradient.
f ′(x) is below the xaxis when −2 < x < 1, so f (x) has a negative gradient when −2 < x < 1.
5
Sketch the curve.
The graph could be any of the family of graphs formed by vertical translations of the graph shown. f(x)
−2
422
Maths Quest 12 Mathematical Methods CAS for the TINspire
0
1
x
REMEMBER
1.
d f ( x ) = f ′( x ) dx
2. f ( x ) = ∫ f ′( x ) dx 3. ∫ a dx = ax + c 4. ∫ ax n dx =
ax n + 1 + c, n ≠ − 1 n +1
5. ∫ [ f ( x ) ± g( x )] dx =
∫ f ( x) dx ± ∫ g( x) dx
6. ∫ k f ( x ) dx = k ∫ f ( x ) dx 7. ∫ (ax + b) n dx =
(ax + b) n + 1 + c, n ≠ − 1 a( n + 1)
1 dx = log e x + c, x ≠ 0 or ∫ x −1 dx = log e x + c x
8. ∫
1 1 1 dx = log e ax + b + c or ∫ (ax + b)−1 dx = log e ax + b + c ax + b a a 10. Write the answer in the same form as the question unless otherwise stated. 11. To draw the graph of the antiderivative function f (x) from the graph of the gradient function f ′(x), use the xintercepts as the xcoordinates of the turning points, and make the gradient positive when f ′(x) is above the xaxis and negative when f ′(x) is below the xaxis. 12. For a polynomial function, the graph of f (x) is one degree greater than the graph of f ′(x). 9. ∫
Exercise
9a
Antidifferentiation 1 WE1 a x
Antidifferentiate each of the following, giving answers with positive indices. c x7 d 3x5 b x4
−
e 5x 2 x4 i 5
f
x3 j 2
2
3
m x 3 q
n 4 x 4
9 x2
2 WE2
−
−2x4
r
− 10
x6
g −6x 4 − x 4 k 3
h 2 x
o x
p
s
l
−3 7
8 x
t
x 5 x3 −6
(x x )
Find the following indefinite integrals.
a ∫ (2 x + 5) dx
b ∫ (3 x 2 + 4 x − 10) dx
c
∫ (10 x 4 + 6 x3 + 2) dx
d ∫ ( − 4 x 5 + x 3 − 6 x 2 + 2 x ) dx
e
∫ ( x3 + 12 − x 2 ) dx
f
∫ ( x + 3)( x − 7) dx
g ∫ 5( x 2 + 2 x − 1) dx
h ∫ ( x 2 + 4)( x − 7) dx
i
∫ x( x − 1)( x + 4) dx Chapter 9 Integration
423
3 MC
∫ ( x 2 + x + 2) dx is equal to:
A ∫ x 2 dx + x + 2
B ∫ x 2 dx + ∫ x dx + ∫ 2 dx
C ∫ (x 2 + x ) dx + 2
D x 2 + ∫ ( x + 2) dx
E x 2 + x + ∫ 2 dx 4 MC
∫ x( x + 3) dx is equal to:
A ∫ x dx ∫ ( x + 3) dx
B x ∫ ( x + 3) dx
C ( x + 1) ∫ x dx
D ∫ x dx + ∫ ( x + 3) dx
E
∫ ( x 2 + 3x) dx
5 WE3 Antidifferentiate each of the following by using ∫ (ax + b) n dx = a (x + 3)2 c 2(2x + 1)4 e (6x + 5)4 g (4 − x)3 i 4(8 − 3x)4 − k (2x + 3) 2 − m 6(4x − 7) 4 − o (6 − 5x) 3 6 MC
b d f h j l n p
(ax + b) n + 1 + c. a( n + 1)
(x − 5)3 −2(3x − 4)5 3(4x − 1)2 (7 − x)4 −3(8 − 9x)10 − (6x + 5) 3 − (3x − 8) 6 −10(7 − 5x)−4
∫ 3( x + 2)4 dx is equal to:
A 3 + ∫ ( x + 2)4 dx
B ∫ 3 dx + ∫ ( x + 2)4 dx
C 3 ∫ ( x + 2)4 dx
D 3 ∫ dx × ∫ ( x + 2)4 dx
E ( x + 2)4 ∫ 3 dx 7 WE4, 5 Antidifferentiate the following.
424
a
∫ x dx
3
b
∫ x dx
8
c
∫ 5x dx
d
∫ 3x dx
7
e
∫ 7x dx
4
f
∫ x + 3 dx
g
∫ x + 3 dx
3
h
∫ x + 4 dx
i
∫ x + 5 dx
j
∫ 3x + 2 dx
4
k
∫ 5x + 6 dx
8
l
∫ 2 x − 5 dx
o
∫ 5 − x dx
−5
m
∫ 3 + 2x
p
3 ∫ 6 − 11x dx
dx
n q
−2
−2
∫ 6 + 7x dx −2
∫ 4 − 3x
dx
Maths Quest 12 Mathematical Methods CAS for the TINspire
r
6
1
−6
3
1
−8
∫ 5 − 2x dx
6
∫ x + 5 dx is equal to:
8 mC A 6∫
1 dx x+5
D
∫ 6dx
B E
∫ ( x + 5) dx
9 We6 Find ∫
1
∫ 6dx ∫ x + 5 dx
∫ 3(4 x + 1)
j
∫
m eBook plus Digital doc
SkillSHEET 9.1 Substitution and evaluation
∫
x+
∫ 6dx + ∫ x + 5 dx
c
∫
f
∫ 4(2 x − 5)5 dx
i
∫
( x − 5)( x + 3) dx x3
l
∫
x2 + x4 dx x
6
∫ ( x + 5) dx
(2 x + 7) dx. x
10 For the following mixed sets, find: 1 a ∫ x4 + 2x + b ∫ (3 x + 1)5 dx x −5 3 d ∫ e ∫ dx dx 2x + 1 6 − 10x g
1
C
−3
( x + 4) 2 dx 2x −1 3 k ∫ 5 x 2 − 2 x + 3 x 3 dx
h
dx
2 dx 3 − x
x2 + 2x − 1 x
n
dx
∫
3x 2 + 2 x − 1 dx x2
10 − x + 2 x 4 dx ∫ x3
11 We7 Find the equation of the curve f (x) given that: a f ′(x) = 4x + 1 and the curve passes through (0, 2) b f ′(x) = 5 − 2x and the curve passes through (1, −1) − c f ′(x) = x 2 + 3 and the curve passes through (1, 4) d f ′(x) = x + x and f (4) = 10 1
e f ′(x) = x 3 − 3 x 2 + 50 and f (8) = −100 1 f f ′(x) = − 2 x and f (1) = −5 x g f ′(x) = (x + 4)3 and the curve passes through (−2, 5) − h f ′(x) = 8(1 − 2x) 5 and f (1) = 3 −1 i f ′(x) = (x + 5) and the curve passes through (−4, 2) 8 j f ′′(x) and f (3) = 7 ( x) = 7 − 2x 12 We8 If a curve has a stationary point (1, 5), and a gradient of 8x + k, where k is a constant, find: a the value of k b y when x = −2. 13 A curve g(x) has g ′( x ) = kx +2 x , where k is a constant, and a stationary point (1, 2). Find: x a the value of k b g(x) c g(4). 14 We9 Sketch the graph of the antiderivative functions from each of the following graphs. a b c f ′(x) f′(x) f ′(x)
(−1, 0)
(1, 0) x
0
(−2, 0) 0
x
0
(0, 2)
(0, −3)
x
(0, −1)
Chapter 9
Integration
425
f ′(x)
d
e
f
f ′(x)
f ′(x)
(0, 1) (0, 0)
( 1–2 , 0) 0 (1, 0)
0
x
0
x
x
(0, −1)
g
f ′(x)
(−2, 0) (0, 0)
9B
(1, 0)
x
Integration of e x, sin (x) and cos (x) Integration of the exponential function e x Since
d x (e ) = e x dx
then
∫ e x ddxx = e x + c
and or
Therefore,
d kx (e ) = kke kx , where k is a constant dx
∫ kekx dx = ekx + c k ∫ e kx dx = e kx + c 1 ∫ ekx ddx = k ekx + c.
Worked examPle 10
Antidifferentiate each of the following. − e 5x 4x a 3e b c (e x − 1)2 4 ThInk a
b
426
WrITe
1
Integrate by rule where k = 4.
2
Simplify.
1
Rewrite the function to be integrated so that the coefficient of the e term is clear.
2
Integrate by rule where k = −5.
maths Quest 12 mathematical methods CaS for the TInspire
a
b
3e 4 x +c 4 3 = 4 e4 x + c
∫ 3e4 x ddxx = ∫
e
−5 x
4
dx = =
1
−
∫ 4 e 5 x ddx 1 4
e −
−5 x
5
+c
3
1 = 4e
Simplify the antiderivative.
−5x 5x
1 = − 20 e
c
1
Expand the function to be integrated.
2
Integrate each term by the rule.
c
×
−5 x
1 −5
+c
+c
∫ (e x − 1)2 dx = ∫ (e2 x − 2e x + 1) dx = 12 e 2 x − 2e x + x + c
Integration of trigonometric functions Since
d [sin (ax)] = a cos (ax) dx
and
d [cos (ax )] = − a sin (ax ) dx
it follows that
∫ sin ( ax) dx =
−
1 cos ( ax ) + c a
1
∫ cos ( ax) dx = a sin ( ax) + c Worked examPle 11
Antidifferentiate the following. a sin (6x)
b 8 cos (4x)
c 3 sin
− x 2
ThInk
WrITe
a
Integrate by rule.
a
b
1
Integrate by rule.
b
2
Simplify the result.
1
Integrate by rule.
2
Simplify the result.
c
−
1
∫ sin (6 x)dx = 6 cos (6 x) + c 8 ∫ 8 cos (4 x)dx = 4 sin (4 x) + c = 2 sin (4x) + c
c
− x dx = 2
∫ 3 sin
−
3 1 2
−
− x cos + c 2
− x = 6 cos + c 2
Worked examPle 12
Find ∫ [ 2 e 4 x − 5 sin ( 2 x ) + 4 x ] dx. ThInk
WrITe
Method 1: Technologyfree 1
Integrate each term separately.
∫ [2e4 x − 5 sin (2 x) + 4 x]dx = 42 e 4 x −
2
Simplify each term where appropriate.
−5 2
cos (2 x ) + 42 x 2 + c
= 12 e 4 x + 25 cos (2 x ) + 2 x 2 + c
Chapter 9
Integration
427
Method 2: Technologyenabled 1
On a Calculator page, press: • MENU b • 4:Calculus 4 • 2:Integral 2 Complete the entry line as:
∫ (2e4 x − 5 sin (2 x) + 4 x) dx Press ENTER ·. Note: When integrating trigonometric functions, ensure the CAS is set to radian mode.
2
∴∫ (2e 4 x − 5 sin (2 x ) + 4 x ) dx
Write the answer and put in the constant of integration.
e 4 x 5 cos (2 x ) + + 2x2 + c 2 2
=
REMEMBER
Exercise
9b
1
1.
∫ e x dx = e x + c and ∫ ekx dx = k ekx + c
2.
∫ sin (ax) dx =
−1
a
cos (ax ) + c and ∫ cos (ax ) dx =
1 sin (ax ) + c a
Integration of e x, sin (x) and cos (x) 1 WE10 a e2x − d e 3x e6 x g 2 j
−8e−2x x
m 3e 2 e x − e− x p 2
Antidifferentiate each of the following. b e4x e 5e5x 2e 3 x h 3 x
−x 3
i
−3e6x x
l 0.1e 4
k e 3 n 3e
c e−x f 7e4x
o e x + e−x
2 Find an antiderivative of (1 + ex)2. 3 Find an antiderivative of (ex − 1)3. 4 Find an antiderivative of x3 − 3x2 + 6e3x. 5 MC If f ′(x) = e2x + k and f (x) has a stationary point (0, 2), where k is a constant, then: a k is equal to: A e B e2 C 1 D −1 E −e b f (1) is equal to: A e2 − 1 B e 2 + 12 C 12 e 2 + 12 D e4 E 1 12
428
Maths Quest 12 Mathematical Methods CAS for the TINspire
6 WE11 Antidifferentiate the following. a sin (3x) b sin (4x) cos ( 2 x ) d e sin (−2x) 3 4 sin (6 x ) g h 8 cos (4x) 3 x j −2 cos (−x) k sin 3 − x x m 3 sin n − 2 sin 5 4 − x p − 6 cos 2 s
− 2 sin
5x 2
t
− 3 cos
−6
sin (3x) x l cos 2 x o 4 cos 4
3x r 6 cos 4
7x 4
u 5 sin (πx) − 4x s x − sin π
b [sin (2x ) − cos ( x )] dx ∫
∫ [cos (4 x) + sin (2 x)] dx e ∫ [4 cos (4 x ) − 13 sin (2 x)] dx c
i
π x w − 2 cos 3
7 WE12 Find: a [sin ( x ) + cos ( x )] dx ∫
∫ 3 sin ( π2x
f cos (−3x)
2x q 4 sin 3
π x v 3 cos 2
g
c cos (7x)
) + 2 cos ( π3x ) dx
) − cos (2x) dx
d
∫ sin ( 2x
f
∫ [5x + 2 sin ( x)] dx
h ∫ [3e6 x − 4 sin (8 x ) + 7] dx
8 Find the antiderivative of e4x + sin (2x) + x3. 9 Find an antiderivative of 3x2 − 2 cos (2x) + 6e3x. 10 Antidifferentiate each of the following. 1 a x 3 − b x2 + 4 cos (2x) − e−x + e2 x 2x + 3 x x c sin + e 2 − (3 x − 1)4 3
d
x x x e 3 sin − 2 cos − e 5 3 2 −
f
1 x + e 4 x + cos 5 3x − 2 π x x + 2 x − 2 sin + 5 3
11 In each of the following, find f (x) if: π a f ′(x) = cos (x) and f = 5 b f ′(x) = 4 sin (2x) and f (0) = −1 2 x x x c f ′( x ) = 3 cos and f (π ) = 9 2 d f ′( x ) = cos − sin and f (2π) = −2. 2 4 4 dy π x 12 If = sin + k, where k is a constant, and y has a stationary point (3, 4), find: 6 dx a the value of k b the equation of the curve c y when x = 6. 13 A curve has a gradient function f ′(x) = 4 cos (2x) + ke x, where k is a constant, and a stationary point (0, −1). Find: a the value of k b the equation of the curve f (x) π c f correct to 2 decimal places. 6
Chapter 9 Integration
429
9C
Integration by recognition As we have seen, if d [ f ( x )] = g( x ) dx then
∫ g( x) dx = f ( x) + c, where g( x) = f '( x).
This result can be used to determine integrals of functions that are too difficult to antidifferentiate, via differentiation of a related function.
Worked examPle 13
eBook plus
a Find the derivative of the function y = (5x + 1)3. b Use this result to deduce the antiderivative of 3(5x + 1)2.
Tutorial
int0564
ThInk a
b
WrITe
1
Write the function and recognise that the chain rule can be used.
2
Let u equal the function inside the brackets. du Find . dx
3 4
Express y in terms of u.
5
Find
6
Write the chain rule.
7
Find
8
Replace u with the expression inside the brackets and simplify where applicable.
1
2
3
Worked example 13
a y = (5x + 1)3
Let u = 5x + 1 du =5 dx y = u3
dy . du
dy = 3u 2 du dy dy du = × dx du dx dy = 3u 2 × 5 dx
dy using the chain rule. dx
du dx = y + c1, express the relationship dx in integral notation. dy Remove a factor from so that it resembles dx the integral required. Since ∫
Divide both sides by the factor in order to obtain the integral required.
= 15(5x + 1)2 b
∫ 15(5x + 1)2 ddxx = (5x + 1)3 + c1 5 ∫ 3(5 x + 1)2 dx = (5 x + 1)3 + c1
∫ 3(5x + 1)2 dx = 15 (5x + 1)3 + c, where c =
c1 5
Therefore, the antiderivative of 3(5x + 1)2 is 15 (5x ( 5x ++ 11))33 + cc..
Note that the shorter form of the chain rule below can be used to differentiate. If f ( x ) = [ g( x ))]]n then f ′ ( x ) = ng ′ ( x )[ g( x ))]]n − 1 .
430
maths Quest 12 mathematical methods CaS for the TInspire
Worked examPle 14 a Differentiate e x . 3
b Hence, antidifferentiate 6x2 e x . 3
ThInk a
1
Write the equation and apply the chain rule to differentiate y.
2
Let u equal the index of e.
3
Find
4
Express y in terms of u.
5
Find
6
b
WrITe a
y = ex
3
Let u = x3
du . dx
du = 3x 2 dx y = eu
dy . du dy Find using the chain rule and dx replace u.
dy = eu du dy = 3 x 2 eu dx = 3x 2 e x
dy in integral notation. dx
b
1
Express
2
Multiply both sides by a constant to obtain the integral required.
3
∫ 3x 2 e x dx = e x 3
3
+ c1
2∫ 3 x 2 e x dx = 2e 2e x + 2c1 3
3
2e x ∫ 6 x 2 e x dx = 2e 3
3
+ c, c where c = 2c1.
Therefore, the antiderivative of 6 x 2 e x is 2e x + c. c 3
3
Note that the shorter form of the chain rule below can be used to differentiate. If y = e f ( x ) then
dy = f ′( x)e f ( x ) . dx
Worked examPle 15 a Find the derivative of sin (2 x + 1) and use this result to deduce the antiderivative of 8 cos (2 x + 1). b Differentiate loge(5x2 − 2) and hence antidifferentiate ThInk a
b
x . 5 x2 − 2 WrITe a f (x) = sin (2x + 1)
1
Define f (x).
2
Differentiate using f ′(x) = g′(x) cos [g(x)] where f (x) = sin [g(x)].
3
Express f (x) using integral notation.
4
Multiply both sides by whatever is necessary for it to resemble the integral required.
4 ∫ 2 cos (2 x + 1) 1)dx = 4 sin (2 x + 1) + c
5
Write the integral in the form in which the question is asked.
∫ 8 cos (2 x + 1)dx = 4 ssinin (2 x + 1) + c
1
Define f (x).
f ′(x) = 2 cos (2x + 1)
∫ 2 cos (2 x + 1)dx = sin (2 x + 1) + c1
The antiderivative of 8 cos (2x + 1) is 4 sin (2x + 1) + c. b f (x) = loge (5x2 − 2)
Chapter 9
Integration
431
2
Differentiate using f ′( x ) = where f (x) = loge [g(x)].
g ′( x ) g( x )
3
Express f (x) using integral notation.
4
Take out a factor so that f ′(x) resembles the integral required.
5
Divide both sides by the factor to obtain the required integral.
10 x 5x 2 − 2
f ′( x ) =
10 10x
∫ 5x 2 − 2 dx = loge 10 ∫
5 x 2 − 2 + c1
x dx = log e 5 x 2 − 2 + c1 −2
5x 2
x
∫ 5x 2 − 2 dx = 101 loge
5x 2 − 2 + c
The antiderivative of x is 1 log e 5 x 2 − 2 + c. 5 x 2 − 2 10
Worked examPle 16
Differentiate x cos (x) and hence find an antiderivative of x sin (x). ThInk
WrITe
Let y = x cos (x)
1
Write the rule.
2
Apply the product rule to differentiate x cos (x).
3
Express the result in integral notation. (Do not add c, as an antiderivative is required.)
∴ ∫ [cos ( x ) − x sin sin ( x )] ddxx = x cos ( x )
4
Express the integral as two separate integrals.
5
Simplify by integrating. (Do not add c.)
6
Make the expression to be integrated the subject of the equation.
cos ( x ) dx − ∫ x ssin in ( x ) ddxx = x cos ( x ) ∫ cos sin ( x ) − ∫ x ssin in ( x ) ddxx = x cos ( x ) − x sin ( x ) dx = x cos ( x ) − sin ( x ) ∫
7
Simplify.
dy = x [−sin (x)] + [cos (x)](1) dx = −x sin (x) + cos (x) = cos (x) − x sin (x)
x ) dx = ssin in ( x ) − x cos ( x ) ∫ x sin ( x) Therefore, an antiderivative of x sin (x) is sin (x) − x cos (x).
Worked examPle 17 a Show that
5x + 1 4 = 5− . x+1 x+1
b Hence, find
ThInk a
432
Use algebraic long division to divide the numerator into the denominator.
∫
5x + 1 dx. x+1 WrITe
)
5
a x + 1 5x + 1
maths Quest 12 mathematical methods CaS for the TInspire
5x + 5 −4
So
5x + 1 4 = 5− x +1 x +1
b
b
∫
5x + 1 4 dx = ∫ 5 − x +1 x + 1
1
Write the expression using integral notation.
2
Express as two separate integrals.
= ∫ 5 dx − ∫
3
Antidifferentiate each part.
= 5 x − 4 log e x + 1 + c
4 dx x +1
rememBer
1. To differentiate using the chain rule, use one of the following rules. (a) If f (x) = [g(x)]n then f ′(x) = ng′(x)[g(x)]n − 1 (b) If y = e f (x),
dy = f ′( x )e f ( x ) dx
dy dy du = × dx du dx (d) f ′(x) = g′(x) cos [g(x)] where f (x) = sin [g(x)] f ′(x) = −g′(x) sin [g(x)] where f (x) = cos [g(x)] (c)
(e) f ′( x ) =
g ′( x) where f ( x ) = logg e g( x ) . g( x )
2. To antidifferentiate, use ∫ g( x ) dx = f ( x ) + c where g(x) = f ′(x).
exerCISe
9C
Integration by recognition 1 We13 For each of the following find the derivative of the function in i and use this result to deduce the antiderivative of the function in ii. a i (3x − 2)8 ii 12(3x − 2)7 2 5 b i (x + 1) ii 5x(x2 + 1)4 1 c i 2x − 5 ii 2x − 5 d i
4x + 3
+ 3x − 1 f i 2 x −1 e i
(x2
ii 7)4
2 4x + 3
ii (2x + 3)(x2 + 3x − 7)3 4x ii 2 ( x − 1)2
2 mC The derivative of (x + 7)4 is 4(x + 7)3. a Therefore, the antiderivative of 4(x + 7)3 is: A (x + 7)4 + c B 14 (x + 7)4 + c D 3(x + 7)4 + c E 12(x + 7)4 + c 3 b The antiderivative of (x + 7) is: A (x + 7)4 + c B 14 (x + 7)4 + c D 3(x + 7)4 + c
C 4(x + 7)4 + c
C 4(x + 7)4 + c
E 12(x + 7)4 + c
Chapter 9
Integration
433
3 mC If the derivative of (2x − 3)6 is 12(2x − 3)5, then ∫ 6(2 x − 3)5 dx is: A 2(2x − 3)6 + c B 4(2x − 3)6 + c C (2x − 3)6 + c 1 6 6 D 6(2x − 3) + c E 2 (2x − 3) + c 4 We14
For each of the following differentiate i and hence antidifferentiate ii.
e4x − 5
a i c i ex
ii 2e4x − 5 ii x e x
2
b i e6 − 5x d i ex − x
2
ii 10e6 − 5x ii (1 − 2 x )e x − x
2
2
5 We15 For each of the following find the derivative of the function in i and use this result to deduce the antiderivative of the function in ii. a i sin (x − 5) ii cos (x − 5) b i sin (3x + 2) ii 6 cos (3x + 2) c i cos (4x − 7) ii sin (4x − 7) d i cos (6x − 3) ii 3 sin (6x − 3) e i sin (2 − 5x) ii 10 cos (2 − 5x) f i cos (3 − 4x) ii −2 sin (3 − 4x) 12 x 20 g i loge (5x + 2) ii h i loge (x2 + 3) ii 2 5x + 2 x +3 i i loge (x2 − 4x) ii
x−2 x2 − 4x
6 We16
Differentiate i and hence find an antiderivative of ii. 2[ x cos cos ( x ) − sin ( x )] sin ( x ) a i x cos (x) + 2 sin (x) ii x sin (x) b i ii x x2 c i e x sin (x) ii 3e x [sin (x) + cos (x)] d i x sin (x) ii x cos (x) e i x ex ii x ex
7 For each of the following differentiate i and use this result to antidifferentiate ii. a i (2x − 3x2)6 ii 6x5(1 − 3x)(2 − 3x)5 x3 + 2x
b i 8 We17
a Show that
9 a Show that 10
x3 + 2x
3x − 2 1 =3+ . x −1 x −1
8x − 7 5 =4+ . 2x − 3 2x − 3
6x − 5 − 4 = 3+ . 3 − 2x 3 − 2x 2
b Hence, find ∫
3x − 2 dx. x −1
b Hence, find ∫
5x + 8 dx. x+2
b Hence, find ∫
8x − 7 dx. 2x − 3
b Hence, find ∫
6x − 5 dx. 3 − 2x
If y = loge[cos (x)]: a find
dy . dx
13 Differentiate 14
3x 2 + 2
5x + 8 2 =5− . x+2 x+2
a Show that
11 a Show that 12
ii
b Hence, find ∫ ta tan ( x ) dx. cos ( x ) 1 and hence find an antiderivative of 2 . sin ( x ) sinn ( x )
Differentiate loge (3x2 − 4) and hence find an antiderivative of
x . −4
3x 2
15 Differentiate sin (ax + b) and hence find an antiderivative of cos (ax + b). (Here, a and b are constants.) 16 Differentiate cos (ax + b) and hence find an antiderivative of sin (ax + b). (Here, a and b are constants.)
434
maths Quest 12 mathematical methods CaS for the TInspire
17 Differentiate eax + b and hence find an antiderivative of eax + b. (Here, a and b are constants.) 18 Antidifferentiate each of the following. a sin (3π x + 1) b cos (1 − 4π x) π x πx d sinn 2 + e 3 cos + 5 2 3
9d
eBook plus
c eπ x + 3
Digital doc
f cos (x)esin (x)
WorkSHEET 9.1
approximating areas enclosed by functions
eBook plus Interactivity
int0254 Approximating areas enclosed by functions
There are several ways of finding an approximation to the area between a graph and the xaxis. We will look at two methods: 1. the left rectangle method 2. the right rectangle method
The left rectangle method Consider the area between the curve f (x) shown at right, the xaxis y f(x) and the lines x = 1 and x = 5. If the area is approximated by ‘left’ rectangles of width 1 unit then the top left corner of each rectangle touches the curve at one point. So, the height of rectangle R1 is f (1) units and the area of R1 = 1 × f (1) square units (area of a R1 R2 R3 R4 rectangle = height × width). Similarly, the area of R2 = 1 × f (2) square units, 0 1 2 3 4 5 x the area of R3 = 1 × f (3) square units, the area of R4 = 1 × f (4) square units. Therefore, the approximate area under the graph between the curve f (x), the xaxis and the lines x = 1 to x = 5 is 1[ f (1) + f (2) + f (3) + f (4)] square units, (the sum of the area of the four rectangles). If the same area was approximated using rectangle widths of 0.5 there would be 8 rectangles and the sum of their areas would be: 0.5[ f (1) + f (1.5) + f (2) + f (2.5) + f (3) + f (3.5) + f (4) + f (4.5)] square units. From the diagram it can be seen that the left rectangle approximation is less than the actual area under the curve. Worked examPle 18
Find an approximation for the area between the curve f (x) shown and the xaxis from x = 1 to x = 3 using left rectangles of width 0.5 units. The graph shown has the equation f (x) = 0.2x2 + 3.
eBook plus
y
f(x)
Tutorial
int0565 Worked example 18
3
0 0.5 1 1.5 2 2.5 3 x ThInk
WrITe
Method 1: Technologyfree 1
Write the number of rectangles and their width.
There are 4 rectangles of width 0.5 units.
Chapter 9
Integration
435
2
Find the height of each rectangle (left) by substituting the appropriate xvalue into the f (x) equation.
h1 = f (1) = 0.2(1)2 + 3 = 3.2 h2 = f (1.5) = 0.2(1.5)2 + 3 = 3.45 h3 = f (2) = 0.2(2)2 + 3 = 3.8 h4 = f (2.5) = 0.2(2.5)2 + 3 = 4.25
3
Area equals the width multiplied by the sum of the heights.
Area = width × (sum of heights of 4 rectangles) = 0.5(3.2 + 3.45 + 3.8 + 4.25) = 0.5(14.7)
4
Calculate this area.
5
State the solution.
The approximate area is 7.35 square units.
= 7.35
Method 2: Technologyenabled 1
Store the function to f (x). On a Calculator page, press: • MENU b • 1:Actions 1 • 1:Define 1 Complete the entry line as: Define f (x) = 0.2x2 + 3 Press ENTER ·.
2
To find the approximate area, complete the entry line as: 0.5( f (1) + f (1.5) + f (2) + f (2.5)) Press ENTER ·.
3
Write the answer with the correct units for area.
∴ The approximate area is 7.35 square units.
The right rectangle method y f(x) Consider the area between the curve f (x) shown at right, the xaxis and the lines x = 1 and x = 5. If the area is approximated by ‘right rectangles’ of width 1 unit then the top right corner of each rectangle touches the curve at one point. So, the height of R1 is f (2) units and the area of R1 is 1 × f (2) square units. R1 R2 R3 R4 Similarly, the area of R2 = 1 × f (3) square units, 0 1 2 3 4 5 x the area of R3 = 1 × f (4) square units the area of R4 = 1 × f (5) square units. Therefore, the approximate area between the curve f (x), the xaxis and the lines x = 1 to x = 5 is (R1 + R2 + R3 + R4) = 1[ f (2) + f (3) + f (4) + f (5)] square units. If the same area was approximated with upper rectangle widths of 0.5 units, the sum of their areas would equal: 0.5[ f (1.5) + f (2) + f (2.5) + f (3) + f (3.5) + f (4) + f (4.5) + f (5)] square units. From the diagram it can be seen that the right rectangle approximation is greater than the actual area under the curve.
For an increasing function, left rectangle approximation ≤ actual area ≤ right rectangle approximation. For a decreasing function, left rectangle approximation ≥ actual area ≥ right rectangle approximation.
436
Maths Quest 12 Mathematical Methods CAS for the TINspire
Worked Example 19
Find an approximation for the area in the diagram in worked example 18 using right rectangles of width 0.5 units. f (x) = 0.2x2 + 3 Think
Write
1
Find the number of rectangles and the height of each one (from left to right).
There are 4 rectangles: h1 = f (1.5) = 0.2(1.5)2 + 3 = 3.45 h2 = f (2) = 0.2(2)2 + 3 = 3.8 h3 = f (2.5) = 0.2(2.5)2 + 3 = 4.25 h4 = f (3) = 0.2(3)2 + 3 = 4.8
2
Area is the width of the interval multiplied by the sum of the heights.
Area = 0.5(3.45 + 3.8 + 4.25 + 4.8)
3
Calculate the area.
4
State the solution.
The approximate area is 8.15 square units.
= 0.5(16.3) = 8.15
It can be seen that the left rectangle approximation (7.35 units) is less than the right rectangle approximation (8.15 units). If the area is divided into narrower strips, the estimate of the area would be closer to the true value. Worked Example 20
With width intervals of 1 unit, calculate an approximation for the area between the graph of f (x) = x2 + 2 and the xaxis from x = −2 to x = 3 using: a left rectangles b right rectangles c averaging of the left and right rectangle areas. Think
Write
1
Sketch the graph of f (x) over a domain which exceeds the width of the required area
2
Draw the left and right rectangles.
y
y = x2 + 2
2 −2 −1
0 1 2 3 x = Left rectangles = Right rectangles
a
1
2
Calculate the height of the left rectangles by substituting the appropriate values of x into the equation for f (x). Note the two rectangles to the right and left of the origin have the same height and are equal in area. Find the area by multiplying the width by the sum of the heights.
a Left rectangle heights:
f (−2) = (−2)2 + 2 = 6 f (−1) = (−1)2 + 2 = 3 f (0) = 02 + 2 = 2 f (1) = 12 + 2 = 3 f (2) = 22 + 2 = 6
Area = 1(6 + 3 + 2 + 3 + 6) = 20 Using left rectangles, the approximate area is 20 square units.
Chapter 9 Integration
437
b
1
2
Calculate the height of the right rectangles by substituting th