Lang Short Calculus solutions

Serge Lang : Short Calculus Solutions to exercices

1

Numbers and Functions

1.2

Inequalities

1. For |x| = x, we have x < 3. For |x| = −x, we have −x < 3 and −3 < x. So, for this inequality, we have : −3 < x < 3 or (3, 3). 2. For |2x + 1| = (2x + 1), we have 2x + 1 ≤ 1 and x ≤ 0. For |2x + 1| = −(2x + 1), we have −2x − 1 ≤ 1 and x ≥ −1. So, for this inequality, we have : −1 ≤ x ≤ 0 or [−1, 0]. √ √ √ 3. For |x2 − 2| = (x2 − 2), we have x2 − 2 ≤ 1 and x ≤ 3 or x ≥ − 3 (since x2 = |x| by theorem 1, p. 10). For |x2 − 2| √ = −(x2 − 2), we have −x2 √ + 2 ≤ 1 and x ≤ −1. So, √ x ≥ 1 or √ for this inequality, we have : − 3 ≤ x ≤ −1 and 1 ≤ x ≤ 3 or [− 3, −1] ∪ [1, 3]. 4. 5. We have (x + 1)(x − 2) = 0 for x = −1 or x = 2. We verify the negativity of each interval between (−∞, −1), (−1, 2) and (2, ∞). So, for this inequality, we have −1 < x and x < 2 or (−1, 2). 6. We have (x − 1)(x + 1) = 0 for x = 1 or x = −1. We verify the positivity of each interval between (−∞, −1), (−1, 1) and (1, ∞). So, for this inequality, we have x < −1 and x > 1 or (−∞, −1) ∪ (1, ∞). 7. We have (x − 5)(x + 5) = 0 for x = 5 or x = −5. We verify the negativity of each interval between (−∞, −5), (−5, 5) and (5, ∞). So, for this inequality, we have −5 < x and x < 5 or (−5, 5). 8. We have x(x + 1) = 0 for x = 0 or x = −1. We verify the negativity of each interval between (−∞, −1], [−1, 0] and [0, ∞). So, for this inequality, we have −1 ≤ x and x ≤ 0 or [−1, 0]. 9. We have x2 (x − 1) = 0 for x = 0 or x = 1. We verify the positivity of each interval between (−∞, 0], [0, 1] and [1, ∞). So, for this inequality, we have 0 and x ≥ 1 or [0] ∪ [1, ∞). 10. We have (x − 5)2 (x + 10) = 0 for x = 5 or x = −10. We verify the negativity of each interval between (−∞, −10], [−10, 5] and [5, ∞). So, for this inequality, we have x ≤ −10 and x = 5 or (−∞, −10] ∪ [5]. 11. We have (x − 5)4 (x + 10) = 0 for x = 5 or x = −10. We verify the negativity of each interval between (−∞, −10], [−10, 5] and [5, ∞). So, for this inequality, we have x ≤ −10 and x = 5 or (−∞, −10] ∪ [5]. 12. We have (2x + 1)6 (x − 1) = 0 for x = − 21 or x = 1. We verify the positivity of each interval between (−∞, − 12 ], [− 21 , 1] and [1, ∞). So, for this inequality, we have x = − 21 or x ≥ 1 or [− 21 ] ∪ [1, ∞).

1

13. We have (4x + 7)20 (2x + 8) = 0 for x = − 74 or x = −4. We verify the negativity of each interval between (−∞, −4), (−4, − 47 ) and (− 47 , ∞). So, for this inequality, we have x < −4 or (−∞, −4) 14. We have to show that |x + y| ≥ |x| − |y|. Proof. |x| = |x + y − y| = |(x + y) + (−y)|

(Associativity)

≤ |x + y| + | − y|

(By theorem 3, p. 11)

≤ |x + y| + |y|

(Absolute value property)

|x| − |y| ≤ |x + y|

15. We have to show that |x − y| ≥ |x| − |y|. Proof. |x| = |x − y + y| = |(x − y) + (y)|

(Associativity)

≤ |x − y| + |y|


|x| − |y| ≤ |x − y|

16. We have to show that |x − y| ≤ |x| + |y|. Proof. |x − y| = |(x) + (−y)|

(Associativity)

≤ |x| + | − y|


≤ |x| + |y|

1.3

(Absolute value property)

Functions

1. f ( 34 ) = 34 , f (− 23 ) = − 32 2. f (2x + 1) =

1 2x+1

for x 6= − 12

3. g(1) = 0, g(−1) = 2, g(−54) = 108 4. f (z) = 2z − z 2 , f (w) = 2w − w2 √ 5. The function x21−2 is defined for x 6= ± 2. We have f (5) = √ 6. f (x) = 3 x is defined for all x. We have f (27) = 3. 2

1 23 .

7. (a) f (1) = 1 (b) f (2) = 1 (c) f (−3) = −1 (d) f (− 43 ) = −1 8. (a) f ( 12 ) = 1 (b) f (2) = 4 (c) f (−4) = 0 (d) f (− 34 ) = 0 9. (a) f (1) = −2 (b) f (−1) = −6 (c) f (x + 1) = 2(x + 1) + (x + 1)2 − 5 = x2 + 4x − 2 √ 10. f (x) = 4 x is defined for x ≥ 0. We have f (16) = 2.

1.4

Powers

1. 23 = 8 and 32 = 9 2. 5−1 =

1 5

and (−1)5 = −1

3. ( 12 )4 =

1 16

4. ( 13 )2 =

1 9

5. (− 12 )4 =

1

and 4 2 = 2 1

and 2 3 1 16

1

and 4− 2 =

1 2

6. 32 = 9 and 23 = 8 7. −3−1 = − 31 and −1−3 = −1 8. −2−2 =

1 4

and −2−2 =

1 4

9. −1−4 = 1 and −4−1 = − 14 1

1 10. (− 21 )9 = − 512 and 9− 2 =

2

1 3

Graphs and Curves

2.1

Coordinates

1.

3

2.

3. x is negative and y is positive. 4. x is negative and y is negative. 5.

6.

4

7.

2.2

Graphs

1.

2.

5

3.

4.

5.

6

6.

7.

8.

7

9.

10.

11.

8

12.

13.

14.

9

15.

16.

17.

10

18.

19.

20.

11

21.

22.

23.

12

24.

25.

26.

13

27.

28.

29.

14

30.

31.

32.

15

33.

34.

35.

16

36.

For other values of x, we define f (x) = x − n if n < x ≤ n + 1 for all n.

2.3

The straight line

1.

2. 17

3.

4.

−7−1 5. We have y2 − y1 = a(x2 − x1 ), then a = 2−(−1) = − 38 . With y = − 83 x + b and taking (−1, 1), we have (1) = − 38 (−1) + b, then b = − 53 . The equation of the line is : y = − 83 x − 53 . −1− 1

6. We have y2 − y1 = a(x2 − x1 ), then a = 4−32 = − 32 . With y = − 23 x + b and taking (4, −1), we have (−1) = − 23 (4) + b, then b = 5. The equation of the line is : y = − 32 x + 5. 18

√ 7. Here, x2 = x1 = 2, so this is not a straight line and √ the y-coordinate of any point can be arbitrary. It’s a vertical line whose equation is x = 2. 9 9 8. We have y2 − y1 = a(x2 − x1 ), then a = √4−(−5) = √3+3 . With y = √3+3 x + b and taking 3−(−3) √ √ √ 3 9 ( 3, 4), we have 4 = √3+3 ( 3) + b, then b = 4 − √93+3 . The equation of the line is :

y=

√9 x 3+3

+4−

√ √9 3 . 3+3

9. With a = 4, we have y = 4x + b. With (1, 1), we have 1 = 4(1) + b and b = −3. The equation of the line is : y = 4x − 3. 10. With a = −2, we have y = −2x + b. With ( 21 , 1), we have 1 = −2( 12 ) + b and b = 2. The equation of the line is : y = −2x + 2. √ √ √ 11. With a = − 21 , we have y = − 12 x + b. With ( 2, 3), we have 3 = − 21 ( 2) + b and b = 3 + 22 . √

The equation of the line is : y = − 12 x + 3 + 22 . √ √ √ √ 5), we have 5 = 2(−1) + b and b = 5 + 3. 12. With a = 3, we have y = 3x +√b. With (−1, √ The equation of the line is : y = 3x + 5 + 3. 13.

14.

19

15.

16.

17.

18.

20

19. We have y2 − y1 = a(x2 − x1 ), then a =

1− 12 −1−1

= − 14 .


−1−1 1 − 14 2

= −8. √


√1−3 2−2


√

2( 2+2) 2 √ = − √2−2 = − (√2−2)( = −2 2+2)

2−1 √ 3− 3)

=

1√ 3− 3

=

√ (3+ 3) √ √ (3− 3)(3+ 3)

=

√ 3+ 3 9−3

2+4 −2

=

=

√ 2 + 2.

√ 3+ 3 6 .

√3−1 = √ 2 . With y = √ 2 x + b and taking (π, 1), 2−π 2−π 2−π √ √ √2−3π . The equation of the line is : y = √ 2 x + √2−3π = 2−π 2−π 2−π

23. We have y2 − y1 = a(x2 − x1 ), then a = √ 2 (π) + b, then b = 2−π √ √−2π + √2−π = √ 2 (x 2−2π 2−π 2−π

we have 1 = √2 x 2−π

+

− π) + 1.

π−2 √ . With y = π−2 √ x + b and taking (1, π), we 1− 2 1− 2 √ π(1− 2)−(π−2) π−2 √ (1) + b, then b = √ √ 2 . The equation of the line is : y = have π = 1− = 2−π 1− √2 1− √ 2 √ √2 √ √ √ 2(1− 2) 2(π−2) 2−π 2 π 2−2 2 2−2 2 π−2 π−2 π−2 π−2 √ x+ √ = √ x− √ √ = √ x− √ √ √ (x− 2)+2. + + = 1− 1− 2 1− 2 1− 2 1− 2 1− 2 1− 2 1− 2 1− 2 2

24. We have y2 − y1 = a(x2 − x1 ), then a = √

√−3 . With y = √−3 x + b and 2+1 √ 2+1 2−1 + b, then b = 2√2+1 . The equation of the line is : √ 2√ 2+2 3 3 = √2+1 (−x − 1) + 2 = −(x + 1)( √2+1 ) + 2. 2+1 √ √ √ √ −3− 2 −3− 2 = = 3 + 2. With y = (3 + 2)x + b −1 −2−(−1)

25. We have y2 − y1 = a(x2 − x1 ), then a = taking (−1, 2), we have 2 = y=

√−3 x 2+1

+

√ 2√ 2−1 2+1

=

√−3 x 2+1

√−3 (−1) 2+1 + √−3 + 2+1


√−1−2 2−(−1)

=

√ √ and taking (−2, −3), then b√= 3 + 2 2. √ we have −3 √ = (3 + √2)(−2) + b, √ √ The equation √ of the line is : y = (3 + 2)x + 3 + 2 2 = (3 + 2)x + 3 + 2 + 2 = (3 + 2)(x + 1) = 2.

27. (a)

21

(b)

(c)

(d)

22

(e)

28. We have y = ax + b and y = cx + d with b 6= d. (a) We have to show that if the lines are parallel, they have no points in common. Proof. Suppose there is a common point on the two lines; at that point, the lines have the same y-coordinate. For x = 0, we have y = b = d which contradicts the fact that y−d c 6= d. For x 6= 0, we have a = y−b x and c = x ; because the lines are parallel, they have the same slope, then a = c, and : y−b y−d = x x y−b=y−d b=d

(Which contradicts the fact that b 6= d)

Therefore, the lines cannot have the same y-coordinate and we conclude that they have no point in common. (b) We have to show that, if the lines are not parallel, they have exactly one point in common.

23

Proof. Because the lines are not parallel, a 6= c. When the two lines have the same y-coordinate, we have : ax + b = cx + d ax − cx = d − b x(a − c) = d − b d−b x= a−c

(With a 6= c)

Considering that a, b, c, d are contants, the x-coordinate has a unique value. We conclude that there is only one point in common between the lines. 29. (a) 3x + 5 = 2x + 1, then x = −4. By substitution in y = 3x + 5, we find y = −7. (b) 3x − 2 = −x + 4, then x = 32 . By substitution in y = 3x − 2, we find y = 52 . (c) 2x + 3 = −x + 2, then x = − 13 . By substitution in y = 2x + 3, we find y = 73 . (d) x + 1 = 2x + 7, then x = −6. By substitution in y = x + 1, we find have y = −5.

2.4

Distance between two points √ (1 − (−3))2 + (4 − (−5))2 = 97 p √ L = (0 − 1)2 + (2 − 1))2 = 2 p √ L = (3 − (−1))2 + (−2 − 4))2 = 52 p √ L = (−1 − 1))2 + (2 − (−1))2 = 13 q q √ L = (1 − 12 ))2 + (1 − 2)2 = 54 = 25

1. L = 2. 3. 4. 5.

p

6. Let A(−1, 2), B(4, 2), C(−1, −3) and D(x, y) be the four vertices of the rectangle. Since A and B has the same y-coordinate, it means that AB is parallel to CD and C and D has the same y-coordinate. Since A and C has the same x-coordinate, then B and D has the same x-coordinate. We conclude that D(4, −3) is the fourth vertex. p 7. We have AB = CD = (4 − (−1))2 + (2 − 2)2 = 5. To check which segment is perpendicu2−2 lar, we have to consider the slopes : aAB = aCD = 4−(−1)) = 0 so AB and CD are horizontal lines; the slopes are not defined for aBD and aAC , so BD and ACpare vertical lines that are perpendicular to AB and CD. We can conclude that BD = AC = (−3 − 2)2 + (−1 − (−1))2 = 5. 8. Let A(−2, −2), B(3, −2), C(3, 5) and D(x, y) be the four vertices of the rectangle. Since A and B has the same y-coordinate, it means that AB is parallel to CD and C and D has the same y-coordinate. Since B and C has the same x-coordinate, then A and D has the same x-coordinate. We can conclude that D(−2, 5) is the fourth vertex. p 9. We have AB = CD = (−2 − (−2))2 + (3 − (−2))2 = 5. To check which segment is perpendicular, we have to consider the slopes : aAB = aCD = −2−(−2) 3−(−2)) = 0 so AB and CD are horizontal lines; the slopes are not defined for aBC or aAD , so BC and AD are vertical lines that are perpendicular to AB and CD. We can conclude that BC = AD = p (5 − (−2))2 + (3 − 3)2 = 7. 24

10. We have to show that, if the distance between numbers x and y is defined to be |x − y|, this is the same as the distance between the points (x, 0) and (y, 0) on the plane. Proof. Let L be the distance between the points (x, 0) and (y, 0) on the plane. We have : p p L = (0 − 0)2 + (y − x)2 = (y − x)2 = |y − x|


11.* With d(x, y) as the distance between numbers x and y, we have to show that if x, y, z are numbers, then d(x, z) ≤ d(x, y) + d(y, z) and d(x, y) = d(y, x). √ √ √ Proof. Let’s remind a square root property : a + b ≤ a + b. √ a + b ≤ a + b + 2 ab √ √ ≤ ( a + b)2 q √ √ √ a + b ≤ ( a + b)2 √ √ √ a + b ≤ | a + b| (Definition of absolute value) √ √ √ a+b≤ a+ b (Since a, b > 0)

Then we have : d(x, y) =

p p (x1 − x2 )2 + (y1 − y2 )2 = (y1 − y2 )2 + (x1 − x2 )2 = d(y, x)

d(x, z)2 = (x1 − x2 )2 + (z1 − z2 )2 ≤ (x1 − x2 )2 + (y1 − y2 )2 + (y1 − y2 )2 + (z1 − z2 )2 p

2.6

≤ d(x, y)2 + d(y, z)2 p d(x, z)2 ≤ d(x, y)2 + d(y, z)2 p p d(x, z) ≤ d(x, y)2 + d(y, z)2 = d(x, y) + d(y, z)

The circle

1. (a) We have center (2, −1) and radius 5.

25

(Since

√

a+b≤

√

a+

√

b)

(b) We have center (2, −1) and radius 2.

(c) We have center (2, −1) and radius 1.

(d) We have center (2, −1) and radius 3.

26

2. (a) We have center (0, 1) and radius 3.

(b) We have center (0, 1) and radius 2.

(c) We have center (0, 1) and radius 5.

27

(d) We have center (0, 1) and radius 1.

3. (a) We have center (−1, 0) and radius 1.

(b) We have center (−1, 0) and radius 2.

28

(c) We have center (−1, 0) and radius 3.

(d) We have center (−1, 0) and radius 5.

29

4. x2 + y 2 − 2x + 3y − 10 = 0 9 9 x2 − 2x + 1 − 1 + y 2 + 3y + − − 10 = 0 4 4 3 9 2 (x − 1) + (y + )2 = 10 + 1 + 2 4 3 53 (x − 1)2 + (y + )2 = 2 4 The center is (1, − 32 ) and the radius is

(Completing the square)

√

53 2 .

5. x2 + y 2 + 2x − 3y − 15 = 0 9 9 x2 + 2x + 1 − 1 + y 2 − 3y + − − 15 = 0 4 4 3 9 (x + 1)2 + (y − )2 = 15 + 1 + 2 4 3 2 73 2 (x + 1) + (y − ) = 2 4 The center is (−1, 23 ) and the radius is

√

73 2 .

30


6. x2 + y 2 + x − 2y − 16 = 0 x2 + x +

1 1 − + y 2 − 2y + 1 − 1 − 16 = 0 4 4 1 1 (x + )2 + (y − 1)2 = 16 + + 1 2 4 1 2 69 (x + ) + (y − 1)2 = 2 4

The center is (− 21 , 1) and the radius is


√

69 2 .

7. x2 + y 2 − x + 2y − 25 = 0 x2 − x +

1 1 − + y 2 + 2y + 1 − 1 − 25 = 0 4 4 1 1 (x − )2 + (y + 1)2 = 25 + + 1 2 4 105 1 2 (x − ) + (y + 1)2 = 2 4 31


The center is ( 12 , −1) and the radius is

2.7

√

105 2 .

Dilations and the ellipse

1. We have center (0, 0). For x = 0, y = ±4; for y = 0, x = ±3, then the vertices of the ellipse are (0, 4), (0, −4), (3, 0), (−3, 0).

2. We have center (0, 0) For x = 0, y = ±3; for y = 0, x = ±2, then the vertices of the ellipse are : (0, 3), (0, −3), (2, 0), (−2, 0).

32

√ 3. We have center (0, 0)√For x = 0,√y = ±3; for y = 0, x = ± 5, then the vertices of the ellipse are : (0, 3), (0, −3), ( 5, 0), (− 5, 0).

4. We have center (0, 0) For x = 0, y = ±5; for y = 0, x = ±2, then the vertices of the ellipse are : (0, 5), (0, −5), (2, 0), (−2, 0).

33

5. We have center (1, −2) For x = 1, y 2 + 4y + 4 = 16, then (y + 6)(y − 2) = 0, and y = −6 and 2; for y = −2, x2 − 2x + 1 = 9, then (x − 4)(x + 2), then x = 4 and −2. The vertices of the ellipse are : (1, 2), (1, −6), (4, −2), (−2, −2).

6. We have : 4x2 + 25y 2 = 100 25 x2 + y 2 = 25 4 x2 y 2 + =1 25 4 So we have center (0, 0) For x = 0, y = ±2; for y = 0, x = ±5, then the vertices of the ellipse are : (0, 2), (0, −2), (5, 0), (−5, 0). 34

7. We have center (−1, −2) For x = −1, y 2 + 4y + 4 = 4, then y(y + 4) = 0, and y = 0 and −4; for y√= −2, x2 + 2x + 1 = 3. Solving the equation with the quadratic formula, we have : √ √ b2 −4ac x = −b± 2a , then x = 3 − 1 and − 3 − 1. The vertices of the ellipse are : (−1, 0), √ √ (−1, −4), ( 3 − 1, −2), (− 3 − 1, −2).

8. We have : 25x2 + 16y 2 = 400 16 x2 + y 2 = 16 25 x2 y2 + =1 16 25 So we have center (0, 0) For x = 0, y = ±5; for y = 0, x = ±5, then the vertices of the ellipse are : (0, 5), (0, −5), (4, 0), (−4, 0).

35

9. We have center (1, −3) For x = 1, y 2 + 6y + 9 = 4, then (y + 5)(y + 1) = 0, and y = −5 and −1; for y = −3, x2 − 2x + 1 = 1, then x(x − 2), then x = 0 and 2. The vertices of the ellipse are : (1, −1), (1, −5), (2, −3), (0, −3).

2.8

The parabola

1. 2. 3. 4.

36

5. It’s a circle. x2 + y 2 − 4x + 2y − 20 = 0 x2 − 4x + 4 − 4 + y 2 + 2y + 1 − 1 − 20 = 0 (x − 2)2 + (y + 1)2 = 20 + 4 + 1 (x − 2)2 + (y + 1)2 = 25 6. It’s a circle. x2 + y 2 − 2y − 8 = 0 x2 − y 2 − 2y + 1 − 1 − 8 = 0 x2 + (y − 1)2 = 8 + 1 x2 + (y − 1)2 = 9 7. It’s a circle. x2 + y 2 + 2x − 2 = 0 x2 + 2x + 1 − 1 + y 2 − 2 = 0 (x + 1)2 + y 2 = 2 + 1 (x + 1)2 + y 2 = 3 8. It’s a parabola. y − 2x2 − x + 3 = 0 y 1 3 = x2 + x − 2 2 2 y 1 1 1 3 2 =x + x+ − − 2 2 16 16 2 y 1 2 1 3 = (x + ) − − 2 4 16 2 1 2 25 = 2(x + ) y+ 8 4 9. It’s a parabola. y − x2 − 4x − 5 = 0 y − 5 = x2 + 4x y − 5 = x2 + 4x + 4 − 4 y − 1 = (x + 2)2 10. It’s a parabola. y − x2 + 2x + 3 = 0 y + 3 = x2 + 2x y + 3 = x2 + 2x + 1 − 1 y + 3 = (x + 1)2 y + 4 = (x + 1)2 37

11. It’s a circle. x2 + y 2 + 2x − 4y = −3 x2 + 2x + y 2 − 4y = −3 x2 + 2x + 1 − 1 + y 2 − 4y + 4 − 4 = −3 (x + 1)2 − 1 + (y − 2)2 − 4 = −3 (x + 1)2 + (y − 2)2 = 2 12. It’s a circle. x2 + y 2 − 4y − 2y = −3 x2 − 4y + y 2 − 2y = −3 x2 − 4y + 4 − 4 + y 2 − 2y + 1 − 1 = −3 (x − 2)2 − 4 + (y − 1)2 − 1 = −3 (x − 2)2 + (y − 1)2 = 2 13. It’s a parabola. x − 2y 2 − y + 3 = 0 x = 2y 2 + y − 3 y 3 x = y2 + − 2 2 2 y 1 1 3 x 2 =y + + − − 2 2 16 16 2 x 1 1 3 = (y + )2 − − 2 4 16 2 x 1 2 1 3 = (y + ) − − 2 4 16 2 25 1 2 x+ = 2(y + ) 8 4 14. It’s a parabola. x − y 2 − 4y = 5 x − 5 = y 2 + 4y x − 5 = y 2 + 4y + 4 − 4 x − 5 = (y + 2)2 − 4 x − 1 = (y + 2)2

2.9

The hyperbola

1. 2. 3. 38

4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17.

3

The derivative

3.1

The slope of a curve

1. We have y = 2x2 and look for the slope at (1, 2). We have f (1 + h) = 2(1 + h)2 = 2 + 4h + 2h2 ; 2 2 −2 = 4h+2h = so the closest point is (1 + h, 2 + 4h + 2h2 ). Finding the slope, we have 2+4h+2h 1+h−1 h 4 + 2h. When h approches 0, the slope is 4. 2. We have y = x2 + 1 and look for the slope at (−1, 2). We have f (−1 + h) = (−1 + h)2 + 1 = 2 − 2h + h2 ; so the closest point is (−1 + h, 2 − 2h + h2 ). Finding the slope, we have 2−2h+h2 −2 −2h+h2 = −2 + h. When h approches 0, the slope is −2. h −1+h−(−1) = 3. We have y = 2x − 7 and look for the slope at (2, −3), but we already know that a = 2 because it’s a straight line. Let’s verify : we have f (2 + h) = 2(2 + h) − 7; so the closest point is (2 + h, 2h − 3). Finding the slope, we have 2h−3−(−3) = 2h 2+h−2 h = 2. Whatever h, the slope is 2. 2

3h 4. We have y = x3 and look for the slope at ( 12 , 18 ). We have f ( 12 +h) = ( 12 +h)3 = 18 +h3 + 3h 4 + 2 ; 2

3h so the closest point is ( 21 + h, 18 + h3 + 3h 4 + 2 ). Finding the slope, we have

h2 +

3 4

+

3h 2 .

5. We have y =

When h approches 0, the slope is 43 . 1 x

and look for the slope at (2, 12 ). We have f (2+h) =

1 (2 + h, 2+h ). Finding the slope, we have When h approches 0, the slope is − 41 .

1 − 12 2+h

2+h−2

39

=

1 − 12 2+h

h

=

2−(2+h) 2(2+h)

h

1 2+h ;

=

2 1 +h3 + 3h + 3h2 − 81 8 4 1 1 +h− 2 2

=

so the closest point is

−h 4+h

h

−h 1 = h1 ( 4+h ) = − 4+h .

6. We have y = x2 + 2x and look for the slope at (−1, −1). We have f (−1 + h) = (−1 + h)2 + 2(−1 + h) = h2 − 1; so the closest point is (−1 + h, h2 − 1). Finding the slope, we have h2 −1−(−1) h2 h−1−(−1) = h = h. When h approches 0, the slope is 0. 7. We have y = x2 and look for the slope at (2, 4). We have f (2 + h) = (2 + h)2 = h2 + 4h + 4; so 2 2 = h +4h = h+4. the closest point is (2+h, h2 +4h+4). Finding the slope, we have h +4h+4−4 2+h−2 h When h approches 0, the slope is 4. 8. We have y = x2 and look for the slope at (3, 9). We have f (3 + h) = (3 + h)2 = h2 + 6h + 9; so 2 2 the closest point is (3+h, h2 +6h+9). Finding the slope, we have h +6h+9−9 = h +6h = h+6. 3+h−3 h When h approches 0, the slope is 6. 9. We have y = x3 and look for the slope at (1, 1). We have f (1 + h) = (1 + h)3 = h2 + h3 + 2h + 2h2 + 1 + h = h3 + 3h2 + 3h + 1; so the closest point is (1 + h, h3 + 3h2 + 3h + 1). Finding 3 2 +3h+1−1 3 2 the slope, we have h +3h1+h−1 = h +3hh +3h = h2 + 3h + 3. When h approches 0, the slope is 3. 10. We have y = x3 and look for the slope at (2, 8). We have f (2 + h) = (2 + h)3 = 2h2 + h3 + 8h + 4h2 + 8 + 4h = h3 + 6h2 + 12h + 8; so the closest point is (2 + h, h3 + 6h2 + 12h + 8). 3 2 3 2 +12h+8−8 = h +6hh +12h = h2 + 6h + 12. When h approches Finding the slope, we have h +6h2+h−2 0, the slope is 12. 11. We have y = 2x + 3 and look for the slope at (2, 7), since y=2(2)+3=7. We have f (2 + h) = 2(2 + h) + 3 = 2h + 7; so the closest point is (2 + h, 2h + 7). Finding the slope, we have 2h+7−7 2h 2+h−2 = h = 2. Whatever h, the slope is 2. 12. We have y = 3x − 5 and look for the slope at (1, −2), since y=3(1)-5=-2. We have f (1 + h) = 3(1 + h) − 5 = 3h − 2; so the closest point is (1 + h, 3h − 2). Finding the slope, we have 3h−2−(−2) = 3h 1+h−1 h = 3. Whatever h, the slope is 3. 13. We have y = ax + b and look for the slope at (x, ax + b). We have f (x + h) = a(x + h) + b = ax + ah + b; so the closest point is (x + h, ax + ah + b). Finding the slope, we have ax+ah+b−(ax+b) = ah x+h−x h = a. Whatever h, the slope is a.

3.2

The derivative

1. We have f (x) = x2 + 1 and f (x + h) = (x + h)2 + 1 = x2 + 2xh + h2 + 1. Then f 0 (x) = f (x + h) − f (x) x2 + 2xh + h2 + 1 − (x2 + 1) = lim = lim (2x + h) = 2x. lim h→0 h→0 h→0 h h 2. We have f (x) = x3 and f (x + h) = (x + h)3 = x3 + 3x2 h + 3xh2 + h3 . Then f 0 (x) = x3 + 3x2 h + 3xh2 + h3 − x3 lim = lim (3x2 + 3xh + h2 ) = 3x2 . h→0 h→0 h 3. We have f (x) = 2x3 and f (x + h) = 2(x + h)3 = 2x3 + 6x2 h + 6xh2 + 2h3 . Then f 0 (x) = 2x3 + 6x2 h + 6xh2 + 2h3 − 2x3 lim = lim (6x2 + 6xh + 2h2 ) = 6x2 . h→0 h→0 h 4. We have f (x) = 3x2 and f (x + h) = 3(x + h)2 = 3x2 + 6xh + 3h2 . 3x2 + 6xh + 3h2 − 3x2 lim = lim (6x + 3h) = 6x. h→0 h→0 h 40

Then f 0 (x) =

5. We have f (x) = x2 − 5 and f (x + h) = (x + h)2 − 5 = x2 + 2xh + h2 − 5. Then f 0 (x) = x2 + 2xh + h2 − 5 − (x2 − 5) lim = lim (2x + h) = 2x. h→0 h→0 h 6. We have f (x) = 2x2 + x and f (x + h) = 2(x + h)2 + x + h = 2x2 + 4xh + 2h2 + x + h. Then 2x2 + 4xh + 2h2 + x + h − (2x2 + x) f 0 (x) = lim = lim (4x + 2h + 1) = 4x + 1. h→0 h→0 h 7. We have f (x) = 2x2 − 3x and f (x + h) = 2(x + h)2 − 3(x + h) = 2x2 + 4xh + 2h2 − 3x − 3h. 2x2 + 4xh + 2h2 − 3x − 3h − (2x2 − 3x) = lim (4x + 2h − 3) = 4x − 3. Then f 0 (x) = lim h→0 h→0 h 8. We have f (x) = 12 x3 + 2x and f (x + h) = 12 x3 + 23 x2 h + 23 xh2 + 21 h3 + 2x + 2h. Then 1 3 x + 32 x2 h + 23 xh2 + 12 h3 + 2x + 2h − ( 21 x3 + 2x) 3 3 1 lim 2 = lim ( x2 + xh+ h2 +2) = h→0 h→0 2 h 2 2 9. We have f (x) =

1 x+1

and f (x+h) =

1 x+h+1 .

Then

f 0 (x)

1 x+h+1

= lim

h→0

(x + 1) − (x + h + 1) −1 −1 = lim = . h→0 h(x + h + 1)(x + 1) h→0 (x + h + 1)(x + 1) (x + 1)2

− h

1 x+1

= lim

h→0

f 0 (x) = 3 2 x +2. 2

(x+1)−(x+h+1) (x+h+1)(x+1)

h

=

lim

10. We have f (x) = lim

h→0

2 x+1

and f (x+h) =

2 x+h+1 .

Then

f 0 (x)

2 x+h+1

= lim

h→0

2(x + 1) − 2(x + h + 1) −2 −2 = lim = . h→0 (x + h + 1)(x + 1) h(x + h + 1)(x + 1) (x + 1)2

− h

2 x+1

= lim

h→0

2(x+1)−2(x+h+1) (x+h+1)(x+1)

h

*11. For each case in Exercice 1 : 1. Finding the tangent line at (2, 5), the slope is 4 = line is y = 4x − 3.

y−5 x−2 ,

and the equation of the tangent

2. Finding the tangent line at (2, 8), the slope is 12 = line is y = 12x − 16.

y−8 x−2 ,


3. Finding the tangent line at (2, 16), the slope is 24 = tangent line is y = 24x − 32.

y−16 x−2 ,

and the equation of the

4. Finding the tangent line at (2, 12), the slope is 12 = tangent line is y = 12x − 12.

y−12 x−2 ,

and the equation of the

5. Finding the tangent line at (2, −1), the slope is 4 = line is y = 4x − 9.

y+1 x−2 ,



y−10 x−2 ,



y−2 x−2 ,



y−8 x−2 ,


9. Finding the tangent line at (2, 13 ), the slope is − 19 = line is y = − 91 x + 59 . 10. Finding the tangent line at (2, 23 ), the slope is − 29 = line is y = − 92 x + 10 9 . 41

y− 31 x−2 ,


y− 32 x−2 ,


=

*12. We have f (x) = −x if x ≤ 0 and f (x) = 2 if x > 0. We look for f 0 (x) when x = −1 −(x + h) − (−x) = −1. For x = 0, we first look for the left and we have f 0 (x) = lim h→0 h derivative : since 0 + h ≤ 0 for h ≤ 0, we have f (0 + h) = f (h) = −h and f (0) = 0, then f (0 + h) − f (0) −h f 0 (x) = lim = lim = −1. For the right derivative : since 0 + h > 0 for h→0 h→0 h h f (0 + h) − f (0) 2 h > 0, we have f (0 + h) = 2 and f (0) = 0, then f 0 (x) = lim = lim , which h→0 h→0 h h is impossible, so there is no right derivative. *13. We have f (x) = |x| + x. For h > 0, we take the right derivative for x = 0 : f (0 + h) = 2h f (0 + h) − f (0) = lim = 2. f (h) = h + h = 2h and f (0) = 0, and we have : f 0 (x) = lim h→0 h h→0 h For h < 0, we take the left derivative for x = 0 : f (0 + h) = f (h) = 0 and f (0) = 0, f (0 + h) − f (0) so : f 0 (x) = lim = 0. Since the right and the left derivatives are not h→0 h equal, the derivative is not defined for x = 0. If x > 0, then x + h > 0, so f 0 (x) = 2x + 2h − 2x 2h f (x + h) − f (x) = lim lim = 2; also, if x < 0, then x + h < 0, so lim h→0 h→0 h h→0 h h f (x + h) − f (x) −x − h + x + h − (−x + x) f 0 (x) = lim = lim = 0. Then, for x 6= 0, the x→0 h→0 h h left and right derivatives are equal for each value of x, so the derivative is defined for these values. *14. Let f (x) = 0 if x ≤ 1 and f (x) = x if x > 1. Finding the left derivative for x = 1, we have h < 0, so (1 + h) ≤ 1; then we havef (1 + h) = 0 and f 0 (1 + h) = 0. Finding the right derivative, we have h > 0, so (1 + h) > 1; then we have f (1 + h) = 1 + h and f 0 (1 + h) = f (1 + h) − f (1) 1+h lim = lim , which is impossible, so there is no right derivative; we h→0 h→0 h h conclude that the derivative is not defined for x = 1. If x 6= 1, then if x > 1 and x + h > 1, x+h−x h f (x + h) − f (x) = lim lim = 1; also, if x < 0, then x + h < 0, so so f 0 (x) = lim h→0 h→0 h h→0 h h 0 f (x + h) − f (x) 0 = lim = 0. f (x) = lim x→0 h h→0 h *15. (a) Let f (x) = x|x|. At x = 0, we find the right derivative : for h > 0, we have f (0 + f (0 + h) − f (0) h) = f (h) = h2 and f (0) = 0, so f 0 (x) = lim = 0. We find the left h→0 h derivative : for h < 0, we have f (0 + h) = f (h) = −h2 and f (0) = 0, so f 0 (x) = f (0 + h) − f (0) lim = 0. The derivative exists and is equal to 0. h→0 h (b) Let f (x) = x2 |x|. At x = 0, we find the right derivative : for h > 0, we have f (0 + f (0 + h) − f (0) h) = f (h) = h3 and f (0) = 0, so f 0 (x) = lim = 0. We find the left h→0 h 3 derivative : for h < 0, we have f (0 + h) = f (h) = −h and f (0) = 0, so f 0 (x) = f (0 + h) − f (0) lim = 0. The derivative exists and is equal to 0. h→0 h (c) Let f (x) = x3 |x|. At x = 0, we find the right derivative : for h > 0, we have f (0 + f (0 + h) − f (0) h) = f (h) = h4 and f (0) = 0, so f 0 (x) = lim = 0. We find the left h→0 h derivative : for h < 0, we have f (0 + h) = f (h) = −h4 and f (0) = 0, so f 0 (x) = f (0 + h) − f (0) = 0. The derivative exists and is equal to 0. lim h→0 h 42

3.3

Limits

1. Let f (x) = 2x2 + 3x, then : 2(x + h)2 + 3(x + h) − (2x2 + 3x) h→0 h 2 2 2x + 4xh + 2h + 3x + 3h − 2x2 − 3x = lim h→0 h = lim (4x + 2h + 3)

f 0 (x) = lim

h→0

= lim 4x + lim 2h + lim 3 h→0

h→0

(Property 1)

h→0

= 4x + 3 2. Let f (x) =

1 2x+1 ,

f 0 (x) = lim

then : 1 2(x+h)+1

−

1 2x+1

h

h→0

= lim

h→0 4x2

−2 + 4x + 4xh + 2h + 1

Taking the numerator, we have lim −2 = −2, and taking de denominator, we have : h→0

lim (4x2 + 4x + 4xh + 2h + 1) = lim 4x2 + lim 4x + lim 4xh + lim 2h + lim 1 (Property 1)

h→0

h→0 2

h→0

h→0

h→0

h→0

= 4x + 4x + 1 = (2x + 1)2 2 By using Property 3, f 0 (x) = − (2x+1) 2.

3. Let f (x) =

x x+1 ,

then :

f 0 (x) = lim

x+h x+h+1

h→0

= lim

h→0 x2

− h

x x+1

1 + 2x + xh + h + 1

Taking the numerator, we have lim 1 = 1, and taking de denominator, we have : h→0

lim (4x2 + 4x + 4xh + 2h + 1) = lim x2 + lim 2x + lim xh + lim h + lim 1 (Property 1)

h→0

h→0

h→0

= x2 + 2x + 1 = (x + 1)2 By using Property 3, f 0 (x) =

1 . (x+1)2

43

h→0

h→0

h→0

4. Let f (x) = x(x + 1) = x2 + x, then : (x + h)2 + x + h − (x2 + x) h→0 h x2 + 2xh + h2 + x + h − x2 − x = lim h→0 h = lim (2x + h + 1)

f 0 (x) = lim

h→0

= lim 2x + lim h + lim 1 h→0

h→0

(Property 1)

h→0

= 2x + 1 5. Let f (x) =

x 2x−1 ,

0

f (x) = lim

then : x+h 2x+2h−1

h→0

x 2x−1

h

h→0

= lim

−

−1 4x2 + 4xh − 4x − 2h + 1


lim (4x2 + 4xh − 4x − 2h + 1) = lim 4x2 + lim 4xh − lim 4x − lim 2h + lim 1 (Property 1)

h→0

h→0 2

h→0

h→0

h→0

= 4x − 4x + 1 = (2x − 1)2 1 By using Property 3, f 0 (x) = − (2x−1) 2.

6. Let f (x) = 3x3 , then : 3(x + h)3 − 3x3 h→0 h 2 h + 9xh2 + 3h3 9x f 0 (x) = lim h→0 h f 0 (x) = lim (9x2 + 9xh + 3h2 )

f 0 (x) = lim

h→0

0

f (x) = lim 9x2 + lim 9xh + lim 3h2 h→0 2

h→0

(Property 1)

h→0

f 0 (x) = 9x

44

h→0

7. Let f (x) = x4 , then : (x + h)4 − x4 h→0 h 4 + 4x3 h + 6x2 h2 + 4xh3 + h4 − x4 x f 0 (x) = lim h→0 h 3 h + 6x2 h2 + 4xh3 + h4 4x f 0 (x) = lim h→0 h 0 3 2 f (x) = lim (4x + 6x h + 4xh2 + h3 ) f 0 (x) = lim

h→0

f 0 (x) = lim 4x3 + lim 6x2 h + lim 4xh2 + lim h3 h→0 3

h→0

h→0

(Property 1)

h→0

f 0 (x) = 4x

8. Let f (x) = x5 , then : (x + h)5 − x5 h→0 h 5 x + 5x4 h + 10x3 h2 + 10x2 h3 + 5xh4 + h5 − x5 f 0 (x) = lim h→0 h 0 4 3 2 2 f (x) = lim (5x + 10x h + 10x h + 5xh3 + h4 )

f 0 (x) = lim

h→0

f 0 (x) = lim 5x4 + lim 10x3 h + lim 10x2 h2 + lim 5xh3 + lim h4 h→0 4

h→0

h→0

h→0

h→0

(Property 1)

f 0 (x) = 5x

9. Let f (x) = 2x3 , then : 2(x + h)3 − 2x3 h→0 h 3 2x + 6x2 h + 6xh2 + 2h3 − 2x3 f 0 (x) = lim h→0 h 0 2 f (x) = lim (6x + 6xh + 2h2 ) f 0 (x) = lim

h→0

0

f (x) = lim 6x2 + lim 6xh + lim 2h2 h→0 2

h→0

(Property 1)

h→0

f 0 (x) = 6x

10. Let f (x) = 21 x3 + x, then : + h)3 + (x + h) − 12 x3 − x h→0 h 1 3 3 2 3 1 3 1 3 2 x + x h + 2 2 xh + 2 h + x + h − 2 x − x f 0 (x) = lim 2 h→0 h 3 3 1 f 0 (x) = lim ( x2 + xh + h2 + 1) h→0 2 2 2 3 3 1 f 0 (x) = lim x2 + lim xh + lim h2 + lim 1 h→0 2 h→0 2 h→0 h→0 2 3 f 0 (x) = x2 + 1 2 f 0 (x) = lim

1 2 (x

45

(Property 1)

11. Let f (x) = x2 , then : 2 x+h

− x2 f (x) = lim h→0 h −2 = lim 2 h→0 x + xh 0


lim (x2 + xh) = lim x2 + lim xh = x2

h→0

h→0

(Property 1)

h→0

By using Property 3, f 0 (x) = − x22 . 12. Let f (x) = x3 , then : − x3 h→0 h −3 = lim 2 h→0 x + xh

f 0 (x) = lim

3 x+h


lim (x2 + xh) = lim x2 + lim xh = x2

h→0

h→0

(Property 1)

h→0

By using Property 3, f 0 (x) = − x32 . 13. Let f (x) =

1 2x−3 ,

0

f (x) = lim

h→0

= lim

h→0

then : 1 2(x+h)−3

−

1 2x−3

h −2h (2x+2h−3)(2x−3)

h

−2 h→0 (2x + 2h − 3)(2x − 3)

= lim


lim [(2x + 2h − 3)(2x − 3)] = lim (2x + 2h − 3) · lim (2x − 3)

h→0

h→0

h→0

= (2x − 3)(2x − 3) = (2x − 3)2 2 By using Property 3, f 0 (x) = − (2x−3) 2.

46

(Property 2)

14. Let f (x) =

1 3x+1 ,

0

f (x) = lim

then : 1 3(x+h)+1

1 3x+1

h

h→0

= lim

−

−3h (3x+3h+1)(3x+1)

h

h→0

−3 h→0 (3x + 3h + 1)(3x + 1)

= lim


lim [(3x + 3h + 1)(3x + 1)] = lim (3x + 3h + 1) · lim (3x + 1)

h→0

h→0

h→0

(Property 2)

= (3x + 1)(3x + 1) = (3x + 1)2 3 By using Property 3, f 0 (x) = − (3x+1) 2.

15. Let f (x) =

1 x+5 ,

then :

f 0 (x) = lim

1 x+h+5

− h

h→0

f 0 (x) = lim

1 x+5

−h (x+h+5)(x+5)

h

h→0

−1 h→0 (x + h + 5)(x + 5)

f 0 (x) = lim


lim [(x + h + 5)(x + 5)] = lim (x + h + 5) · lim (x + 5)

h→0

h→0

h→0

= (x + 5)(x + 5) = (x + 5)2 1 By using Property 3, f 0 (x) = − (x+5) 2.

16. Let f (x) =

1 x−2 ,

then :

0

f (x) = lim

h→0

0

f (x) = lim

h→0

0

f (x) = lim

h→0

1 x+h−2

− h

1 x−2

−h (x+h−2)(x−2)

h −1 (x + h − 2)(x − 2)

47

(Property 2)


lim [(x + h − 2)(x − 2)] = lim (x + h − 2) · lim (x − 2)

h→0

h→0

(Property 2)

h→0

= (x − 2)(x − 2) = (x − 2)2 1 By using Property 3, f 0 (x) = − (x−2) 2.

17. Let f (x) =

1 , x2

then :

0

f (x) = lim

h→0

f 0 (x) = lim

1 (x+h)2

−

1 x2

h −2xh−h2 (x+h)2 x2

h −2x −h f 0 (x) = lim h→0 (x + h)2 x2 h→0

Taking the numerator, we have lim (−2x − h) = − lim 2x − lim h = −2x, and taking de h→0

h→0

h→0

denominator, we have : lim [(x + h)2 x2 ] = lim (x + h)2 · lim x2

h→0

h→0 2 2

(Property 2)

h→0

=x ·x = x4 By using Property 3, f 0 (x) = 18. Let f (x) =

1 , (x+1)2

0

f (x) = lim

h→0

f 0 (x) = lim

−2x x4

= − x23 .

then : 1 (x+h+1)2

−

1 (x+1)2

h −2xh−h2 −2h (x+h+1)2 (x+1)2

h −2x −h−2 f 0 (x) = lim h→0 (x + h + 1)2 (x + 1)2 h→0

Taking the numerator, we have lim (−2x − h − 2) = − lim 2x − lim h − lim 2 = −2x − 2, and h→0

h→0

h→0

h→0

taking de denominator, we have : lim [(x + h + 1)2 (x + 1)2 ] = lim (x + h + 1)2 · lim (x + 1)2

h→0

h→0

h→0

= (x + 1)2 (x + 1)2 = (x + 1)4 By using Property 3, f 0 (x) =

−2x−2 (x+1)4

=

−2(x+1) (x+1)4

48

2 = − (x+1) 3.

(Property 2)

3.4

Powers

1. We have : (x + h)4 = x(x + h)(x + h)(x + h) + h(x + h)(x + h)(x + h) = x[x(x + h)(x + h) + h(x + h)(x + h)] + h[x(x + h)(x + h) + h(x + h)(x + h)] = x[x(x2 + 2xh + h2 ) + h(x2 + 2xh + h2 )] + h[x(x2 + 2xh + h2 ) + h(x2 + 2xh + h2 )] = x[x3 + 2x2 h + xh2 + x2 h + 2xh2 + h3 ] + h[x3 + 2x2 h + xh2 + x2 h + 2xh2 + h3 ] = (x4 + 3x3 h + 3x2 h2 + xh3 ) + (x3 h + 3x2 h2 + 3xh3 + h4 ) = x4 + 4x3 h + 6x2 h2 + 4xh3 + h4 2. We have

df dx

= nxn−1 . For f (x) = x4 , then f 0 (x) = 4x3 . 2

3. (a) For f (x) = x 3 , f 0 (x) = 3

2 . 3x3

(b) For f (x) = x− 2 , f 0 (x) = −

3 2

.

2x 5 7

1

(c) For f (x) = x 6 , f 0 (x) = − 76 x 6 . 4. We have f (x) = x9 , then f 0 (x) = 9x8 . The slope at f 0 (1) = 9, so the equation at (1, 1) is y−1 9 = x−1 , and y = 9x − 8. 2

5. We have f (x) = x 3 , then f 0 (x) = 1 3

=

y−4 x−8 ,

and y = 13 x + 43 .

2 1

3x 3

3

6. We have f (x) = x− 4 , then f 0 (x) = − so the equation at (16, 18 ) is − 239 =

. The slope at f 0 (8) = 13 , so the equation at (8, 4) is

3 7

. The slope at f 0 (16) = − 239 . We have f (16) = 18 ,

4x 4 y− 81 x−16 , and

y = − 239 x +

7 32 .

√ 1 1 7. We have f (x) = x = x 2 , then f 0 (x) = 11 . The slope at f 0 (3) = 2√ = 3 2 2x √ √ √ √ √ √ 3 3 3 f (3) = 3, so the equation at (3, 3) is 63 = y− x−3 , and y = 6 x + 2 . 1

8. (a) For f (x) = x 4 , we have f 0 (x) =

1 3

, so f 0 (5) =

4x 4

1 3

.

4·5 4

, we have f 0 (x) = − 15 , so f 0 (7) = − 1 5 . 4x 4 4·7 4 √ √ √2−1 √ 2 0 0 (c) For f (x) = x , we have f (x) = 2x , so f (10) = 2(10 2−1 ).

(b) For f (x) =

1

1

x4 √

(d) For f (x) = xπ , we have f 0 (x) = πxπ−1 , so f 0 (7) = π7π−1 .

3.5

Sum, products, and quotients

1.

49

√

3 6 .

We have

Lang Short Calculus solutions

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