Lab 4

Lab 4

12/8/15, 7� 17 PM

Chris Ramos Math 2413.WS1 F2015, Fall 2015 Instructor: Denise Brown

WebAssign Lab 4 (Lab) Current Score : 80 / 100

Due : Monday, October 26 2015 11:59 PM CDT

The due date for this assignment is past. Your work can be viewed below, but no changes can be made.

grant you an extension if you Important! Before you view the answer key, decide whether or not you plan to request an extension. Your Instructor may not grant have viewed the answer key. Automatic extensions are not granted if you have viewed the answer key. Request Extension

AnswersSCalcET8 4.TF.002. 1. 10/10 points | Previous AnswersSCalcET8

Determine whether the statement is true or fa lse. If f has has an absolute minimum minimum value value at c , then

f ' (c )

= 0.

True False

Solution or Explanation Explanation Click to View Solution Solution

Show My Work (Required)

What steps or reasoning did you use? Your work counts towards your score. The Fermat's Theorem says that if

f

has a local maximum or minimum at c, and f'(c), exists, then f'(c)=0.

This states absolute minimum, so the theory doesn't apply. apply. Uploaded File (10 file maximum)

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Lab 4

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AnswersSCalcET8 4.TF.008. 2. 10/10 points | Previous AnswersSCalcET8

Determine whether the statement is true or fa lse. There exists a function

f such

that f (1) = !4,

f (5)

= 0 and

f ' f ' ( x x )

> 1 for all x .

True False

Solution or Explanation Click to View Solution


What steps or reasoning did you use? Your work counts towards your score. ["f",{"tmpl":["\\left({","}\\right)"],"args":[["5"]],"inline":0,"arity":1},"-","f",{"tmpl":["\\left({","}\\right)"],"args": [["1"]],"inline":0,"arity":1},"=","f","'",{"tmpl":["\\left({","}\\right)"],"args":[["x"]],"inline":0,"arity":1},{"tmpl": ["\\left({","}\\right)"],"args":[["5","-","1"]],"inline":0,"arity":1}," \\\\& ","4","=","f","'",{"tmpl":["\\left({","}\\right)"],"args": [["x"]],"inline":0,"arity":1},{"tmpl":["\\left({","}\\right)"],"args":[["4"]],"inline":0,"arity":1}," \\\\& ","f","'",{"tmpl": ["\\left({","}\\right)"],"args":[["x"]],"inline":0,"arity":1},"=","1"," \\\\& "," \\\\& ","f ","'",{"tmpl":["\\left({","}\\right)"],"args": [["x"]],"inline":0,"arity":1},"\\geq","1","\\;","\\neq","f","'",{"tmpl":["\\left({","}\\right)"],"args": [["x"]],"inline":0,"arity":1},"\\gt","1"] []

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Lab 4

12/8/15, 7� 17 PM

AnswersSCalcET8 4.1.054. 3. 10/10 points | Previous AnswersSCalcET8

Find the absolute maximum and absolute minimum values of f on on the given interval. f ( x x )

=

x x 2

+ 1 ! x +

, [0, 3] $$0

absolute minimum value

$$1 absolute maximum value

Solution or Explanation ( x + 1) ! x (2 (2 x ! 1) + 1 ! 2 x 2 + x 1 ! x 2 x 2 ! x + x 2 ! x + = = = 0  x = = !1, but x = = !1 is + 1 x 2 ! x + ( x + 1)2 ( x + 1)2 ( x + 1)2 x 2 ! x + x 2 ! x + x 2 ! x + 3 not in the given interval, [0, 3]. f (0) (0) = 0, f (1) = 1, and f (3) = . So f (1) = 1 is the absolute maximum value and f (0) (0) = 0 is the 7 absolute minimum value. f ( x x )

=

x

, [0, 3].

f ' ( x x )

=


What steps or reasoning did you use? Your work counts towards your score. Uploaded File (10 file maximum)

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Lab 4

12/8/15, 7� 17 PM


After the consumption of an a lcoholic beverage, the concentration of alcohol in the bloodstream (blood alcohol concentration, or BAC) surges as the alcohol is absorbed, followed by a gradual decline as the alcohol is metabolized. The function C (t )

= 1.35te!2.802t †

models the average BAC, measured in mg/mL, of a group of eight male subjects t hours hours after rapid consumption of 15 mL of ethanol (corresponding to one alcoholic drink). What is the maximum average BAC during the first 2 hours? (Round your answer to three decimal places.) -.0229 -.0229

0.177 mg/mL

When does it occur? (Round your answer to two decimal places.) 14.55 14.55

0.36 h

Solution or Explanation Let

a =

C' (t )

1.35 and

=0



b =

+ bt +

!2.802. Then C (t ) = atebt

1=0



= t =

!

1 b

" 0.36

h.

C' (t )

(0) C (0)

= 0,

= a(tebt ! b + C (!1/b)

=!

ebt ! 1). a b

! e 1 =

!

a be

" 0.177,

and

C (2)

= 2ae2b

"

0.010.. The 0.010

maximum average BAC during the first three hours is about 0.177 mg/mL and it occurs at approximately 0.36 h (21.4 min).




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Lab 4

12/8/15, 7� 17 PM


After an antibiotic tablet is taken, the concen tration of the antibiotic in the bloodstream is modeled by the function C (t )

= 3(e!0.4t !

! e 0.6t )

where the time t is is measured in hours and C is is measured in "g/mL. What is the maximum concentration of the a ntibiotic during the first 12 hours? (Round your answer to four decimal places.) .0085 .0085

0.4444 "g/mL

Solution or Explanation C (t )

= 3(e!0.4t !

! e 0.6t ),

[0, 12].

C' (t )

= 3(!0.4e!0.4t + 0.6e!0.6t ).

ln(3/2) ln(3/2) 4 (0) = 0, C = C (0) 0.2 0.2 9 the first 12 hours is 0.4444 "g/mL. = t =

"

0.4444,, and 0.4444

(12) C (12)

C' (n)

= 3(e!4.8 +

=0



! e 7.2) "

3(!0.4e!0.4t + 0.6e!0.6t ) = 0



3 = e0.2t 2



0.0224.. So the maximum concentration during 0.0224




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Lab 4

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AnswersSCalcET8 4.1.512.XP. 6. 10/10 points | Previous AnswersSCalcET8

Use a graph to estimate the critical numbers of f ( x x ) = | x x 3 ! 5 x 2 + 4| correct to one decimal place. -.8 -0.8 (smallest value) = -.8 x = 0 0 = x = 1 3.3 = 3.3 x = 4.8 = 4.8 x = = x =

1 3.3 4.8 (largest value)

Solution or Explanation

We see that

)=0 f ' f ' ( x x )=0

at about x = = 0.0 and x = = 3.3 3.3,, and that

f ' f ' ( x x )

does not exist at about

= x =

0.8,, 1.0, and 4.8 4.8,, so the critical !0.8

numbers of f are are about !0.8 0.8,, 0, 1, 3.3 3.3,, and 4.8 4.8..



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AnswersSCalcET8 4.2.510.XP. 7. 10/10 points | Previous AnswersSCalcET8

Does the function satisfy the hypotheses of the Mean Value Theorem on Theorem on the given interval?


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Lab 4

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f ( x x )

=

x

+ x +

2

,

[1, 4]

No, f is is continuous on [1, 4] but not differentiable on (1, 4). Yes, it does not matter if f is is continuous or differentiable; every function satisfies the Mean Value Theorem. Yes, f is is continuous on [1, 4] and differentiable on (1, 4). No, f is is not continuous on [1, 4]. There is not enough information to verify if this func tion satisfies the Mean Value Theorem.

If it satisfies the hypotheses, find all numbers c that that satisfy the conclusion of the Mean Value Theorem. (Enter your answers a s a comma-separated list. If it does not satisfy the hypotheses, enter DNE). = c = $$!2+" 18 18



What steps or reasoning did you use? Your work counts towards your score. ["f",{"tmpl":["\\left({","}\\right)"],"args":[["x"]],"inline":0,"arity":1},"=",{"args":[["x"], ["x","+","2"]],"name":"\\frac","arity":2,"up":0,"down":1}," \\\\& ","f","'",{"tmpl":["\\left({","}\\right)"],"args": [["x"]],"inline":0,"arity":1},"=",{"args":[[{"tmpl":["\\left({","}\\right)"],"args":[["x","+","2"]],"inline":0,"arity":1},{"tmpl": ["\\left({","}\\right)"],"args":[["1"]],"inline":0,"arity":1},"-","x",{"tmpl":["\\left({","}\\right)"],"args": [["1"]],"inline":0,"arity":1}],[{"tmpl":["{","}^{","}"],"arity":2,"args":[[{"tmpl":["\\left({","}\\right)"],"args": [["x","+","2"]],"inline":0,"arity":1}],["2"]],"name":" {`}^{`}","inline":0,"up":1}]],"name":"\\frac","arity":2,"up":0,"down":1},"=",{"args":[["2"],[{"tmpl":[" {","}^{","}"],"arity":2,"args":[[{"tmpl":["\\left({","}\\right)"],"args":[["x","+","2"]],"inline":0,"arity":1}],["2"]],"name":" {`}^{`}","inline":0,"up":1}]],"name":"\\frac","arity":2,"up":0,"down":1}," \\\\& "," \\\\& ","f",{"tmpl": ["\\left({","}\\right)"],"args":[["4"]],"inline":0,"arity":1},"=",{"args":[["2"], ["3"]],"name":"\\frac","arity":2,"up":0,"down":1},"\\;","\\;","\\;","f",{"tmpl":["\\left({","}\\right)"],"args": [["1"]],"inline":0,"arity":1},"=",{"args":[["1"],["3"]],"name":"\\frac","arity":2,"up":0,"down":1}," \\\\& "," \\\\& ","f",{"tmpl": ["\\left({","}\\right)"],"args":[["4"]],"inline":0,"arity":1},"-","f",{"tmpl":["\\left({","}\\right)"],"args": [["1"]],"inline":0,"arity":1},"=","f","'",{"tmpl":["\\left({","}\\right)"],"args":[["c"]],"inline":0,"arity":1},{"tmpl": ["\\left({","}\\right)"],"args":[["4","-","1"]],"inline":0,"arity":1}," \\\\& ",{"args":[["2"], ["3"]],"name":"\\frac","arity":2,"up":0,"down":1},"-",{"args":[["1"],["3"]],"name":"\\frac","arity":2,"up":0,"down":1},"=", {"args":[["2"],[{"tmpl":["{","}^{","}"],"arity":2,"args":[[{"tmpl":["\\left({","}\\right)"],"args": [["c","+","2"]],"inline":0,"arity":1}],["2"]],"name":"{`}^{`}","inline":0,"up":1}]],"name":"\\frac","arity":2,"up":0,"down":1}, {"tmpl":["\\left({","}\\right)"],"args":[["3"]],"inline":0,"arity":1}," \\\\& ",{"args":[["1"], ["3"]],"name":"\\frac","arity":2,"up":0,"down":1},"=",{"args":[["6"],[{"tmpl":["\\left({","}\\right)"],"args": [["c","+","2"]],"inline":0,"arity":1},{"arity":1,"args":[["2"]],"name":"^","up":0}]],"name":"\\frac","arity":2,"up":0,"down":1}," \\\\& ",{"tmpl":["\\left({","}\\right)"],"args":[["c","+","2"]],"inline":0,"arity":1},{"arity":1,"args": [["2"]],"name":"^","up":0},"=","1","8"," \\\\& ","c","+","2","=","\\pm",{"arity":1,"args":


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[["1","8"]],"name":"\\sqrt","inline":0}," \\\\& ","c","=","\\pm",{"arity":1,"args":[["1","8"]],"name":"\\sqrt","inline":0},"","2","\\;","\\;","\\;","\\;","\\;",{"tmpl":["\\left[{","}\\right]"],"args":[["1",",","\\;","4"]],"inline":0,"arity":1}," \\\\& ","c","=", {"arity":1,"args":[["1","8"]],"name":"\\sqrt","inline":0},"-","2"] [] Uploaded File (10 file maximum)


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Suppose that 2 #

f ' ( x x )

# 3 for all values of x . What are the minimum and maximum possible values of f (5) ! f (3)?

4 # f (5) ! f (3) # 6

4

6



What steps or reasoning did you use? Your work counts towards your score. ["2","\\leq","f","'",{"tmpl":["\\left({","}\\right)"],"args":[["x"]],"inline":0,"arity":1},"\\leq","3"," \\\\& ","2","\\leq",{"args": [["f",{"tmpl":["\\left({","}\\right)"],"args":[["5"]],"inline":0,"arity":1},"-","f",{"tmpl":["\\left({","}\\right)"],"args": [["3"]],"inline":0,"arity":1}],["5","-","3"]],"name":"\\frac","arity":2,"up":0,"down":1},"\\leq","3"," \\\\& ","2","\\leq",{"args": [["f",{"tmpl":["\\left({","}\\right)"],"args":[["5"]],"inline":0,"arity":1},"-","f",{"tmpl":["\\left({","}\\right)"],"args": [["3"]],"inline":0,"arity":1}],["2"]],"name":"\\frac","arity":2,"up":0,"down":1},"\\leq","3"," \\\\& ","4","\\leq","f",{"tmpl": ["\\left({","}\\right)"],"args":[["5"]],"inline":0,"arity":1},"-","f",{"tmpl":["\\left({","}\\right)"],"args": [["3"]],"inline":0,"arity":1},"\\leq","6"] [] Uploaded File (10 file maximum)


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9. 10/10 points | Previous AnswersSCalcET8 AnswersSCalcET8 4.2.506.XP.

(a) In the viewing rectangle [!3, 3] by [!5, 5], graph the function f ( x x ) = x 3 ! 2 x and its secant line through the points (!2, !4)


Page 8 of 11

Lab 4

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and (2, 4).

(b) Find the exact values of the numbers c that that satisfy the conclusion of the Mean Value Theorem for the interval [!2, 2]. (Enter your answers as a comma-separated list.) = c = $$4" 12, 12, !4" 12 12

Solution or Explanation


Page 9 of 11

Lab 4

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Click to View Solution


What steps or reasoning did you use? Your work counts towards your score. ["f",{"tmpl":["\\left({","}\\right)"],"args":[["x"]],"inline":0,"arity":1},"=","x",{"arity":1,"args":[["3"]],"name":"^","up":0},"","2","x"," \\\\& ","f","'",{"tmpl":["\\left({","}\\right)"],"args":[["x"]],"inline":0,"arity":1},"=","3","x",{"arit ","f","'",{"tmpl":["\\left({","}\\right)"],"args":[["x"]],"inline":0,"arity":1},"=","3","x",{"arity":1,"args": y":1,"args": [["2"]],"name":"^","up":0},"-","2"," \\\\& ","f",{"tmpl":["\\left({","}\\right)"],"args": [["2"]],"inline":0,"arity":1},"=","4","\\;","\\;","\\;","\\;","\\;","\\;","\\;","\\;","f",{"tmpl":["\\left({","}\\right)"],"args":[["","2"]],"inline":0,"arity":1},"=","-","4"," \\\\& "," \\\\& ","f",{"tmpl":["\\left({","}\\right)"],"args":[["2"]],"inline":0,"arity":1},"","f",{"tmpl":["\\left({","}\\right)"],"args":[["-","2"]],"inline":0,"arity":1},"=","f","'",{"tmpl":["\\left({","}\\right)"],"args": [["c"]],"inline":0,"arity":1},{"tmpl":["\\left({","}\\right)"],"args":[["2","-",{"tmpl":["\\left({","}\\right)"],"args":[["","2"]],"inline":0,"arity":1}]],"inline":0,"arity":1}," \\\\& ","4","+","4","=","3","c",{"arity":1,"args": [["2"]],"name":"^","up":0},"-","2",{"tmpl":["\\left({","}\\right)"],"args":[["2","+","2"]],"inline":0,"arity":1}," \\\\& ","8","=","1","2","c",{"arity":1,"args":[["2"]],"name":"^","up":0},"-","8"," \\\\& ","1","6","=","1","2","c",{"arity":1,"args": [["2"]],"name":"^","up":0}," \\\\& ",{"args":[["1","6"],["1","2"]],"name":"\\frac","arity":2,"up":0,"down":1},"=","c", {"arity":1,"args": [["2"]],"name":"^","up":0},"\\;","\\;","\\;","\\;","\\;","\\;","\\;","\\;","\\;","\\;","\\;","\\;","\\;","\\;","\\;","c","=","\\pm", {"args":[["4"],[{"arity":1,"args":[["1","2"]],"name":"\\sqrt","inline":0}]],"name":"\\frac","arity":2,"up":0,"down":1},"\\;","\\;", {"tmpl":["\\left[{","}\\right]"],"args":[["-","2",",","2"]],"inline":0,"arity":1}," \\\\& "] [] Uploaded File (10 file maximum)


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Page 10 of 11

Lab 4

12/8/15, 7� 17 PM

AnswersSCalcET8 4.2.509.XP. 10.10/10 points | Previous AnswersSCalcET8

Does the function satisfy the hypotheses of the Mean Value Theorem on Theorem on the given interval? f ( x x )

=

! e 2 x ,

[0, 1]

No, f is is not continuous on [0, 1]. Yes, it does not matter if f is is continuous or differentiable; every function satisfies the Mean Value Theorem. Yes, f is is continuous continuous and differentiable differentiable on

, so it is continuous on [0, 1] and differentiable on on (0, 1) .

There is not enough information to verify if this func tion satisfies the Mean Value Theorem. No, f is is continuous on [0, 1] but not differentiable on (0, 1).

If it satisfies the hypotheses, find all numbers c that that satisfy the conclusion of the Mean Value Theorem. (Enter your answers a s a comma-separated list. If it does not satisfy the hypotheses, enter DNE). = c = $$!ln(12!12e2)2




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Page 11 of 11

Lab 4

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