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Question 1234567891011121314151617181920212223242526272829303132333435363738394041

1. Question DetailsLarCalc9 3.1.004.MI. [1373972]


3. Question DetailsLarCalc9 3.1.012. [1196826]



Practice Exam 2 (1479895)

Find the value of the derivative (if it exists) at the indicated extremum. (If an answer does not exist, enter DNE.)

f(x) = 3x x+ 1

Tutorial

Solution or Explanation

Find any critical numbers of the function. (Enter your answers as a comma-separated list.)

f(x) =x8 8x7

x=

Tutorial






Find the absolute extrema of the function on the closed interval.

minimum (x, y) = ( 1, -2 )

maximum (x, y) = ( 4, 52 )


,23

2

3

3

f' = 023

f(x) = 3x

f'(x) = 3x (x+ 1)1/2 + (3)

= (x+ 1)1/2[x+ 2(x+ 1)]

= (x+ 1)1/2(3x+ 2)

f' = 0

x+ 1

12

x+ 1

3232

23

Critical numbers:x= 0, 7

f(x) = x8 8x7

f'(x) = 8x7 56x6= 8x6(x 7)

g(x) =x5 10x3

Critical numbers:x= 0,

g(x) = x5 10x3

g'(x) = 5x4 30x2= 5x2(x2 6)

6

(sin(x))2+ cos(x) 0

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Use a graphing utility to graph the function. Locate the absolute extrema of the function on the given interval. (If an answer does not exist, enter DNE.)

minimum (x, y) = ( 6, 9/2 )

maximum (x, y) = ( DNE )


Find the twox-intercepts of the function fand show that f'(x) = 0 at some point between the twox-intercepts.

(x, y) = ( ) (smallerx-value)

(x, y) = ( ) (largerx-value)

Find a value ofxsuch that f '(x) = 0.

x= 1.5


Find the twox-intercepts of the function fand show that f '(x) = 0 at some point between the twox-intercepts.

(x, y) = ( -7, 0 ) (smallerx-value)

(x, y) = ( 0, 0 ) (largerx-value)

Find a value ofxsuch that f '(x) = 0.

x= -14/3


The graph of fis shown. Apply Rolle's Theorem and find all values of csuch that f '(c) = 0 at some point between the labeled intercepts. (Enter your answers as a comma-separated l ist.)

c= -3

Left endpoint: (0, 0)Critical number: (1, -2) MinimumRight endpoint: (4, 52) MaximumNote:x= 1is not in the interval.

f(x) = x3 3x, [0, 4]

f'(x) = 3x2 3= 3(x2 1)

f(x) = , (4, 6]

Right endpoint: 6, Minimum

9x 4

92

f(x) =x2 3x 10

f(x) =x2 3x 10= (x 5)(x+ 2)x-intercepts: (2, 0), (5, 0)

f'(x) = 2x 3= 0 atx= .32

f(x) =x x+ 7

f(x) =x

x-intercepts: (7, 0), (0, 0)

x+ 7

f'(x) = x (x+ 7)1/2+ (x+ 7)1/2

= (x+ 7)1/2 + (x+ 7)

f'(x) = x+ 7 (x+ 7)1/2= 0 atx=

12

x2

32

143

f(x) =x2+ 6x 27

ignment Previewer http://www.webassign.net/v4cgijeff.downs@wnc/assignments/preview....

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The height of a ball tseconds after it is thrown upward from a height of 6feet and with an initial velocity of 80feet per second is f(t) = -16t2+ 80t+ 6.(a) Verify that f(2) = f(3).

f(2) = 102 ft

f(3) = 102 ft

(b) According to Rolle's Theorem, what must be the velocity at some time in the interval (2, 3)? 0 ft/sec

Find that time.

t= 5/2 s


(a) f(2) = f(3) = 102

(b) v= f '(t) must be 0 at some time in (2, 3).

Determine whether the Mean Value theorem can be applied to fon the closed interval [a, b]. (Select all that apply.)

If the Mean Value Theorem can be applied, find all values of cin the open interval (a, b) such that . (Enter your answers as a comma-separated list. If

the Mean Value Theorem cannot be applied, enter NA.)

c=


Determine whether the Mean Value theorem can be applied to fon the closed interval [a, b]. (Select all that apply.)

f(x) =x2+ 6x 27= (x+ 9)(x 3)f(9) = f(3) = 0f'(x) = 2x+ 6= 0 atx= 3c= 3and f'(3) = 0.

f(t) = 16t2+ 80t+ 6

f'(t) = 32t+ 80= 0

t= sec52

f(x) =x9, [0,1]

Yes, the Mean Value Theorem can be applied.

No, fis not continuous on [a, b].

No, fis not differentiable on (a, b).

None of the above.

f'(c) = f(b) f(a)b a

f(x) =x9is continuous on [0, 1] and differentiable on (0, 1).

= = 1

In the interval (0, 1): c=

f(1) f(0)1 0

1 01

f'(x) = 9x8 = 1

x= 9

98

9

98

f(x) = cosx+tanx, [0, ]


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If the Mean Value Theorem can be applied, find all values of cin the open interval (a, b) such that If the Mean Value Theorem cannot be applied,

explain why not. (Enter your answers as a comma-separated list. If the Mean Value Theorem cannot be appli ed, enter NA.) NA


Use the graph to estimate the open intervals on which the function is increasing or decreasing. Then find the open intervals analytically. (Select all that apply.)

Increasing:

Decreasing:


From the graph, yis increasing on ( , 4) and (4, ), and decreasing on (4, 4).

Analytically,

Critical numbers:x= 4

Test Interva ls :

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Decreasing:


Critical numbers:x= 6

Test Interva ls :

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Increasing:

Decreasing:

(c) Apply the First Derivative Test to identify the relative extremum. (If an answer does not exist, enter DNE.)

relative maximum

relative minimum

(d) Use a graphing utili ty to confirm your results.


(b) Test intervals:

Sign of f '(x): f '> 0 f '> 0 f '< 0 f '< 0

Conclusion: Increasi ng Increasing Decreasing Decreasing

The function s(t) describes the motion of a particle along a line.

(a) Find the velocity function of the particle at any time t 0.

v(t) =

(b) Identify the time interval in which the particle is moving in a positive direction.

(c) Identify the time interval in which the particle is moving in a negative direction.

(- ,

-2)

(-2,

0)

(0, 2)

(2, )

none ofthese

(- , -2)

(-2, 0)

(0, 2)

(2, )

none ofthese

(x, y) =

(x, y) =

Critical number:x= 0Discontinuities:x= 2, 2

(a) f(x) =

f'(x) = =

x2

x2 4

(x2 4)(2x) (x2)(2x)

(x2 4)28x

(x2 4)2

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(d) Identify the time at which the particle changes direction.

t= 8


Determine the open intervals on which the graph is concave upward or concave downward.

Concave upward:

Concave downward:



Concave upward:

Concave downward:

(0, )

(0, 8)

(0, 16)

(16, )

(8, )

None

(b) v(t) = 0 when t= 8.Moving in positive direction for 0 < t< 8because v(t) > 0 on 0 < t< 8.

(c) Moving in negative direction when t> 8.

(d) The particle changes direction at t= 8.

(a) s(t) = 16t t2, t 0v(t) = 16 2t

y= x33x2 3

none of these

( ,1)

(,1)

(1,

)

(1,

)

none of these

( ,1)

(,1)

(1,

)

(1,

)

Concave upward: ( , 1)Concave downward: (1, )

y= x33x2 3

y'= 3x2

6xy''= 6x6

f(x) = x39x2 5x 9

none of these

( ,3)

(,3)

(3,

)

(3,

)


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Concave upward:

Concave downward:



Concave upward:

Concave downward:

none of these

( ,3)

(,3)

(3,

)

(3,

)

Concave upward: ( , 3)Concave downward: (3, )

f(x) = x39x2 5x 9

f'(x) = 3x218x 5f''(x) = 6x18= 6(x+3)

f(x) =12

x2+ 12

( , 2)

(2, 2)

( , 2) (2, )

( , )

none of these

( , 2)

(2, 2)

( , 2) (2, )

( , )

none of these

Concave upward: ( , 2), (2, )Concave downward: (2, 2)

f(x) =

f'=

f''=

12

x2+ 1224x

(x2+ 12)2

72(4x2)

(x2+ 12)3

y= 6x tanx, ,

2

2

none of these

,2

2

,

0

2

0,

2


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Find the point of inflection of the graph of the function. (If an answer does not exist, enter DNE.)

(x, y) = ( 2, 18 )

Describe the concavity.

Concave upward:

Concave downward:


Find all relative extrema. Use the Second Derivative Test where applicable. (If an answer does not exist, enter DNE.)


f ''(0) = 12< 0

none of these

,2

2

,

0

2

0,

2

Concave upward: , 0

Concave downward: 0,

y= 6x tanx, ,

y'= 6 sec2x

y''= 2sec2xtanx

2

2

2

2

f(x) =x3 6x2+ 17x

none of these

(,2)

( ,2)

(2,

)

(2,

)

none of these

(,2)

( ,2)

(2,

)

(2,

)

Concave upward: (2, )Concave downward: ( , 2)Point of inflection: (2, 18)

f(x) = x3 6x2+ 17x

f'(x) = 3x2 12x+ 17f''(x) = 6(x 2) = 0 whenx= 2.

f(x) =x3 6x2+ 1

relative maximum (x, y) = ( )

relative minimum (x, y) = ( )

Critical numbers:x= 0,x= 4

f(x) = x3 6x2+ 1

f'(x) = 3x2 12x= 3x(x 4)f''(x) = 6x 12= 6(x 2)

Therefore, (0, 1) is a relative maximum.


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f ''(4) = 12> 0

Use a graphing utility to complete the table and estimate the limit asxapproaches infinity. Then use a graphing utility to graph the function and estimate your answergraphically. (Round your answers to five decimal places. If you need to use or , enter INFINITY or INFINITY, respectively.)

x 100 101 102 103

f(x) -6.50000 5.31250 4.10714 4.01052

x 104 105 106

f(x) 4.00105 4.00011 4.00001

4


x

f(x) -6.50000 5.31250 4.10714 4.01052 4.00105 4.00011 4.00001

Use a graphing utility to complete the table and estimate the limit asxapproaches infinity. Then use a graphing utility to graph the function and estimate your answer

graphically. (Round your answers to five decimal places. If you need to use or , enter INFINITY or INFINITY, respectively.)

x 100 101 102 103

f(x) 3.88057 3.20513 3.20005 3.20000

x 104 105 106

f(x) 3.20000 3.20000 3.20000

16/5


x

f(x) 3.88057 3.20513 3.20005 3.20000 3.20000 3.20000 3.20000

Therefore, (4, -31) is a relative minimum.

f(x) =limx

f(x) = 8x+ 52x 4

100 101 102 103 104 105 106

f(x)= 4limx

f(x) =limx

f(x) =16x

25x2 8

100 101 102 103 104 105 106

f(x)=limx

165


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y y' y'' Conclusion

+ Increasing, concave down

0 Relative maximum

Decreasing, concave down

0 Undefined Undefined Endpoint

Suppose f(0) = 9and 2 f '(x) 6for allxin the interval [-5, 5]. Determine the greatest and least possible values of f(5).

least 19

greatest 39

Solution or ExplanationIf f '(x) = 2in [5, 5], then f(x) = 2x+ 9and f(5) = 19is the least possible value of f(5).If f '(x) = 6in [5, 5], then f(x) = 6x+ 9and f(5) = 39is the greatest possible value of f(5).

Create a function whose graph has the given characteristics. (There is more than one correct answer.)Vertical asymptote:x= 3Slant asymptote: y= -x


Find two positive numbers that satisfy the given requirements.The sum of the first number squared and the second number is 51and the product is a maximum.

(first number)

y=x ,

Domain: ( , 8]

y'= = 0 whenx= and undefined whenx= 8.

y''= = 0 whenx= and undefined whenx= 8.

Note:x= is not in the domain.

8x

16 3x

2 8x163

3x 32

4(8x)3/2323

323

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(second number)


Find the point on the graph of the function that is closest to the given point.

(x, y) = ( 2, 4 )


Because dis smallest when the expression inside the radical is smallest, you need only find the critical numbers of

By the First Derivative Test, the point nearest to is (2, 4).

On a given day, the flow rate F(in cars per hour) on a congested roadway is given by the equat ion below, where vis the speed of traffic in miles per hour.

What speed will maximize the flow rate on the road? (Round your answer to one decimal place.)

49 mi/h Tutorial


A farmer plans to enclose a rectangular pasture adjacent to a river. (see figure). The pasture must contain 405,000square meters in order to provide enough grass for theherd. What dimensions will require the least amount of fencing if no fencing is needed along the river?

x= 900 m

y= 450 m


Letxand ybe two posi tive numbers such thatx2+ y= 51.

The product is a maximum whenx= and y= 34.

P= xy=x(54 x2) = 51xx3

= 51 3x2= 0 whenx= .

= 6x< 0 whenx= .

dPdx

17

d2P

dx217

17

Function Point

f(x) =x2 16,12

d=

=

(x 16)2+ [x2 (1/2)]2

x4 32x+ (1025/4)

f(x) = x4 32x+ .

f'(x) = 4x3 32= 0x= 2

10254

16, 12

F=v

24+ 0.01v2

By the First Derivative Test, the flow rate on the road is maximized when v 49.0mi/h.

F=

=

= 0 when v= 49.0.

v

24+ 0.01v2

dFdv

24 0.01v2

(24+ 0.01v2)2

2400.000

Sis a minimum whenx= 900m and y= 450m.

xy= 405,000(see figure)

S = x+ 2y

= x+ where Sis the length of fence needed.

= 1 = 0 whenx= 900.

= > 0 whenx= 900.

810,000x

dSdx

810,000

x2

d2S

dx2

1,620,000

x3


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Complete two iterations of Newton's Method for the function using the given initi al guess. (Round your answers to four decimal places.)

n

1 1.9000 0.6100 3.8000 0.1605 1.7395

2 1.7395 0.0258 3.4789 0.0074 1.7321


n

1

2

Complete two iterations of Newton's Method for the function using the given initi al guess. (Round your answers to four decimal places.)

n

1 0.06 0.0601 1.0036 0.0599 0.0001

2 0.0001 0.0001 1.0000 0.0001 0.0000


n

1 0.0600 0.0601 1.0036 0.0599 0.0001

2 0.0001 0.0001 1.0000 0.0001 0.0000

Find the equation of the tangent line Tto the graph of fat t he given point. Use this l inear approximation to complete the table. (Round your answers to four decimalplaces.)

T(x) =

x 3.9 3.99 4 4.01 4.1

f(x) 15.2100 15.9201 16.0000 16.0801 16.8100

T(x) 15.2000 15.9200 16.0000 16.0800 16.8000


x 3.9 3.99 4 4.01 4.1

15.2100 15.9201 16.0000 16.0801 16.8100

15.2000 15.9200 16.0000 16.0800 16.8000

Find the equation of the tangent line Tto the graph of fat t he given point. Use this l inear approximation to complete the table. (Round your answer to four decimalplaces.)

x 2.9 2.99 3 3.01 3.1

f(x) 1.7029 1.7292 1.7321 1.7349 1.7607

T(x) 1.7032 1.7292 1.7321 1.7349 1.7609

f(x) =x2 3, x1= 1.9;

xn f(xn) f'(xn) f(xn)

f'(xn) xnf(xn)f'(xn)

f(x) = x2 3f'(x) = 2x

x1 = 1.9


f'(xn) xn

f(xn)f'(xn)

1.9000 0.6100 3.8000 0.1605 1.7395

1.7395 0.0258 3.4789 0.0074 1.7321

f(x) = tanx, x1= 0.06


f'(xn) xn

f(xn)f'(xn)

f(x) = tanx

f'(x) = sec2xx1 = 0.06

xn f(xn) f'(xn) f(xn)f'(xn)

xn f(xn)f'(xn)

f(x) =x2, (4, 16)

f(x) = x2

f'(x) = 2xTangent line at (4, 16): y f(4) = f'(4)(x 4)

y 16= 8(x 4)y= 8x 16

f(x) =x2

T(x) = 8x 16

f(x) = , (3, )x 3


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Name (AID): Practice Exam 2 (1479895)

Submissions Allowed: 5

Category: Homework

Code:

Locked: No

Author: Downs, Jeff ([email protected])

Last Saved: Oct 13, 2010 05:26 PM PDT

Permission: Protected

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x 2.9 2.99 3 3.01 3.1

1.7029 1.7292 1.7321 1.7349 1.7607

1.7032 1.7292 1.7321 1.7349 1.7609

Use the information to evaluate and compare yand dy. (Round your answers to four decimal places.)

y= -0.3176

dy= -0.3200


The measurement of the edge of a cube is measured to be 11inches, with a possible error of 0.02inch. Use differentials to approximate the possible error and the relativeerror in computing the following values. (Round your answers to three decimal places.)

(a) the volume of the cube

possible error 7.260 in3

relative error 0.005

(b) the surface area of the cube

possible error 2.640 in2

relative error 0.004


Assignment Details

Tangent line at (3, ):

f(x) =

f'(x) =

x

1

2 x

3

y f(3) = f'(3)(x 3)

y = (x 3)

y= +

31

2 3

x

2 3 2

3

f(x) = x

T(x) = +x

2 3 2

3

y=x4+ 5 x= 2 x= dx= 0.01

y= f(x) =x4+ 5, f'(x) = 4x3,x= 2, x= dx= 0.01

y= f(x+ x) f(x) dy= f'(x) dx= f(1.99) f(2) = f'(2)(0.01)

= [(1.99)4+ 5] [(2)4+ 5] 0.3176 = (4 (2)3) (0.01) = 0.32

x= 11inchesx= dx= 0.02inch

relative error = 0.005

relative error = 0.004

(a) V= x3

dV= 3x2dx= 3(11)2(0.02)

= 7.260in3

dVV

7.260

113

(b) S = 6x2

dS= 12xdx= 12(11)(0.02)

= 2.640in2

dSS

2.640

6112


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