Question 1

Use l'Hopital's Rule to evaluate the limit.
-$\lim _ { x \rightarrow \infty } \frac { 8 + 2 x - 4 x ^ { 2 } } { 17 - 6 x - 17 x ^ { 2 } }$

Accepted Answer

A)  $\frac { 8 } { 17 }$ 
B)  1 
C)  $\infty$ 
D)  $\frac { 4 } { 17 }$ 
A)  $\frac { 8 } { 17 }$ 
B)  1 
C)  $\infty$ 
D)  $\frac { 4 } { 17 }$

Question 2

Use l'H^opital's rule to find the limit.
-$\lim _{ x \rightarrow \infty }  x \sin \frac { 16 } { x }$

Accepted Answer

A)  1 
B)  0 
C)  16 
D)  $\frac { 1 } { 16 }$ 
A)  1 
B)  0 
C)  16 
D)  $\frac { 1 } { 16 }$

Question 3

Solve the problem.
-Suppose that $c ( x ) = 3 x ^ { 3 } - 26 x ^ { 2 } + 7664 x$ is the cost of manufacturing $x$ items. Find a production level that will minimize the average cost of making $x$ items.

Accepted Answer

A)  7664 items 
B)  -79 items 
C)  97 items 
D)  There is not a production level that will minimize average cost. 
A)  7664 items 
B)  -79 items 
C)  97 items 
D)  There is not a production level that will minimize average cost.

Question 4

Solve the problem.
-Given the acceleration, initial velocity, and initial position of a body moving along a coordinate line at time t, find the body's position at time t.
A = 16 cos 4t, v(0) = 8, s(0) = 3

Accepted Answer

A)  s = - 1 cos 4t + 8t + 3 
B)  s = 1 sin 4t + 8t + 3 
C)  s = 1 cos 4t - 8t + 3 
D)  s = - 1 sin 4t + 8t + 3 
A)  s = - 1 cos 4t + 8t + 3 
B)  s = 1 sin 4t + 8t + 3 
C)  s = 1 cos 4t - 8t + 3 
D)  s = - 1 sin 4t + 8t + 3

Question 5

Find the absolute extreme values of the function on the interval.
-$$g ( x ) = - x ^ { 2 } + 7 x - 12,4 \leq x \leq 3$$

Accepted Answer

A)  absolute maximum is $\frac { 1 } { 4 }$ at $x = \frac { 9 } { 2 } ;$ absolute minimum is 0 at 3 and 0 at $x = 4$ 
B)  absolute maximum is $\frac { 97 } { 4 }$ at $x = \frac { 7 } { 2 }$; absolute minimum is 0 at 3 and 0 at $x = 4$ 
C)  absolute maximum is $\frac { 1 } { 4 }$ at $x = \frac { 7 } { 2 }$; absolute minimum is 0 at 3 and 0 at $x = 4$ 
D)  absolute maximum is $\frac { 5 } { 4 }$ at $x = \frac { 9 } { 2 }$; absolute minimum is 0 at 3 and 0 at $x = 4$ 
A)  absolute maximum is $\frac { 1 } { 4 }$ at $x = \frac { 9 } { 2 } ;$ absolute minimum is 0 at 3 and 0 at $x = 4$ 
B)  absolute maximum is $\frac { 97 } { 4 }$ at $x = \frac { 7 } { 2 }$; absolute minimum is 0 at 3 and 0 at $x = 4$ 
C)  absolute maximum is $\frac { 1 } { 4 }$ at $x = \frac { 7 } { 2 }$; absolute minimum is 0 at 3 and 0 at $x = 4$ 
D)  absolute maximum is $\frac { 5 } { 4 }$ at $x = \frac { 9 } { 2 }$; absolute minimum is 0 at 3 and 0 at $x = 4$

Question 6

Find an antiderivative of the given function.
-$$27 x ^ { 2 } - 6 x - 3$$

Accepted Answer

A)  $9 x ^ { 3 } - 3 x ^ { 2 } - 2 x$ 
B)  $9 x ^ { 3 } - 2 x ^ { 2 } - 3 x$ 
C)  $10 x ^ { 3 } - 3 x ^ { 2 } - 3 x$ 
D)  $9 x ^ { 3 } - 3 x ^ { 2 } - 3 x$ 
A)  $9 x ^ { 3 } - 3 x ^ { 2 } - 2 x$ 
B)  $9 x ^ { 3 } - 2 x ^ { 2 } - 3 x$ 
C)  $10 x ^ { 3 } - 3 x ^ { 2 } - 3 x$ 
D)  $9 x ^ { 3 } - 3 x ^ { 2 } - 3 x$

Question 7

Solve the problem.
-The strength $\mathrm { S }$ of a rectangular wooden beam is proportional to its width times the square of its depth. Find the dimensions of the strongest beam that can be cut from a 10 -in.-diameter cylindrical log. (Round answers to the nearest tenth.)

Accepted Answer

A)  w = 5.8 in.; d = 8.2 in. 
B)  w = 6.8 in.; d = 7.2 in. 
C)  w = 6.8 in.; d = 9.2 in. 
D)  w = 4.8 in.; d = 9.2 in. 
A)  w = 5.8 in.; d = 8.2 in. 
B)  w = 6.8 in.; d = 7.2 in. 
C)  w = 6.8 in.; d = 9.2 in. 
D)  w = 4.8 in.; d = 9.2 in.

Question 8

Use the maximum/minimum finder on a graphing calculator to determine the approximate location of all local extrema.
-$$f ( x ) = 0.01 x ^ { 5 } - x ^ { 4 } + x ^ { 3 } + 8 x ^ { 2 } - 7 x + 81$$

Accepted Answer

A)  Approximate local maxima at -1.926 and 2.312; approximate local minima at 0.335 and 79.257 
B)  Approximate local maxima at -1.778 and 2.322; approximate local minima at 0.384 and 79.267 
C)  Approximate local maxima at -1.861 and 2.247; approximate local minimum at 0.423 
D)  Approximate local maxima at -1.861 and 2.247; approximate local minima at 0.423 and 79.192 
A)  Approximate local maxima at -1.926 and 2.312; approximate local minima at 0.335 and 79.257 
B)  Approximate local maxima at -1.778 and 2.322; approximate local minima at 0.384 and 79.267 
C)  Approximate local maxima at -1.861 and 2.247; approximate local minimum at 0.423 
D)  Approximate local maxima at -1.861 and 2.247; approximate local minima at 0.423 and 79.192

Question 9

Use Newton's method to estimate the requested solution of the equation. Start with given value of x0 and then give x2 as
the estimated solution.
-$$x ^ { 4 } - 6 x + 3 = 0 ; x _ { 0 } = 2 ; \text { Find the right-hand solution. }$$

Accepted Answer

A)  1.600 
B)  1.604 
C)  1.607 
D)  1.602 
A)  1.600 
B)  1.604 
C)  1.607 
D)  1.602

Question 10

The graph below shows solution curves of a differential equation. Find an equation for the curve through the given point.
- 
$$\frac { d y } { d x } = 1 - \frac { 6 } { 5 } x ^ { 1 / 5 }$$

Accepted Answer

A)  $y = x + x ^ { 6 / 5 } + \frac { 3 } { 5 }$ 
B)  $y = 1 - x ^ { 6 / 5 } - \frac { 3 } { 5 }$ 
C)  $y = x - x ^ { 6 / 5 } + \frac { 5 } { 3 }$ 
D)  $y = x - x ^ { 6 / 5 } + \frac { 3 } { 5 }$ 
A)  $y = x + x ^ { 6 / 5 } + \frac { 3 } { 5 }$ 
B)  $y = 1 - x ^ { 6 / 5 } - \frac { 3 } { 5 }$ 
C)  $y = x - x ^ { 6 / 5 } + \frac { 5 } { 3 }$ 
D)  $y = x - x ^ { 6 / 5 } + \frac { 3 } { 5 }$

Question 11

Solve the initial value problem.
-$$\frac { \mathrm { d } ^ { 2 } \mathrm { r } } { \mathrm { dt } ^ { 2 } } = \frac { 4 } { \mathrm { t } ^ { 3 } } ; \left. \frac { \mathrm { dr } } { \mathrm { dt } } \right| _ { \mathrm { t } = 1 } = 2 , \quad \mathrm { r } ( 1 ) = 5$$

Accepted Answer

A)  $r = \frac { 2 } { t } + 4 t - 1$ 
B)  $r = 2 t + 4 t + 11$ 
C)  $r = \frac { 2 } { t } + 4 t + 11$ 
D)  $r = \frac { 4 } { - 5 t ^ { 5 } } + \frac { 14 } { 5 } t - 1$ 
A)  $r = \frac { 2 } { t } + 4 t - 1$ 
B)  $r = 2 t + 4 t + 11$ 
C)  $r = \frac { 2 } { t } + 4 t + 11$ 
D)  $r = \frac { 4 } { - 5 t ^ { 5 } } + \frac { 14 } { 5 } t - 1$

Question 12

Find the limit.
-$\lim _ { x \rightarrow \infty } \left( 1 + \frac { 2 } { x ^ { 4 } } \right) ^ { x }$

Accepted Answer

A)  $\infty$ 
B)  2 
C)  0 
D)  1 
A)  $\infty$ 
B)  2 
C)  0 
D)  1

Question 13

Find the most general antiderivative.
-$\int ( \sqrt { t } - \sqrt [ 6 ] { t } ) d t$

Accepted Answer

A)  $\frac { 2 } { 3 } t ^ { 3 / 2 } - \frac { 6 } { 7 } t ^ { 7 / 6 } + C$ 
B)  $\sqrt { t } - \sqrt [ 5 ] { t } + C$ 
C)  $\frac { 3 } { 2 } t ^ { 3 / 2 } - \frac { 7 } { 6 } t ^ { 7 / 6 } + C$ 
D)  $\frac { - 1 } { 2 } t ^ { 1 / 2 } - \frac { 1 } { 6 } t ^ { - 5 / 6 } + C$ 
A)  $\frac { 2 } { 3 } t ^ { 3 / 2 } - \frac { 6 } { 7 } t ^ { 7 / 6 } + C$ 
B)  $\sqrt { t } - \sqrt [ 5 ] { t } + C$ 
C)  $\frac { 3 } { 2 } t ^ { 3 / 2 } - \frac { 7 } { 6 } t ^ { 7 / 6 } + C$ 
D)  $\frac { - 1 } { 2 } t ^ { 1 / 2 } - \frac { 1 } { 6 } t ^ { - 5 / 6 } + C$

Question 14

Find the absolute extreme values of the function on the interval.
-$$f ( x ) = \tan x , - \frac { \pi } { 6 } \leq x \leq \frac { \pi } { 6 }$$

Accepted Answer

A)  absolute maximum is $\frac { \sqrt { 3 } } { 3 }$ at $x = \frac { \pi } { 6 }$ and $- \frac { \pi } { 6 }$; absolute minimum does not exist 
B)  absolute maximum is $- \frac { \sqrt { 3 } } { 3 }$ at $x = \frac { \pi } { 6 }$; absolute minimum is $\frac { \sqrt { 3 } } { 3 }$ at $x = - \frac { \pi } { 6 }$ 
C)  absolute maximum is $\frac { \sqrt { 3 } } { 3 }$ at $x = \frac { 2 \pi } { 18 }$; absolute minimum is $- \frac { \sqrt { 3 } } { 3 }$ at $x = - \frac { \pi } { 12 }$ 
D)  absolute maximum is $\frac { \sqrt { 3 } } { 3 }$ at $x = \frac { \pi } { 6 }$; absolute minimum is $- \frac { \sqrt { 3 } } { 3 }$ at $x = - \frac { \pi } { 6 }$ 
A)  absolute maximum is $\frac { \sqrt { 3 } } { 3 }$ at $x = \frac { \pi } { 6 }$ and $- \frac { \pi } { 6 }$; absolute minimum does not exist 
B)  absolute maximum is $- \frac { \sqrt { 3 } } { 3 }$ at $x = \frac { \pi } { 6 }$; absolute minimum is $\frac { \sqrt { 3 } } { 3 }$ at $x = - \frac { \pi } { 6 }$ 
C)  absolute maximum is $\frac { \sqrt { 3 } } { 3 }$ at $x = \frac { 2 \pi } { 18 }$; absolute minimum is $- \frac { \sqrt { 3 } } { 3 }$ at $x = - \frac { \pi } { 12 }$ 
D)  absolute maximum is $\frac { \sqrt { 3 } } { 3 }$ at $x = \frac { \pi } { 6 }$; absolute minimum is $- \frac { \sqrt { 3 } } { 3 }$ at $x = - \frac { \pi } { 6 }$

Question 15

Find the function with the given derivative whose graph passes through the point P.
-$g ^ { \prime } ( x ) = \frac { 4 } { x ^ { 2 } } + 6 x , P ( - 4,5 )$

Accepted Answer

A)  $g ( x ) = 4 x ^ { - 2 } + 6 x - 44$ 
B)  $g ( x ) = - 4 x ^ { - 1 } + 3 x ^ { 2 } - 44$ 
C)  $g ( x ) = 4 x ^ { 2 } + 3 x ^ { 2 } - 44$ 
D)  $g ( x ) = - 4 x ^ { - 1 } + 3 x ^ { 2 }$ 
A)  $g ( x ) = 4 x ^ { - 2 } + 6 x - 44$ 
B)  $g ( x ) = - 4 x ^ { - 1 } + 3 x ^ { 2 } - 44$ 
C)  $g ( x ) = 4 x ^ { 2 } + 3 x ^ { 2 } - 44$ 
D)  $g ( x ) = - 4 x ^ { - 1 } + 3 x ^ { 2 }$

Question 16

Find the absolute extreme values of the function on the interval.
-$$f ( x ) = | x - 5 | , 4 \leq x \leq 9$$

Accepted Answer

A)  absolute maximum is 4 at x = 9; absolute minimum is 0 at x = 5 
B)  absolute maximum is 1 at x = 4; absolute minimum is 0 at x = 5 
C)  absolute maximum is -1 at x = 4; absolute minimum is 4 at x = 9 
D)  absolute maximum is 4 at x = 9; absolute minimum is 1 at x = 4 
A)  absolute maximum is 4 at x = 9; absolute minimum is 0 at x = 5 
B)  absolute maximum is 1 at x = 4; absolute minimum is 0 at x = 5 
C)  absolute maximum is -1 at x = 4; absolute minimum is 4 at x = 9 
D)  absolute maximum is 4 at x = 9; absolute minimum is 1 at x = 4

Question 17

Find the function with the given derivative whose graph passes through the point P.
-$\mathrm { r } ^ { \prime } ( \theta ) = 5 - \csc ^ { 2 } \theta , \mathrm { P } \left( \frac { \pi } { 4 } , 0 \right)$

Accepted Answer

A)  $r ( \theta ) = 5 \theta + \cot \theta - \frac { 5 } { 4 } \pi - 1$ 
B)  $r ( \theta ) = 5 \theta - \cot \theta - \frac { 5 } { 4 } \pi + 2$ 
C)  $r ( \theta ) = 5 \theta - \cot \theta - \frac { 5 } { 4 } \pi - 2$ 
D)  $r ( \theta ) = 5 \theta + \cot \theta - \frac { 5 } { 4 } \pi + 1$ 
A)  $r ( \theta ) = 5 \theta + \cot \theta - \frac { 5 } { 4 } \pi - 1$ 
B)  $r ( \theta ) = 5 \theta - \cot \theta - \frac { 5 } { 4 } \pi + 2$ 
C)  $r ( \theta ) = 5 \theta - \cot \theta - \frac { 5 } { 4 } \pi - 2$ 
D)  $r ( \theta ) = 5 \theta + \cot \theta - \frac { 5 } { 4 } \pi + 1$

Question 18

Find the most general antiderivative.
-$$\int \left( \frac { 1 } { x ^ { 3 } } - x ^ { 3 } - \frac { 1 } { 2 } \right) d x$$

Accepted Answer

A)  $\frac { 1 } { 3 x ^ { 4 } } - \frac { x ^ { 4 } } { 4 } - \frac { 1 } { 2 x } + C$ 
B)  $\frac { - 1 } { 2 x ^ { 2 } } - \frac { x ^ { 4 } } { 4 } - \frac { x } { 2 } + C$ 
C)  $\frac { 1 } { 4 x ^ { 4 } } - \frac { x ^ { 2 } } { 2 } + \frac { 1 } { 4 } + C$ 
D)  $- 3 x ^ { 2 } - 3 x ^ { 3 } + C$ 
A)  $\frac { 1 } { 3 x ^ { 4 } } - \frac { x ^ { 4 } } { 4 } - \frac { 1 } { 2 x } + C$ 
B)  $\frac { - 1 } { 2 x ^ { 2 } } - \frac { x ^ { 4 } } { 4 } - \frac { x } { 2 } + C$ 
C)  $\frac { 1 } { 4 x ^ { 4 } } - \frac { x ^ { 2 } } { 2 } + \frac { 1 } { 4 } + C$ 
D)  $- 3 x ^ { 2 } - 3 x ^ { 3 } + C$

Question 19

Solve the problem.
-Suppose $c ( x ) = x ^ { 3 } - 22 x ^ { 2 } + 20,000 x$ is the cost of manufacturing $x$ items. Find a production level that will minimize the average cost of making $x$ items.

Accepted Answer

A)  13 items 
B)  11 items 
C)  10 items 
D)  12 items 
A)  13 items 
B)  11 items 
C)  10 items 
D)  12 items

Question 20

Use the maximum/minimum finder on a graphing calculator to determine the approximate location of all local extrema.
-$$f ( x ) = x ^ { 4 } - 4 x ^ { 3 } - 53 x ^ { 2 } - 86 x - 49$$

Accepted Answer

A)  Approximate local maximum at -0.944; approximate local minima at -3.192 and 7.136 
B)  Approximate local maximum at 0.871; approximate local minima at -3.229 and 7.085 
C)  Approximate local maximum at 0.943; approximate local minima at -3.113 and 7.216 
D)  Approximate local maximum at 1.001; approximate local minima at -3.14 and 7.159 
A)  Approximate local maximum at -0.944; approximate local minima at -3.192 and 7.136 
B)  Approximate local maximum at 0.871; approximate local minima at -3.229 and 7.085 
C)  Approximate local maximum at 0.943; approximate local minima at -3.113 and 7.216 
D)  Approximate local maximum at 1.001; approximate local minima at -3.14 and 7.159

Use l'Hopital's Rule to evaluate the limit. - $\lim _ { x \rightarrow \infty } \frac { 8 + 2 x - 4 x ^ { 2 } } { 17 - 6 x - 17 x ^ { 2 } }$

Use l'H^opital's rule to find the limit. - $\lim _{ x \rightarrow \infty } x \sin \frac { 16 } { x }$

Solve the problem. -Suppose that $c ( x ) = 3 x ^ { 3 } - 26 x ^ { 2 } + 7664 x$ is the cost of manufacturing $x$ items. Find a production level that will minimize the average cost of making $x$ items.

Solve the problem. -Given the acceleration, initial velocity, and initial position of a body moving along a coordinate line at time t, find the body's position at time t. A = 16 cos 4t, v(0) = 8, s(0) = 3

Find the absolute extreme values of the function on the interval. - $g ( x ) = - x ^ { 2 } + 7 x - 12,4 \leq x \leq 3$

Find an antiderivative of the given function. - $27 x ^ { 2 } - 6 x - 3$

Use the maximum/minimum finder on a graphing calculator to determine the approximate location of all local extrema. - $f ( x ) = 0.01 x ^ { 5 } - x ^ { 4 } + x ^ { 3 } + 8 x ^ { 2 } - 7 x + 81$

Use Newton's method to estimate the requested solution of the equation. Start with given value of x0 and then give x2 as the estimated solution. - $x ^ { 4 } - 6 x + 3 = 0 ; x _ { 0 } = 2 ; \text { Find the right-hand solution. }$

Solve the initial value problem. - $\frac { \mathrm { d } ^ { 2 } \mathrm { r } } { \mathrm { dt } ^ { 2 } } = \frac { 4 } { \mathrm { t } ^ { 3 } } ; \left. \frac { \mathrm { dr } } { \mathrm { dt } } \right| _ { \mathrm { t } = 1 } = 2 , \quad \mathrm { r } ( 1 ) = 5$

Find the limit. - $\lim _ { x \rightarrow \infty } \left( 1 + \frac { 2 } { x ^ { 4 } } \right) ^ { x }$

Find the most general antiderivative. - $\int ( \sqrt { t } - \sqrt [ 6 ] { t } ) d t$

Find the absolute extreme values of the function on the interval. - $f ( x ) = \tan x , - \frac { \pi } { 6 } \leq x \leq \frac { \pi } { 6 }$

Find the function with the given derivative whose graph passes through the point P. - $g ^ { \prime } ( x ) = \frac { 4 } { x ^ { 2 } } + 6 x , P ( - 4,5 )$

Find the absolute extreme values of the function on the interval. - $f ( x ) = | x - 5 | , 4 \leq x \leq 9$

Find the function with the given derivative whose graph passes through the point P. - $\mathrm { r } ^ { \prime } ( \theta ) = 5 - \csc ^ { 2 } \theta , \mathrm { P } \left( \frac { \pi } { 4 } , 0 \right)$

Find the most general antiderivative. - $\int \left( \frac { 1 } { x ^ { 3 } } - x ^ { 3 } - \frac { 1 } { 2 } \right) d x$

Solve the problem. -Suppose $c ( x ) = x ^ { 3 } - 22 x ^ { 2 } + 20,000 x$ is the cost of manufacturing $x$ items. Find a production level that will minimize the average cost of making $x$ items.

Use the maximum/minimum finder on a graphing calculator to determine the approximate location of all local extrema. - $f ( x ) = x ^ { 4 } - 4 x ^ { 3 } - 53 x ^ { 2 } - 86 x - 49$

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Exam 5: Applications of Derivatives

Use l'Hopital's Rule to evaluate the limit. - lim⁡x→∞8+2x−4x217−6x−17x2\lim _ { x \rightarrow \infty } \frac { 8 + 2 x - 4 x ^ { 2 } } { 17 - 6 x - 17 x ^ { 2 } }limx→∞​17−6x−17x28+2x−4x2​

Use l'H^opital's rule to find the limit. - lim⁡x→∞xsin⁡16x\lim _{ x \rightarrow \infty } x \sin \frac { 16 } { x }limx→∞​xsinx16​

Solve the problem. -Suppose that c(x)=3x3−26x2+7664xc ( x ) = 3 x ^ { 3 } - 26 x ^ { 2 } + 7664 xc(x)=3x3−26x2+7664x is the cost of manufacturing xxx items. Find a production level that will minimize the average cost of making xxx items.

Solve the problem. -Given the acceleration, initial velocity, and initial position of a body moving along a coordinate line at time t, find the body's position at time t. A = 16 cos 4t, v(0) = 8, s(0) = 3

Find the absolute extreme values of the function on the interval. - g(x)=−x2+7x−12,4≤x≤3g ( x ) = - x ^ { 2 } + 7 x - 12,4 \leq x \leq 3g(x)=−x2+7x−12,4≤x≤3

Find an antiderivative of the given function. - 27x2−6x−327 x ^ { 2 } - 6 x - 327x2−6x−3

Solve the problem. -The strength S\mathrm { S }S of a rectangular wooden beam is proportional to its width times the square of its depth. Find the dimensions of the strongest beam that can be cut from a 10 -in.-diameter cylindrical log. (Round answers to the nearest tenth.)

Use the maximum/minimum finder on a graphing calculator to determine the approximate location of all local extrema. - f(x)=0.01x5−x4+x3+8x2−7x+81f ( x ) = 0.01 x ^ { 5 } - x ^ { 4 } + x ^ { 3 } + 8 x ^ { 2 } - 7 x + 81f(x)=0.01x5−x4+x3+8x2−7x+81

The graph below shows solution curves of a differential equation. Find an equation for the curve through the given point. - dydx=1−65x1/5\frac { d y } { d x } = 1 - \frac { 6 } { 5 } x ^ { 1 / 5 }dxdy​=1−56​x1/5

Find the limit. - lim⁡x→∞(1+2x4)x\lim _ { x \rightarrow \infty } \left( 1 + \frac { 2 } { x ^ { 4 } } \right) ^ { x }limx→∞​(1+x42​)x

Find the most general antiderivative. - ∫(t−t6)dt\int ( \sqrt { t } - \sqrt [ 6 ] { t } ) d t∫(t​−6t​)dt

Find the absolute extreme values of the function on the interval. - f(x)=tan⁡x,−π6≤x≤π6f ( x ) = \tan x , - \frac { \pi } { 6 } \leq x \leq \frac { \pi } { 6 }f(x)=tanx,−6π​≤x≤6π​

Find the function with the given derivative whose graph passes through the point P. - g′(x)=4x2+6x,P(−4,5)g ^ { \prime } ( x ) = \frac { 4 } { x ^ { 2 } } + 6 x , P ( - 4,5 )g′(x)=x24​+6x,P(−4,5)

Find the absolute extreme values of the function on the interval. - f(x)=∣x−5∣,4≤x≤9f ( x ) = | x - 5 | , 4 \leq x \leq 9f(x)=∣x−5∣,4≤x≤9

Find the function with the given derivative whose graph passes through the point P. - r′(θ)=5−csc⁡2θ,P(π4,0)\mathrm { r } ^ { \prime } ( \theta ) = 5 - \csc ^ { 2 } \theta , \mathrm { P } \left( \frac { \pi } { 4 } , 0 \right)r′(θ)=5−csc2θ,P(4π​,0)

Find the most general antiderivative. - ∫(1x3−x3−12)dx\int \left( \frac { 1 } { x ^ { 3 } } - x ^ { 3 } - \frac { 1 } { 2 } \right) d x∫(x31​−x3−21​)dx

Solve the problem. -Suppose c(x)=x3−22x2+20,000xc ( x ) = x ^ { 3 } - 22 x ^ { 2 } + 20,000 xc(x)=x3−22x2+20,000x is the cost of manufacturing xxx items. Find a production level that will minimize the average cost of making xxx items.

Use the maximum/minimum finder on a graphing calculator to determine the approximate location of all local extrema. - f(x)=x4−4x3−53x2−86x−49f ( x ) = x ^ { 4 } - 4 x ^ { 3 } - 53 x ^ { 2 } - 86 x - 49f(x)=x4−4x3−53x2−86x−49