Question 1

Using the derivative of f(x) given below, determine the critical points of f(x).
-$f ^ { \prime } ( x ) = ( x - 1 ) ^ { 2 } ( x + 4 )$

Accepted Answer

A)  -1, 0, 4 
B)  -1, 4 
C)  -4, -1, 1 
D)  -4, 1 
A)  -1, 0, 4 
B)  -1, 4 
C)  -4, -1, 1 
D)  -4, 1

Question 2

Find all possible functions with the given derivative.
-$$y ^ { \prime } = 8 x ^ { 2 } - 3 x$$

Accepted Answer

A)  $\frac { 8 } { 3 } x ^ { 3 } - \frac { 3 } { 2 } x ^ { 2 } + C$ 
B)  $\frac { 8 } { 3 } x ^ { 3 } + C$ 
C)  $- \frac { 8 } { 3 } x ^ { 3 } - \frac { 3 } { 2 } x ^ { 2 } + C$ 
D)  $\frac { 8 } { 3 } x ^ { 2 } + \frac { 3 } { 2 } x + C$ 
A)  $\frac { 8 } { 3 } x ^ { 3 } - \frac { 3 } { 2 } x ^ { 2 } + C$ 
B)  $\frac { 8 } { 3 } x ^ { 3 } + C$ 
C)  $- \frac { 8 } { 3 } x ^ { 3 } - \frac { 3 } { 2 } x ^ { 2 } + C$ 
D)  $\frac { 8 } { 3 } x ^ { 2 } + \frac { 3 } { 2 } x + C$

Question 3

Find the derivative at each critical point and determine the local extreme values.
-$$y = x ^ { 2 / 3 } \left( x ^ { 2 } - 16 \right) ; x \geq 0$$

Accepted Answer

A) 
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=0 & \text { Undefined } & \text { local max } & 4 \
x=2 & 0 & \text { minimum } & -19.049
\end{array}$ 
B) 
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=0 & \text { Undefined } & \text { local max } & 0 \
x=2 & 0 & \text { minimum } & 31.748
\end{array}$ 
C) 
$\begin{array}{l|l|l|c}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=0 & \text { Undefined } & \text { local max } & 0 \
x=2 & 0 & \text { minimum } & -19.049
\end{array}$ 
D) 
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=0 & 0 & \text { maximum } & 0 \
x=2 & 0 & \text { minimum } & -19.049
\end{array}$ 
A) 
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=0 & \text { Undefined } & \text { local max } & 4 \
x=2 & 0 & \text { minimum } & -19.049
\end{array}$ 
B) 
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=0 & \text { Undefined } & \text { local max } & 0 \
x=2 & 0 & \text { minimum } & 31.748
\end{array}$ 
C) 
$\begin{array}{l|l|l|c}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=0 & \text { Undefined } & \text { local max } & 0 \
x=2 & 0 & \text { minimum } & -19.049
\end{array}$ 
D) 
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=0 & 0 & \text { maximum } & 0 \
x=2 & 0 & \text { minimum } & -19.049
\end{array}$

Question 4

Determine whether the function satisfies the hypotheses of the Mean Value Theorem for the given interval.
-$g ( x ) = x ^ { 3 / 4 }$ , $[0,5]$

Accepted Answer

A) True 
 B)False

Question 5

Find all possible functions with the given derivative.
-$$r ^ { \prime } = 3 + \frac { 1 } { \theta ^ { 4 } }$$

Accepted Answer

A)  $r = 3 \theta + \theta ^ { 4 }$ 
B)  $r = 3 \theta + \frac { 1 } { 5 \theta ^ { 5 } }$ 
C)  $r = 3 \theta - \theta ^ { 5 }$ 
D)  $r = 3 \theta - \frac { 1 } { 3 \theta ^ { 3 } }$ 
A)  $r = 3 \theta + \theta ^ { 4 }$ 
B)  $r = 3 \theta + \frac { 1 } { 5 \theta ^ { 5 } }$ 
C)  $r = 3 \theta - \theta ^ { 5 }$ 
D)  $r = 3 \theta - \frac { 1 } { 3 \theta ^ { 3 } }$

Question 6

Using the derivative of f(x) given below, determine the critical points of f(x).
-f'(x) = (x + 2)(x + 9)

Accepted Answer

A)  -11 
B)  -9, -2 
C)  0, -9, -2 
D)  2, 9 
A)  -11 
B)  -9, -2 
C)  0, -9, -2 
D)  2, 9

Question 7

Find the extreme values of the function and where they occur.
-$$y = \frac { 7 } { \sqrt { 1 - 9 x ^ { 2 } } }$$

Accepted Answer

A)  The maximum is 7 at x = 2. 
B)  The minimum is 7 at x = 0. 
C)  The minimum is 0 at x = 1. 
D)  The maximum is 7 at x = -2. 
A)  The maximum is 7 at x = 2. 
B)  The minimum is 7 at x = 0. 
C)  The minimum is 0 at x = 1. 
D)  The maximum is 7 at x = -2.

Question 8

Find the derivative at each critical point and determine the local extreme values.
-$$y = \left\{ \begin{array} { l } 
6 - 2 x , x \leq 1 \
x + 3 , x > 1
\end{array} \right.$$

Accepted Answer

A) 
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=1 & \text { undefined } & \text { minimum } & 4
\end{array}$ 
B) 
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=2 & \text { undefined } & \text { minimum } & 5
\end{array}$ 
C) 
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=1 & 0 & \text { minimum } & 4
\end{array}$ 
D) 
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=0 & 0 & \text { minimum } & 5
\end{array}$ 
A) 
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=1 & \text { undefined } & \text { minimum } & 4
\end{array}$ 
B) 
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=2 & \text { undefined } & \text { minimum } & 5
\end{array}$ 
C) 
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=1 & 0 & \text { minimum } & 4
\end{array}$ 
D) 
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=0 & 0 & \text { minimum } & 5
\end{array}$

Question 9

Solve the problem.
-A particle moves on a coordinate line with acceleration $a = d ^ { 2 } s / d t ^ { 2 } = ( 5 / \sqrt { t } ) + 9 \sqrt { t }$, subject to the conditions that $\mathrm { ds } / \mathrm { dt } = 3$ and $\mathrm { s } = 1$ when $\mathrm { t } = 1$. Find the velocity $\mathrm { v } = \mathrm { ds } / \mathrm { dt }$ in terms of $\mathrm { t }$ and the position $s$ in terms of $t$.

Accepted Answer

A)  $v = 10 \sqrt [ 3 ] { t } + 6 \sqrt [ 3 ] { t } - 13 ; s = \frac { 20 } { 3 } \sqrt [ 5 ] { t } + \frac { 12 } { 5 } \sqrt [ 3 ] { t } - 13 t + \frac { 74 } { 15 }$ 
B)  $v = 10 \sqrt { t } - 6 \sqrt [ 3 ] { t } + 13 ; s = \frac { 20 } { 3 } \sqrt [ 3 ] { t } - \frac { 12 } { 5 } \sqrt [ 5 ] { t } + 13 t + \frac { 74 } { 15 }$ 
C)  $v = 10 \sqrt { t } + 6 \sqrt [ 3 ] { t } - 13 ; s = \frac { 20 } { 3 } \sqrt [ 3 ] { t } + \frac { 12 } { 5 } \sqrt [ 5 ] { t } - 13 t + \frac { 74 } { 15 }$ 
D)  $v = \frac { 20 } { 3 } \sqrt [ 3 ] { t } + \frac { 12 } { 5 } \sqrt [ 5 ] { t } - 13 t + \frac { 74 } { 15 } ; s = 10 \sqrt { t } + 6 \sqrt [ 3 ] { t } - 13$ 
A)  $v = 10 \sqrt [ 3 ] { t } + 6 \sqrt [ 3 ] { t } - 13 ; s = \frac { 20 } { 3 } \sqrt [ 5 ] { t } + \frac { 12 } { 5 } \sqrt [ 3 ] { t } - 13 t + \frac { 74 } { 15 }$ 
B)  $v = 10 \sqrt { t } - 6 \sqrt [ 3 ] { t } + 13 ; s = \frac { 20 } { 3 } \sqrt [ 3 ] { t } - \frac { 12 } { 5 } \sqrt [ 5 ] { t } + 13 t + \frac { 74 } { 15 }$ 
C)  $v = 10 \sqrt { t } + 6 \sqrt [ 3 ] { t } - 13 ; s = \frac { 20 } { 3 } \sqrt [ 3 ] { t } + \frac { 12 } { 5 } \sqrt [ 5 ] { t } - 13 t + \frac { 74 } { 15 }$ 
D)  $v = \frac { 20 } { 3 } \sqrt [ 3 ] { t } + \frac { 12 } { 5 } \sqrt [ 5 ] { t } - 13 t + \frac { 74 } { 15 } ; s = 10 \sqrt { t } + 6 \sqrt [ 3 ] { t } - 13$

Question 10

Find the derivative at each critical point and determine the local extreme values.
-$y = x \left( 4 - x ^ { 2 } \right)$

Accepted Answer

A) 
$\begin{array}{l|l|l|r}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=-1.15 & 0 & \text { local max } & 3.08 \
x=1.15 & 0 & \text { local min } & -6.16
\end{array}$ 
B) 
$\begin{array}{l|l|l|r}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=1.15 & 0 & \text { local max } & 3.08 \
x=-1.15 & 0 & \text { local min } & -3.08
\end{array}$ 
C) 
$\begin{array}{c|l|l|c}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=-1.15 & 0 & \text { local max } & -6.16 \
x=1.15 & 0 & \text { local min } & 3.08
\end{array}$ 
D) 
$\begin{array}{l|l|l|r}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline \mathrm{x}=1.15 & 0 & \text { local max } & -6.16 \
\mathrm{x}=-1.15 & 0 & \text { local min } & 3.08
\end{array}$ 
A) 
$\begin{array}{l|l|l|r}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=-1.15 & 0 & \text { local max } & 3.08 \
x=1.15 & 0 & \text { local min } & -6.16
\end{array}$ 
B) 
$\begin{array}{l|l|l|r}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=1.15 & 0 & \text { local max } & 3.08 \
x=-1.15 & 0 & \text { local min } & -3.08
\end{array}$ 
C) 
$\begin{array}{c|l|l|c}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=-1.15 & 0 & \text { local max } & -6.16 \
x=1.15 & 0 & \text { local min } & 3.08
\end{array}$ 
D) 
$\begin{array}{l|l|l|r}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline \mathrm{x}=1.15 & 0 & \text { local max } & -6.16 \
\mathrm{x}=-1.15 & 0 & \text { local min } & 3.08
\end{array}$

Question 11

Graph the function, then find the extreme values of the function on the interval and indicate where they occur.
-y = $\mid{x - 1}\mid$ -$\mid{ x - 6}\mid$ on the interval -2 < x < 7

Accepted Answer

A) 
 
Absolute maximum is: 5 on the interval $[ 6,7 )$; absolute minimum is: $- 4$ on the interval $( - 2,2 ]$ 
B) 
 
Absolute maximum is: $- 5$ on the interval $[ 6,7 )$; absolute minimum is: 5 on the interval $( - 2,1 ]$ 
C) 
 
Absolute maximum is: 5 on the interval $[ 6,7 )$; absolute minimum is: $- 5$ on the interval $( - 2,1 ]$
 
D) 
 
Absolute maximum is: 6 on the interval $[ 6,7 )$; absolute minimum is: $- 4$ on the interval $( - 2,1 ]$ 
A) 
 
Absolute maximum is: 5 on the interval $[ 6,7 )$; absolute minimum is: $- 4$ on the interval $( - 2,2 ]$ 
B) 
 
Absolute maximum is: $- 5$ on the interval $[ 6,7 )$; absolute minimum is: 5 on the interval $( - 2,1 ]$ 
C) 
 
Absolute maximum is: 5 on the interval $[ 6,7 )$; absolute minimum is: $- 5$ on the interval $( - 2,1 ]$
 
D) 
 
Absolute maximum is: 6 on the interval $[ 6,7 )$; absolute minimum is: $- 4$ on the interval $( - 2,1 ]$

Question 12

Find the location of the indicated absolute extremum for the function.
-Minimum

Accepted Answer

A)  x = 1 
B)  x = -1 
C)  x = 2 
D)  No minimum 
A)  x = 1 
B)  x = -1 
C)  x = 2 
D)  No minimum

Question 13

Determine all critical points for the function.
-$$f ( x ) = ( x - 9 ) ^ { 3 }$$

Accepted Answer

A)  $x = 9$ 
B)  $x = 0 , x = 9$, and $x = 3$ 
C)  $x = 9$ and $x = 3$ 
D)  $x = 0$ and $x = 9$ 
A)  $x = 9$ 
B)  $x = 0 , x = 9$, and $x = 3$ 
C)  $x = 9$ and $x = 3$ 
D)  $x = 0$ and $x = 9$

Question 14

Find the function with the given derivative whose graph passes through the point P.
-$$g ^ { \prime } ( x ) = \frac { 1 } { x ^ { 2 } } + 2 x , P ( - 4,4 )$$

Accepted Answer

A)  $g ( x ) = \frac { 1 } { x } + x ^ { 2 } + \frac { 49 } { 4 }$ 
B)  $g ( x ) = \frac { 1 } { x } + x ^ { 2 } - \frac { 49 } { 4 }$ 
C)  $g ( x ) = - \frac { 1 } { x } - x ^ { 2 } - \frac { 49 } { 4 }$ 
D)  $g ( x ) = - \frac { 1 } { x } + x ^ { 2 } - \frac { 49 } { 4 }$ 
A)  $g ( x ) = \frac { 1 } { x } + x ^ { 2 } + \frac { 49 } { 4 }$ 
B)  $g ( x ) = \frac { 1 } { x } + x ^ { 2 } - \frac { 49 } { 4 }$ 
C)  $g ( x ) = - \frac { 1 } { x } - x ^ { 2 } - \frac { 49 } { 4 }$ 
D)  $g ( x ) = - \frac { 1 } { x } + x ^ { 2 } - \frac { 49 } { 4 }$

Question 15

Solve the problem.
-Find the table that matches the given graph.

Accepted Answer

A) 
$\begin{array}{l|l}
x & f^{\prime}(x) \
\hline a & \text { does not exist } \
\hline b & 0 \
\hline c & 1
\end{array}$ 
B) 
$\begin{array}{c|c}
\mathrm{x} & \mathrm{f}^{\prime}(\mathrm{x}) \
\hline \mathrm{a} & 0 \
\hline \mathrm{b} & 0 \
\hline \mathrm{c} & -1
\end{array}$ 
C) 
$\begin{array}{l|l}
x & f^{\prime}(x) \
\hline a & \text { does not exist } \
\hline b & 0 \
\hline c & -1
\end{array}$ 
D) 
$\begin{array}{l|l}
x & f^{\prime}(x) \
\hline a & \text { does not exist } \
\hline b & \text { does not exist } \
\hline c & -1
\end{array}$ 
A) 
$\begin{array}{l|l}
x & f^{\prime}(x) \
\hline a & \text { does not exist } \
\hline b & 0 \
\hline c & 1
\end{array}$ 
B) 
$\begin{array}{c|c}
\mathrm{x} & \mathrm{f}^{\prime}(\mathrm{x}) \
\hline \mathrm{a} & 0 \
\hline \mathrm{b} & 0 \
\hline \mathrm{c} & -1
\end{array}$ 
C) 
$\begin{array}{l|l}
x & f^{\prime}(x) \
\hline a & \text { does not exist } \
\hline b & 0 \
\hline c & -1
\end{array}$ 
D) 
$\begin{array}{l|l}
x & f^{\prime}(x) \
\hline a & \text { does not exist } \
\hline b & \text { does not exist } \
\hline c & -1
\end{array}$

Question 16

Find the absolute extreme values of the function on the interval.
-$$f ( x ) = x ^ { 4 / 3 } , - 1 \leq x \leq 8$$

Accepted Answer

A)  absolute maximum is 16 at $x = 8$; absolute minimum is 1 at $x = - 1$ 
B)  absolute maximum is 64 at $x = 8$; absolute minimum is 0 at $x = 0$ 
C)  absolute maximum is 16 at $x = 8$; absolute minimum does not exist 
D)  absolute maximum is 16 at $x = 8$; absolute minimum is 0 at $x = 0$ 
A)  absolute maximum is 16 at $x = 8$; absolute minimum is 1 at $x = - 1$ 
B)  absolute maximum is 64 at $x = 8$; absolute minimum is 0 at $x = 0$ 
C)  absolute maximum is 16 at $x = 8$; absolute minimum does not exist 
D)  absolute maximum is 16 at $x = 8$; absolute minimum is 0 at $x = 0$

Question 17

Identify the function's local and absolute extreme values, if any, saying where they occur.
-f(r) = (r - 7) 3

Accepted Answer

A)  no local extrema 
B)  local minimum: x = 0 
C)  local minimum: x = 7 
D)  local minimum: x = 0; local maximum: x = 7 
A)  no local extrema 
B)  local minimum: x = 0 
C)  local minimum: x = 7 
D)  local minimum: x = 0; local maximum: x = 7

Question 18

Solve the problem.
-Given the velocity and initial position of a body moving along a coordinate line at time t, find the body's positiol $\mathrm { t }$.
$$v = \cos \frac { \pi } { 2 } t , s ( 0 ) = 1$$

Accepted Answer

A)  $s = \sin t + 1$ 
B)  $s = \frac { 2 } { \pi } \sin \frac { \pi } { 2 } t + \pi$ 
C)  $s = \frac { 2 } { \pi } \sin \frac { \pi } { 2 } t + 1$ 
D)  $s = 2 \pi \sin \frac { \pi } { 2 } t$ time 
A)  $s = \sin t + 1$ 
B)  $s = \frac { 2 } { \pi } \sin \frac { \pi } { 2 } t + \pi$ 
C)  $s = \frac { 2 } { \pi } \sin \frac { \pi } { 2 } t + 1$ 
D)  $s = 2 \pi \sin \frac { \pi } { 2 } t$ time

Question 19

Find the absolute extreme values of the function on the interval.
-$$f ( x ) = \ln ( x + 2 ) + \frac { 1 } { x } , 1 \leq x \leq 8$$

Accepted Answer

A)  Minimum value is $\ln 4 + \frac { 1 } { 2 }$ at $x = 2$; maximum value is $\ln 10 + \frac { 1 } { 8 }$ at $x = 8$ 
B)  Minimum value is $- 1$ at $x = - 1$; maximum value is $\ln 10 + \frac { 1 } { 8 }$ at $x = 8$ 
C)  Minimum value is $\ln 3 + 1$ at $x = 1$; maximum value is $\ln 10 + \frac { 1 } { 8 }$ at $x = 8$ 
D)  Minimum value is $\ln 4 + \frac { 1 } { 2 }$ at $x = 2$; maximum value is $\ln 3 + 1$ at $x = 1$ 
A)  Minimum value is $\ln 4 + \frac { 1 } { 2 }$ at $x = 2$; maximum value is $\ln 10 + \frac { 1 } { 8 }$ at $x = 8$ 
B)  Minimum value is $- 1$ at $x = - 1$; maximum value is $\ln 10 + \frac { 1 } { 8 }$ at $x = 8$ 
C)  Minimum value is $\ln 3 + 1$ at $x = 1$; maximum value is $\ln 10 + \frac { 1 } { 8 }$ at $x = 8$ 
D)  Minimum value is $\ln 4 + \frac { 1 } { 2 }$ at $x = 2$; maximum value is $\ln 3 + 1$ at $x = 1$

Using the derivative of f(x) given below, determine the critical points of f(x). - $f ^ { \prime } ( x ) = ( x - 1 ) ^ { 2 } ( x + 4 )$

Find all possible functions with the given derivative. - $y ^ { \prime } = 8 x ^ { 2 } - 3 x$

Find the derivative at each critical point and determine the local extreme values. - $y = x ^ { 2 / 3 } \left( x ^ { 2 } - 16 \right) ; x \geq 0$

Determine whether the function satisfies the hypotheses of the Mean Value Theorem for the given interval. - $g ( x ) = x ^ { 3 / 4 }$ , $[0,5]$

Find all possible functions with the given derivative. - $r ^ { \prime } = 3 + \frac { 1 } { \theta ^ { 4 } }$

Using the derivative of f(x) given below, determine the critical points of f(x). -f'(x) = (x + 2)(x + 9)

Find the extreme values of the function and where they occur. - $y = \frac { 7 } { \sqrt { 1 - 9 x ^ { 2 } } }$

Find the derivative at each critical point and determine the local extreme values. - $y = \left\{ \begin{array} { l } 6 - 2 x , x \leq 1 \\x + 3 , x > 1\end{array} \right.$

Find the derivative at each critical point and determine the local extreme values. - $y = x \left( 4 - x ^ { 2 } \right)$

Graph the function, then find the extreme values of the function on the interval and indicate where they occur. -y = $\mid{x - 1}\mid$ - $\mid{ x - 6}\mid$ on the interval -2 < x < 7

Find the location of the indicated absolute extremum for the function. -Minimum

Determine all critical points for the function. - $f ( x ) = ( x - 9 ) ^ { 3 }$

Find the function with the given derivative whose graph passes through the point P. - $g ^ { \prime } ( x ) = \frac { 1 } { x ^ { 2 } } + 2 x , P ( - 4,4 )$

Solve the problem. -Find the table that matches the given graph.

Find the absolute extreme values of the function on the interval. - $f ( x ) = x ^ { 4 / 3 } , - 1 \leq x \leq 8$

Identify the function's local and absolute extreme values, if any, saying where they occur. -f(r) = (r - 7) 3

Solve the problem. -Given the velocity and initial position of a body moving along a coordinate line at time t, find the body's positiol $\mathrm { t }$ . $v = \cos \frac { \pi } { 2 } t , s ( 0 ) = 1$

Find the absolute extreme values of the function on the interval. - $f ( x ) = \ln ( x + 2 ) + \frac { 1 } { x } , 1 \leq x \leq 8$

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Exam 4: Applications of Derivatives

Using the derivative of f(x) given below, determine the critical points of f(x). - f′(x)=(x−1)2(x+4)f ^ { \prime } ( x ) = ( x - 1 ) ^ { 2 } ( x + 4 )f′(x)=(x−1)2(x+4)

Find all possible functions with the given derivative. - y′=8x2−3xy ^ { \prime } = 8 x ^ { 2 } - 3 xy′=8x2−3x

Find the derivative at each critical point and determine the local extreme values. - y=x2/3(x2−16);x≥0y = x ^ { 2 / 3 } \left( x ^ { 2 } - 16 \right) ; x \geq 0y=x2/3(x2−16);x≥0

Determine whether the function satisfies the hypotheses of the Mean Value Theorem for the given interval. - g(x)=x3/4g ( x ) = x ^ { 3 / 4 }g(x)=x3/4 , [0,5][0,5][0,5]

Find all possible functions with the given derivative. - r′=3+1θ4r ^ { \prime } = 3 + \frac { 1 } { \theta ^ { 4 } }r′=3+θ41​