Question 1

Using the derivative of f(x)  given below, determine the critical points of f(x) .
-f'(x)  = (x - 5)  e-x

Accepted Answer

A)   5
B)   6
C)   -6
D)   -5

Question 2

Determine all critical points for the function.
- f(x) =x3−12x+5f ( x )  = x ^ { 3 } - 12 x + 5f(x) =x3−12x+5

Accepted Answer

A)    x=−2x = - 2x=−2  and  x=2x = 2x=2 
B)    x=2x = 2x=2 
C)    x=−2x = - 2x=−2 
D)    x=−2,x=0x = - 2 , x = 0x=−2,x=0 , and  x=2x = 2x=2

Question 3

Find the absolute extreme values of the function on the interval.
- g(x) =6−7x2,−4≤x≤5g ( x )  = 6 - 7 x ^ { 2 } , - 4 \leq x \leq 5g(x) =6−7x2,−4≤x≤5

Accepted Answer

A)   absolute maximum is 12 at  x=0x = 0x=0 ; absolute minimum is  −106- 106−106  at  x=5x = 5x=5 
B)   absolute maximum is 6 at  x=0x = 0x=0 ; absolute minimum is  −169- 169−169  at  x=5x = 5x=5 
C)   absolute maximum is 7 at  x=0x = 0x=0 ; absolute minimum is  −181- 181−181  at  x=5x = 5x=5 
D)   absolute maximum is 42 at  x=0x = 0x=0 ; absolute minimum is  −106- 106−106  at  x=−4x = - 4x=−4

Question 4

Find the largest open interval where the function is changing as requested.
-Increasing  f(x) =1x2+1\quad f ( x )  = \frac { 1 } { x ^ { 2 } + 1 }f(x) =x2+11​

Accepted Answer

A)    (−∞,1) ( - \infty , 1 ) (−∞,1)  
B)    (0,∞) ( 0 , \infty ) (0,∞)  
C)    (−∞,0) ( - \infty , 0 ) (−∞,0)  
D)    (1,∞) ( 1 , \infty ) (1,∞)

Question 5

Identify the function's local and absolute extreme values, if any, saying where they occur.
- f(x) =x3+9x2+27x−3f ( x )  = x ^ { 3 } + 9 x ^ { 2 } + 27 x - 3f(x) =x3+9x2+27x−3

Accepted Answer

A)   local maximum at  x=−3x = - 3x=−3 
B)   local minimum at  x=−3x = - 3x=−3 
C)   local maximum at  x=−3x = - 3x=−3 ; local minimum at  x=3x = 3x=3 
D)   no local extrema

Question 6

Find the value or values of  ccc  that satisfy the equation  f(b) −f(a) b−a=f′(c) \frac { f ( b )  - f ( a )  } { b - a } = f ^ { \prime } ( c ) b−af(b) −f(a) ​=f′(c)   in the conclusion of the Mean Value Theorem for the function and interval.
-An approximation to the total profit (in thousands of dollars)  from the sale of  xxx  hundred thousand tires is given by  p=−x3+15x2−48x+450,x≥3p = - x ^ { 3 } + 15 x ^ { 2 } - 48 x + 450 , x \geq 3p=−x3+15x2−48x+450,x≥3 . Find the number of hundred thousands of tires that must be sold to maximize profit.

Accepted Answer

A)   5 hundred thousand
B)   3 hundred thousand
C)   10 hundred thousand
D)   8 hundred thousand

Question 7

Determine whether the function satisfies the hypotheses of the Mean Value Theorem for the given interval.
-$$f ( \theta ) = \left\{ \begin{array} { c c } 
\frac { \cos \theta } { \theta } , & - \pi \leq \theta < 0 \
0 , & \theta = 0
\end{array} \right.$$

Accepted Answer

A) True 
 B)False

Question 8

Find the largest open interval where the function is changing as requested.
-Decreasing  y=1x2+7y = \frac { 1 } { x ^ { 2 } } + 7y=x21​+7

Accepted Answer

A)    (−7,0) ( - 7,0 ) (−7,0)  
B)    (0,∞) ( 0 , \infty ) (0,∞)  
C)    (7,∞) ( 7 , \infty ) (7,∞)  
D)    (−7,7) ( - 7,7 ) (−7,7)

Question 9

Find the extreme values of the function and where they occur.
- y=1x2+1y = \frac { 1 } { x ^ { 2 } + 1 }y=x2+11​

Accepted Answer

A)   The minimum value is  −1- 1−1  at  x=0.5x = 0.5x=0.5 .
B)   The maximum value is 1 at  x=0.5x = 0.5x=0.5 , the minimum value is  −1- 1−1  at  x=0.5x = 0.5x=0.5 .
C)   The maximum value is 1 at  x=0.5x = 0.5x=0.5 .
D)   The maximum value is 1 at  x=0x = 0x=0 .

Question 10

Find the absolute extreme values of the function on the interval.
- h(x) =12x+4,−2≤x≤4h ( x )  = \frac { 1 } { 2 } x + 4 , - 2 \leq x \leq 4h(x) =21​x+4,−2≤x≤4

Accepted Answer

A)   absolute maximum is  −2- 2−2  at  x=−4x = - 4x=−4 ; absolute minimum is  −4- 4−4  at  x=2x = 2x=2 
B)   absolute maximum is  −2- 2−2  at  x=−2x = - 2x=−2 ; absolute minimum is 3 at  x=4x = 4x=4 
C)   absolute maximum is  −2- 2−2  at  x=4x = 4x=4 ; absolute minimum is 3 at  x=−2x = - 2x=−2 
D)   absolute maximum is 6 at  x=4x = 4x=4 ; absolute minimum is 3 at  x=−2x = - 2x=−2

Question 11

Use the maximum/minimum finder on a graphing calculator to determine the approximate location of all local extrema.
- f(x) =0.1x4−x3−15x2+59x+8f ( x )  = 0.1 x ^ { 4 } - x ^ { 3 } - 15 x ^ { 2 } + 59 x + 8f(x) =0.1x4−x3−15x2+59x+8

Accepted Answer

A)   Approximate local maximum at 1.805; approximate local minima at -6.72 and 12.449
B)   Approximate local maximum at 1.703; approximate local minima at -6.754 and 12.588
C)   Approximate local maximum at 1.735; approximate local minima at -6.777 and 12.542
D)   Approximate local maximum at 1.77; approximate local minima at -6.771 and 12.53

Question 12

Provide an appropriate response.
-The function $f ( x ) = \left\{ \begin{array} { l l } 7 x & 0 \leq x < 1 \ 0 & x = 1 \end{array} \right.$ is zero at $x = 0$ and $x = 1$ and differentiable on $( 0,1 )$, but its derivative on $( 0,1 )$ is never zero. Does this example contradict Rolle's Theorem?

Accepted Answer

This example does not contradi...

Question 13

Find the derivative at each critical point and determine the local extreme values.
-  1\end{array} ight.">y={−14x2−12x+154,x≤1x3−6x2+8x,x>1y = \left\{ \begin{array} { l l } - \frac { 1 } { 4 } x ^ { 2 } - \frac { 1 } { 2 } x + \frac { 15 } { 4 } , & x \leq 1 \x ^ { 3 } - 6 x ^ { 2 } + 8 x , & x > 1\end{array} ight.y={−41​x2−21​x+415​,x3−6x2+8x,​x≤1x>1​

Accepted Answer

A)    Critical Pt.  derivative  Extremum  Value x=−10 local max 4x=3.150 local min −3.08\begin{array}{l|l|l|l}	ext { Critical Pt. } & 	ext { derivative } & 	ext { Extremum } & 	ext { Value } \\hline x=-1 & 0 & 	ext { local max } & 4 \x=3.15 & 0 & 	ext { local min } & -3.08\end{array} Critical Pt. x=−1x=3.15​ derivative 00​ Extremum  local max  local min ​ Value 4−3.08​​ 
B)    Critical Pt.  derivative  Extremum  Value x=−1 undefined  local min 4x=3.150 local max −3.08\begin{array}{l|l|l|l}	ext { Critical Pt. } & 	ext { derivative } & 	ext { Extremum } & 	ext { Value } \\hline x=-1 & 	ext { undefined } & 	ext { local min } & 4 \x=3.15 & 0 & 	ext { local max } & -3.08\end{array} Critical Pt. x=−1x=3.15​ derivative  undefined 0​ Extremum  local min  local max ​ Value 4−3.08​​ 
C)    Critical Pt.  derivative  Extremum  Value x=−10 local min 4x=3.150 local max −3.08\begin{array}{l|l|l|l}	ext { Critical Pt. } & 	ext { derivative } & 	ext { Extremum } & 	ext { Value } \\hline x=-1 & 0 & 	ext { local min } & 4 \x=3.15 & 0 & 	ext { local max } & -3.08\end{array} Critical Pt. x=−1x=3.15​ derivative 00​ Extremum  local min  local max ​ Value 4−3.08​​ 
D)    Critical Pt.  derivative  Extremum  Value x=10 local max 4x=3.15 undefined  local max −3.08\begin{array}{l|l|l|l}	ext { Critical Pt. } & 	ext { derivative } & 	ext { Extremum } & 	ext { Value } \\hline x=1 & 0 & 	ext { local max } & 4 \x=3.15 & 	ext { undefined } & 	ext { local max } & -3.08\end{array} Critical Pt. x=1x=3.15​ derivative 0 undefined ​ Extremum  local max  local max ​ Value 4−3.08​​

Question 14

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

Accepted Answer

A)   x = -3
B)   x = -4
C)   x = 2
D)   x = 0

Question 15

Show that the function has exactly one zero in the given interval.
-$$r ( \theta ) = 5 \cot \theta + \frac { 1 } { \theta ^ { 2 } } + 6 , ( 0 , \pi )$$

Accepted Answer

The function <img  loading="lazy" src="/assets/images/blured-textbook.svg" class="d-inline m-1" alt="blured image" width="139" height="30"> is continuous on the open ...

Question 16

Find all possible functions with the given derivative.
- y′=x3−6xy ^ { \prime } = x ^ { 3 } - 6 xy′=x3−6x

Accepted Answer

A)    x44−6x22+C\frac { x ^ { 4 } } { 4 } - \frac { 6 x ^ { 2 } } { 2 } + C4x4​−26x2​+C 
B)    x33−6x22+C\frac { x ^ { 3 } } { 3 } - \frac { 6 x ^ { 2 } } { 2 } + C3x3​−26x2​+C 
C)    3x2−6+C3 x ^ { 2 } - 6 + C3x2−6+C 
D)    x44+6x2+C\frac { x ^ { 4 } } { 4 } + 6 x ^ { 2 } + C4x4​+6x2+C

Question 17

Find all possible functions with the given derivative.
- y′=3t−2ty ^ { \prime } = 3 t - \frac { 2 } { \sqrt { t } }y′=3t−t​2​

Accepted Answer

A)    3t2−4t+C3 t ^ { 2 } - 4 \sqrt { t } + C3t2−4tc-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​+C 
B)    t2−2t+Ct ^ { 2 } - 2 \sqrt { t } + Ct2−2tc-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​+C 
C)    3t2−2t+C3 t ^ { 2 } - \frac { 2 } { \sqrt { t } } + C3t2−tc-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​2​+C 
D)    32t2−4t+C\frac { 3 } { 2 } t ^ { 2 } - 4 \sqrt { t } + C23​t2−4tc-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​+C

Question 18

Find all possible functions with the given derivative.
- y′=x5y ^ { \prime } = x ^ { 5 }y′=x5

Accepted Answer

A)    5x4+C5 x 4 + C5x4+C 
B)    6x6+C6 x ^ { 6 } + C6x6+C 
C)    15x4+C\frac { 1 } { 5 } x ^ { 4 } + C51​x4+C 
D)    16x6+C\frac { 1 } { 6 } x ^ { 6 } + C61​x6+C

Question 19

Find the derivative at each critical point and determine the local extreme values.
- y=x21−xy = x ^ { 2 } \sqrt { 1 - x }y=x21−x​

Accepted Answer

A)    Critical Pt.  derivative  Extremum  Value x=450 local max 161255\begin{array}{l|l|l|l}	ext { Critical Pt. } & 	ext { derivative } & 	ext { Extremum } & 	ext { Value } \\hline x=\frac{4}{5} & 0 & 	ext { local max } & \frac{16}{125} \sqrt{5}\end{array} Critical Pt. x=54​​ derivative 0​ Extremum  local max ​ Value 12516​5c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​​ 
B)    Critical Pt.  derivative  Extremum  Value x=00 min 0x=10 min 0x=450 local max⁡161255\begin{array}{l|l|l|l}	ext { Critical Pt. } & 	ext { derivative } & 	ext { Extremum } & 	ext { Value } \\hline x=0 & 0 & 	ext { min } & 0 \x=1 & 0 & 	ext { min } & 0 \x=\frac{4}{5} & 0 & 	ext { local } \max & \frac{16}{125} \sqrt{5}\end{array} Critical Pt. x=0x=1x=54​​ derivative 000​ Extremum  min  min  local max​ Value 0012516​5c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​​ 
C)    Critical Pt.  derivative  Extremum  Value x=00min⁡0x=1 undefined  min 0x=450 local max 161255\begin{array}{l|l|l|l}	ext { Critical Pt. } & 	ext { derivative } & 	ext { Extremum } & 	ext { Value } \\hline x=0 & 0 & \min & 0 \x=1 & 	ext { undefined } & 	ext { min } & 0 \x=\frac{4}{5} & 0 & 	ext { local max } & \frac{16}{125} \sqrt{5}\end{array} Critical Pt. x=0x=1x=54​​ derivative 0 undefined 0​ Extremum min min  local max ​ Value 0012516​5c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​​ 
D)    Critical Pt.  derivative  Extremum  Value x=1 undefined  min 0x=450 local max⁡161255\begin{array}{l|l|l|l}	ext { Critical Pt. } & 	ext { derivative } & 	ext { Extremum } & 	ext { Value } \\hline x=1 & 	ext { undefined } & 	ext { min } & 0 \x=\frac{4}{5} & 0 & 	ext { local } \max & \frac{16}{125} \sqrt{5}\end{array} Critical Pt. x=1x=54​​ derivative  undefined 0​ Extremum  min  local max​ Value 012516​5c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​​

Question 20

Find the largest open interval where the function is changing as requested.
-Increasing  y=(x2−9) 2\quad y = \left( x ^ { 2 } - 9 \right)  ^ { 2 }y=(x2−9) 2

Accepted Answer

A)    (−∞,0) ( - \infty , 0 ) (−∞,0)  
B)    (−3,0) ( - 3,0 ) (−3,0)  
C)    (3,∞) ( 3 , \infty ) (3,∞)  
D)    (−3,3) ( - 3,3 ) (−3,3)

Exam 4: Applications of Derivatives

Find the absolute extreme values of the function on the interval. - $h ( x ) = \frac { 1 } { 2 } x + 4 , - 2 \leq x \leq 4$

Correct Answer
verified

Find the largest open interval where the function is changing as requested. -Increasing $\quad y = \left( x ^ { 2 } - 9 \right) ^ { 2 }$

Correct Answer
verified

Show that the function has exactly one zero in the given interval. - $r ( \theta ) = 5 \cot \theta + \frac { 1 } { \theta ^ { 2 } } + 6 , ( 0 , \pi )$

Correct Answer
verified
The function is continuous on the open ...
View Answer

Determine all critical points for the function. - $f ( x ) = x ^ { 3 } - 12 x + 5$

Correct Answer
verified

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

Correct Answer
verified

Provide an appropriate response. -The function $f ( x ) = \left\{ \begin{array} { l l } 7 x & 0 \leq x < 1 \\ 0 & x = 1 \end{array} \right.$ is zero at $x = 0$ and $x = 1$ and differentiable on $( 0,1 )$ , but its derivative on $( 0,1 )$ is never zero. Does this example contradict Rolle's Theorem?

Correct Answer
verified
This example does not contradi...
View Answer

Find the derivative at each critical point and determine the local extreme values. - $y = \left\{ \begin{array} { l l } - \frac { 1 } { 4 } x ^ { 2 } - \frac { 1 } { 2 } x + \frac { 15 } { 4 } , & x \leq 1 \\x ^ { 3 } - 6 x ^ { 2 } + 8 x , & x > 1\end{array} \right.$

Correct Answer
verified

Find the largest open interval where the function is changing as requested. -Increasing $\quad f ( x ) = \frac { 1 } { x ^ { 2 } + 1 }$

Correct Answer
verified

Use the maximum/minimum finder on a graphing calculator to determine the approximate location of all local extrema. - $f ( x ) = 0.1 x ^ { 4 } - x ^ { 3 } - 15 x ^ { 2 } + 59 x + 8$

Correct Answer
verified

Find the derivative at each critical point and determine the local extreme values. - $y = x ^ { 2 } \sqrt { 1 - x }$

Correct Answer
verified

Find the largest open interval where the function is changing as requested. -Decreasing $y = \frac { 1 } { x ^ { 2 } } + 7$

Correct Answer
verified

Identify the function's local and absolute extreme values, if any, saying where they occur. - $f ( x ) = x ^ { 3 } + 9 x ^ { 2 } + 27 x - 3$

Correct Answer
verified

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

Correct Answer
verified

Find all possible functions with the given derivative. - $y ^ { \prime } = x ^ { 3 } - 6 x$

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Find the absolute extreme values of the function on the interval. - $g ( x ) = 6 - 7 x ^ { 2 } , - 4 \leq x \leq 5$

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

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Using the derivative of f(x) given below, determine the critical points of f(x) . -f'(x) = (x - 5) e^-x

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Determine whether the function satisfies the hypotheses of the Mean Value Theorem for the given interval. - $f ( \theta ) = \left\{ \begin{array} { c c } \frac { \cos \theta } { \theta } , & - \pi \leq \theta < 0 \\0 , & \theta = 0\end{array} \right.$

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Find all possible functions with the given derivative. - $y ^ { \prime } = x ^ { 5 }$

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Exam 4: Applications of Derivatives

Find the absolute extreme values of the function on the interval. - h(x) =12x+4,−2≤x≤4h ( x ) = \frac { 1 } { 2 } x + 4 , - 2 \leq x \leq 4h(x) =21​x+4,−2≤x≤4

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Find the largest open interval where the function is changing as requested. -Increasing y=(x2−9) 2\quad y = \left( x ^ { 2 } - 9 \right) ^ { 2 }y=(x2−9) 2

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Show that the function has exactly one zero in the given interval. - r(θ)=5cot⁡θ+1θ2+6,(0,π)r ( \theta ) = 5 \cot \theta + \frac { 1 } { \theta ^ { 2 } } + 6 , ( 0 , \pi )r(θ)=5cotθ+θ21​+6,(0,π)

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Determine all critical points for the function. - f(x) =x3−12x+5f ( x ) = x ^ { 3 } - 12 x + 5f(x) =x3−12x+5

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Find all possible functions with the given derivative. - y′=3t−2ty ^ { \prime } = 3 t - \frac { 2 } { \sqrt { t } }y′=3t−t​2​

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Correct AnswerverifiedThis example does not contradi...View AnswerShow Answer

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Find the largest open interval where the function is changing as requested. -Increasing f(x) =1x2+1\quad f ( x ) = \frac { 1 } { x ^ { 2 } + 1 }f(x) =x2+11​

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Use the maximum/minimum finder on a graphing calculator to determine the approximate location of all local extrema. - f(x) =0.1x4−x3−15x2+59x+8f ( x ) = 0.1 x ^ { 4 } - x ^ { 3 } - 15 x ^ { 2 } + 59 x + 8f(x) =0.1x4−x3−15x2+59x+8

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Find the derivative at each critical point and determine the local extreme values. - y=x21−xy = x ^ { 2 } \sqrt { 1 - x }y=x21−x​

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Find the largest open interval where the function is changing as requested. -Decreasing y=1x2+7y = \frac { 1 } { x ^ { 2 } } + 7y=x21​+7

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Identify the function's local and absolute extreme values, if any, saying where they occur. - f(x) =x3+9x2+27x−3f ( x ) = x ^ { 3 } + 9 x ^ { 2 } + 27 x - 3f(x) =x3+9x2+27x−3

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Find the extreme values of the function and where they occur. - y=1x2+1y = \frac { 1 } { x ^ { 2 } + 1 }y=x2+11​

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Find all possible functions with the given derivative. - y′=x3−6xy ^ { \prime } = x ^ { 3 } - 6 xy′=x3−6x

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Find the absolute extreme values of the function on the interval. - g(x) =6−7x2,−4≤x≤5g ( x ) = 6 - 7 x ^ { 2 } , - 4 \leq x \leq 5g(x) =6−7x2,−4≤x≤5

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

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Using the derivative of f(x) given below, determine the critical points of f(x) . -f'(x) = (x - 5) e-x

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Find all possible functions with the given derivative. - y′=x5y ^ { \prime } = x ^ { 5 }y′=x5

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Find the absolute extreme values of the function on the interval. - $h ( x ) = \frac { 1 } { 2 } x + 4 , - 2 \leq x \leq 4$

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Find the largest open interval where the function is changing as requested. -Increasing $\quad y = \left( x ^ { 2 } - 9 \right) ^ { 2 }$

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Show that the function has exactly one zero in the given interval. - $r ( \theta ) = 5 \cot \theta + \frac { 1 } { \theta ^ { 2 } } + 6 , ( 0 , \pi )$

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The function is continuous on the open ...
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Determine all critical points for the function. - $f ( x ) = x ^ { 3 } - 12 x + 5$

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

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This example does not contradi...
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Find the largest open interval where the function is changing as requested. -Increasing $\quad f ( x ) = \frac { 1 } { x ^ { 2 } + 1 }$

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Use the maximum/minimum finder on a graphing calculator to determine the approximate location of all local extrema. - $f ( x ) = 0.1 x ^ { 4 } - x ^ { 3 } - 15 x ^ { 2 } + 59 x + 8$

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Find the derivative at each critical point and determine the local extreme values. - $y = x ^ { 2 } \sqrt { 1 - x }$

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Find the largest open interval where the function is changing as requested. -Decreasing $y = \frac { 1 } { x ^ { 2 } } + 7$

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Identify the function's local and absolute extreme values, if any, saying where they occur. - $f ( x ) = x ^ { 3 } + 9 x ^ { 2 } + 27 x - 3$

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

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Find all possible functions with the given derivative. - $y ^ { \prime } = x ^ { 3 } - 6 x$

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Find the absolute extreme values of the function on the interval. - $g ( x ) = 6 - 7 x ^ { 2 } , - 4 \leq x \leq 5$

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Using the derivative of f(x) given below, determine the critical points of f(x) . -f'(x) = (x - 5) e^-x

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Find all possible functions with the given derivative. - $y ^ { \prime } = x ^ { 5 }$

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