Question 1

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 2

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 3

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 4

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 5

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 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 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 8

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 9

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 10

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 11

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 12

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 13

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)

Question 14

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 15

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 16

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 17

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 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 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 20

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

Exam 4: Applications of Derivatives

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

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

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

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

Correct Answer
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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

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

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

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

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

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

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

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

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

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

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

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

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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. - $y = \left\{ \begin{array} { l l } - \frac { 1 } { 4 } x ^ { 2 } - \frac { 1 } { 2 } x + \frac { 15 } { 4 } , & x \leq 1 \\x ^ { 3 } - 6 x ^ { 2 } + 8 x , & x > 1\end{array} \right.$

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Exam 4: Applications of Derivatives

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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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. -Increasing f(x) =1x2+1\quad f ( x ) = \frac { 1 } { x ^ { 2 } + 1 }f(x) =x2+11​

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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′=x3−6xy ^ { \prime } = x ^ { 3 } - 6 xy′=x3−6x

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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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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 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 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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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. -Decreasing y=1x2+7y = \frac { 1 } { x ^ { 2 } } + 7y=x21​+7

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

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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′=3t−2ty ^ { \prime } = 3 t - \frac { 2 } { \sqrt { t } }y′=3t−t​2​

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

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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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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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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. -Increasing $\quad f ( x ) = \frac { 1 } { x ^ { 2 } + 1 }$

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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. - $y ^ { \prime } = x ^ { 3 } - 6 x$

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

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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. -Decreasing $y = \frac { 1 } { x ^ { 2 } } + 7$

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This example does not contradi...
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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. - $y ^ { \prime } = 3 t - \frac { 2 } { \sqrt { t } }$

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

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

Correct Answer
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Correct Answer
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