Question 1

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+94f ( x )  = 0.01 x ^ { 5 } - x ^ { 4 } + x ^ { 3 } + 8 x ^ { 2 } - 7 x + 94f(x) =0.01x5−x4+x3+8x2−7x+94

Accepted Answer

A)   Approximate local maxima at -1.861 and 2.247; approximate local minima at 0.423 and 79.192
B)   Approximate local maxima at -1.878 and 2.17; approximate local minima at 0.466 and 79.115
C)   Approximate local maxima at -1.827 and 2.289; approximate local minima at 0.455 and 79.158
D)   Approximate local maxima at -1.861 and 2.247; approximate local minimum at 0.423

Question 2

Find the function with the given derivative whose graph passes through the point P.
- r′(θ) =8−csc⁡2θ,P(π4,0) \mathrm { r } ^ { \prime } ( 	heta )  = 8 - \csc ^ { 2 } 	heta , \mathrm { P } \left( \frac { \pi } { 4 } , 0 ight) r′(θ) =8−csc2θ,P(4π​,0)

Accepted Answer

A)    r(θ) =8θ+cot⁡θ−2π−1r ( 	heta )  = 8 	heta + \cot 	heta - 2 \pi - 1r(θ) =8θ+cotθ−2π−1 
B)    r(θ) =8θ−cot⁡θ−2π−2r ( 	heta )  = 8 	heta - \cot 	heta - 2 \pi - 2r(θ) =8θ−cotθ−2π−2 
C)    r(θ) =8θ+cot⁡θ−2π+1r ( 	heta )  = 8 	heta + \cot 	heta - 2 \pi + 1r(θ) =8θ+cotθ−2π+1 
D)    r(θ) =8θ−cot⁡θ−2π+2r ( 	heta )  = 8 	heta - \cot 	heta - 2 \pi + 2r(θ) =8θ−cotθ−2π+2

Question 3

Find all possible functions with the given derivative.
- y′=32ty ^ { \prime } = \frac { 3 } { 2 \sqrt { t } }y′=2t​3​

Accepted Answer

A)    3t22+C\frac { 3 t ^ { 2 } } { 2 } + C23t2​+C 
B)    3t+C3 \sqrt { t } + C3tc-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)    t+C\sqrt { t } + Ctc-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 
D)    3t2+C\frac { 3 t } { 2 } + C23t​+C

Question 4

Find the largest open interval where the function is changing as requested.
-Increasing y = 7x - 5

Accepted Answer

A)   (-5, 7) 
B)   (-5, ) 
C)   (-, 7) 
D)   (-, )

Question 5

Find the value or values of $c$ that satisfy the equation $\frac { f ( b ) - f ( a ) } { b - a } = f ^ { \prime } ( c )$ in the conclusion of the Mean Value Theorem for the function and interval.
-$\text { The function } \mathrm { P } ( \mathrm { x } ) = 2 \mathrm { x } + \frac { 200 } { \mathrm { x } } , 0 < \mathrm { x } < \infty$ models the perimeter of a rectangle of dimensions x by $\frac{100}{x}$
(a) Find the extreme values for P.
(b) Give an interpretation in terms of perimeter of the rectangle for any values found in part (a).

Accepted Answer

(a) The only critical point in the inter...

Question 6

Find the function with the given derivative whose graph passes through the point P.
- r′(t) =sec⁡2t−4,P(0,0) r ^ { \prime } ( t )  = \sec ^ { 2 } t - 4 , P ( 0,0 ) r′(t) =sec2t−4,P(0,0)

Accepted Answer

A)    r(t) =tan⁡t−4tr ( t )  = 	an t - 4 tr(t) =tant−4t 
B)    r(t) =sec⁡ttan⁡t−4t−1r ( t )  = \sec t 	an t - 4 t - 1r(t) =secttant−4t−1 
C)    r(t) =sec⁡t−t−6r ( t )  = \sec t - t - 6r(t) =sect−t−6 
D)    r(t) =sec⁡t−4t−4r ( t )  = \sec t - 4 t - 4r(t) =sect−4t−4

Question 7

Identify the function's local and absolute extreme values, if any, saying where they occur.
- h(x) =x−2x2+3x+6h ( x )  = \frac { x - 2 } { x ^ { 2 } + 3 x + 6 }h(x) =x2+3x+6x−2​

Accepted Answer

A)   local minimum at x = -5; local maximum at x = 6
B)   local minimum at x = -2; local maximum at x = 6
C)   local minimum at x = -2; no local maxima
D)   no local extrema

Question 8

Find the function with the given derivative whose graph passes through the point P.
- r′(θ) =4+sec⁡2θ,P(π,0) \mathrm { r } ^ { \prime } ( 	heta )  = 4 + \sec ^ { 2 } 	heta , \mathrm { P } ( \pi , 0 ) r′(θ) =4+sec2θ,P(π,0)

Accepted Answer

A)    r(θ) =4θ+13sec⁡3θr ( 	heta )  = 4 	heta + \frac { 1 } { 3 } \sec ^ { 3 } 	hetar(θ) =4θ+31​sec3θ 
B)    r(θ) =2θ2+tan⁡θ−4πr ( 	heta )  = 2 	heta ^ { 2 } + 	an 	heta - 4 \pir(θ) =2θ2+tanθ−4π 
C)    r(θ) =4θ+tan⁡θ−4πr ( 	heta )  = 4 	heta + 	an 	heta - 4 \pir(θ) =4θ+tanθ−4π 
D)    r′(θ) =2θ2+tan⁡θ+4π\mathrm { r } ^ { \prime } ( 	heta )  = 2 	heta ^ { 2 } + 	an 	heta + 4 \pir′(θ) =2θ2+tanθ+4π

Question 9

Find the derivative at each critical point and determine the local extreme values.
- y={4−x,x<04+3x−x2,x≥0y = \left\{ \begin{array} { l l } 4 - x , & x < 0 \4 + 3 x - x ^ { 2 } , & x \geq 0\end{array} ight.y={4−x,4+3x−x2,​x<0x≥0​

Accepted Answer

A)    Critical Pt.  derivative  Extremum  Value x=4 undefined  local min 4x=00 local max 254\begin{array}{l|l|l|l}	ext { Critical Pt. } & 	ext { derivative } & 	ext { Extremum } & 	ext { Value } \\hline x=4 & 	ext { undefined } & 	ext { local min } & 4 \x=0 & 0 & 	ext { local max } & \frac{25}{4}\end{array} Critical Pt. x=4x=0​ derivative  undefined 0​ Extremum  local min  local max ​ Value 4425​​​ 
B)    Critical Pt.  derivative  Extremum  Value x=0 undefined  local min 4x=520 local max 414\begin{array}{l|l|l|l}	ext { Critical Pt. } & 	ext { derivative } & 	ext { Extremum } & 	ext { Value } \\hline x=0 & 	ext { undefined } & 	ext { local min } & 4 \x=\frac{5}{2} & 0 & 	ext { local max } & \frac{41}{4}\end{array} Critical Pt. x=0x=25​​ derivative  undefined 0​ Extremum  local min  local max ​ Value 4441​​​ 
C)    Critical Pt.  derivative  Extremum  Value x=0 undefined  local min −4x=320 local max 74\begin{array}{l|l|l|l}	ext { Critical Pt. } & 	ext { derivative } & 	ext { Extremum } & 	ext { Value } \\hline x=0 & 	ext { undefined } & 	ext { local min } & -4 \x=\frac{3}{2} & 0 & 	ext { local max } & \frac{7}{4}\end{array} Critical Pt. x=0x=23​​ derivative  undefined 0​ Extremum  local min  local max ​ Value −447​​​ 
D)    Critical Pt.  derivative  Extremum  Value x=0 undefined  local min 4x=320 local max 254\begin{array}{l|l|l|l}	ext { Critical Pt. } & 	ext { derivative } & 	ext { Extremum } & 	ext { Value } \\hline x=0 & 	ext { undefined } & 	ext { local min } & 4 \x=\frac{3}{2} & 0 & 	ext { local max } & \frac{25}{4}\end{array} Critical Pt. x=0x=23​​ derivative  undefined 0​ Extremum  local min  local max ​ Value 4425​​​

Question 10

Find the extreme values of the function and where they occur.
- y=x2exy = x ^ { 2 } e ^ { x }y=x2ex

Accepted Answer

A)   Minimum value is  4e−24 \mathrm { e } ^ { - 2 }4e−2  at  x=−2\mathrm { x } = - 2x=−2 ; no maximum value.
B)   Minimum value is 0 at  x=0x = 0x=0 , maximum value is  4e−24 \mathrm { e } ^ { - 2 }4e−2  at  x=−2x = - 2x=−2 .
C)   Minimum value is 0 at  x=0x = 0x=0 ; no maximum value.
D)   None

Question 11

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.
- f(x) =x2+5x+2,[1,2]f ( x )  = x ^ { 2 } + 5 x + 2 , [ 1,2 ]f(x) =x2+5x+2,[1,2]

Accepted Answer

A)    −32,32- \frac { 3 } { 2 } , \frac { 3 } { 2 }−23​,23​ 
B)   1,2
C)    32\frac { 3 } { 2 }23​ 
D)    0,320 , \frac { 3 } { 2 }0,23​

Question 12

Solve the problem.
-On our moon, the acceleration of gravity is  1.6 m/sec21.6 \mathrm {~m} / \mathrm { sec } ^ { 2 }1.6 m/sec2 . If a rock is dropped into a crevasse, how fast will it be going just before it hits bottom 45 seconds later?

Accepted Answer

A)    3240 m/sec3240 \mathrm {~m} / \mathrm { sec }3240 m/sec 
B)    −36 m/sec- 36 \mathrm {~m} / \mathrm { sec }−36 m/sec 
C)    −72 m/sec- 72 \mathrm {~m} / \mathrm { sec }−72 m/sec 
D)    72 m/sec72 \mathrm {~m} / \mathrm { sec }72 m/sec

Question 13

Find the absolute extreme values of the function on the interval.
- f(x) =ln⁡(−x) ,−7≤x≤−1f ( x )  = \ln ( - x )  , - 7 \leq x \leq - 1f(x) =ln(−x) ,−7≤x≤−1

Accepted Answer

A)   No minimum value; no maximum value
B)   Minimum value is 0 at  x=−1x = - 1x=−1 ; no maximum value
C)   Maximum value is 0 at  x=−1x = - 1x=−1 ; minimum value is  −ln⁡7- \ln 7−ln7  at  x=−7x = - 7x=−7 
D)   Minimum value is 0 at  x=−1x = - 1x=−1 ; maximum value is  ln⁡7\ln 7ln7  at  x=−7x = - 7x=−7

Question 14

Using the derivative of f(x)  given below, determine the intervals on which f(x)  is increasing or decreasing.
- f′(x) =(x+5) 2e−xf ^ { \prime } ( x )  = ( x + 5 )  ^ { 2 } e ^ { - x }f′(x) =(x+5) 2e−x

Accepted Answer

A)   Decreasing on  (−∞,−5) ( - \infty , - 5 ) (−∞,−5)  ; increasing on  (−5,∞) ( - 5 , \infty ) (−5,∞)  
B)   Never decreasing; increasing on  (−∞,−5) ∪(−5,∞) ( - \infty , - 5 )  \cup ( - 5 , \infty ) (−∞,−5) ∪(−5,∞)  
C)   Never increasing; decreasing on  (−∞,−5) ∪(−5,∞) ( - \infty , - 5 )  \cup ( - 5 , \infty ) (−∞,−5) ∪(−5,∞)  
D)   Never decreasing; increasing on  (−∞,5) ∪(5,∞) ( - \infty , 5 )  \cup ( 5 , \infty ) (−∞,5) ∪(5,∞)

Question 15

Find the absolute extreme values of the function on the interval.
- F(x) =−3x2,0.5≤x≤3F ( x )  = - \frac { 3 } { x ^ { 2 } } , 0.5 \leq x \leq 3F(x) =−x23​,0.5≤x≤3

Accepted Answer

A)   absolute maximum is  −13- \frac { 1 } { 3 }−31​  at  x=12x = \frac { 1 } { 2 }x=21​ ; absolute minimum is  −12- 12−12  at  x=−3x = - 3x=−3 
B)   absolute maximum is  −13- \frac { 1 } { 3 }−31​  at  x=3x = 3x=3 ; absolute minimum is  −12- 12−12  at  x=12x = \frac { 1 } { 2 }x=21​ 
C)   absolute maximum is  13\frac { 1 } { 3 }31​  at  x=12x = \frac { 1 } { 2 }x=21​ ; absolute minimum is  −12- 12−12  at  x=3x = 3x=3 
D)   absolute maximum is  −13- \frac { 1 } { 3 }−31​  at  x=3x = 3x=3 ; absolute minimum is  −12- 12−12  at  x=−12x = - \frac { 1 } { 2 }x=−21​

Question 16

Using the derivative of f(x)  given below, determine the intervals on which f(x)  is increasing or decreasing.
- f′(x) =(8−x) (9−x) f ^ { \prime } ( x )  = ( 8 - x )  ( 9 - x ) f′(x) =(8−x) (9−x)

Accepted Answer

A)   Decreasing on  (−∞,8) ( - \infty , 8 ) (−∞,8)  ; increasing on  (9,∞) ( 9 , \infty ) (9,∞)  
B)   Decreasing on  (8,9) ( 8,9 ) (8,9)  ; increasing on  (−∞,8) ∪(9,∞) ( - \infty , 8 )  \cup ( 9 , \infty ) (−∞,8) ∪(9,∞)  
C)   Decreasing on  (−∞,−8) ∪(−9,∞) ( - \infty , - 8 )  \cup ( - 9 , \infty ) (−∞,−8) ∪(−9,∞)  ; increasing on  (−8,−9) ( - 8 , - 9 ) (−8,−9)  
D)   Decreasing on  (−∞,8) ∪(9,∞) ( - \infty , 8 )  \cup ( 9 , \infty ) (−∞,8) ∪(9,∞)  ; increasing on  (8,9) ( 8,9 ) (8,9)

Question 17

Provide an appropriate response.
-Let f have a derivative on an interval I. f' has successive distinct zeros at x = 1 and x = 5. Prove that there can be at most one zero of f on the interval (1, 5).

Accepted Answer

Assume there are 2 (or more) zeros <img  loading="lazy" src="/assets/images/blured-textbook.svg" class="d-inline m-1" alt="blured image" width="139" height="30"> and ...

Question 18

Find the absolute extreme values of the function on the interval.
- f(x) =csc⁡x,−π2≤x≤3π2f ( x )  = \csc x , - \frac { \pi } { 2 } \leq x \leq \frac { 3 \pi } { 2 }f(x) =cscx,−2π​≤x≤23π​

Accepted Answer

A)   absolute maximum is 0 at  x=−π\mathrm { x } = - \pix=−π ; absolute minimum is  −1- 1−1  at  x=π\mathrm { x } = \pix=π 
B)   absolute maximum does not exist; absolute minimum does not exist
C)   absolute maximum is  −1- 1−1  at  x=πx = \pix=π ; absolute minimum is 1 at  x=0x = 0x=0 
D)   absolute maximum is 1 at  x=πx = \pix=π ; absolute minimum is  −1- 1−1  at  x=πx = \pix=π

Question 19

Find the extreme values of the function and where they occur.
- y=(x−5) 2/3y = ( x - 5 )  ^ { 2 / 3 }y=(x−5) 2/3

Accepted Answer

A)   The minimum value is 0 at x = -5.
B)   The maximum value is 0 at x = -5.
C)   There are no definable extrema.
D)   The minimum value is 0 at x = 5.

Question 20

Find the extreme values of the function and where they occur.
- y=x3−12x+2y = x ^ { 3 } - 12 x + 2y=x3−12x+2

Accepted Answer

A)   Local maximum at  (−2,18) ( - 2,18 ) (−2,18)  , local minimum at  (2,−14) ( 2 , - 14 ) (2,−14)  .
B)   Local maximum at  (2,−14) ( 2 , - 14 ) (2,−14)  , local minimum at  (−2,18) ( - 2,18 ) (−2,18)  .
C)   None
D)   Local maximum at  (0,0) ( 0,0 ) (0,0)  .

Exam 4: Applications of Derivatives

Correct Answer
verified
(a) The only critical point in the inter...
View Answer

Find the extreme values of the function and where they occur. - $y = ( x - 5 ) ^ { 2 / 3 }$

Correct Answer
verified

Use the maximum/minimum finder on a graphing calculator to determine the approximate location of all local extrema. - $f ( x ) = 0.01 x ^ { 5 } - x ^ { 4 } + x ^ { 3 } + 8 x ^ { 2 } - 7 x + 94$

Correct Answer
verified

Using the derivative of f(x) given below, determine the intervals on which f(x) is increasing or decreasing. - $f ^ { \prime } ( x ) = ( 8 - x ) ( 9 - x )$

Correct Answer
verified

Find the function with the given derivative whose graph passes through the point P. - $\mathrm { r } ^ { \prime } ( \theta ) = 8 - \csc ^ { 2 } \theta , \mathrm { P } \left( \frac { \pi } { 4 } , 0 \right)$

Correct Answer
verified

Identify the function's local and absolute extreme values, if any, saying where they occur. - $h ( x ) = \frac { x - 2 } { x ^ { 2 } + 3 x + 6 }$

Correct Answer
verified

Find the extreme values of the function and where they occur. - $y = x ^ { 3 } - 12 x + 2$

Correct Answer
verified

Find the largest open interval where the function is changing as requested. -Increasing y = 7x - 5

Correct Answer
verified

Find the derivative at each critical point and determine the local extreme values. - $y = \left\{ \begin{array} { l l } 4 - x , & x < 0 \\4 + 3 x - x ^ { 2 } , & x \geq 0\end{array} \right.$

Correct Answer
verified

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

Correct Answer
verified

Solve the problem. -On our moon, the acceleration of gravity is $1.6 \mathrm {~m} / \mathrm { sec } ^ { 2 }$ . If a rock is dropped into a crevasse, how fast will it be going just before it hits bottom 45 seconds later?

Correct Answer
verified

Find the function with the given derivative whose graph passes through the point P. - $r ^ { \prime } ( t ) = \sec ^ { 2 } t - 4 , P ( 0,0 )$

Correct Answer
verified

Find the absolute extreme values of the function on the interval. - $f ( x ) = \ln ( - x ) , - 7 \leq x \leq - 1$

Correct Answer
verified

Find the value or values of $c$ that satisfy the equation $\frac { f ( b ) - f ( a ) } { b - a } = f ^ { \prime } ( c )$ in the conclusion of the Mean Value Theorem for the function and interval. - $f ( x ) = x ^ { 2 } + 5 x + 2 , [ 1,2 ]$

Correct Answer
verified

Find the absolute extreme values of the function on the interval. - $f ( x ) = \csc x , - \frac { \pi } { 2 } \leq x \leq \frac { 3 \pi } { 2 }$

Correct Answer
verified

Provide an appropriate response. -Let f have a derivative on an interval I. f' has successive distinct zeros at x = 1 and x = 5. Prove that there can be at most one zero of f on the interval (1, 5).

Correct Answer
verified
Assume there are 2 (or more) zeros and ...
View Answer

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

Correct Answer
verified

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

Correct Answer
verified

Find the absolute extreme values of the function on the interval. - $F ( x ) = - \frac { 3 } { x ^ { 2 } } , 0.5 \leq x \leq 3$

Correct Answer
verified

Using the derivative of f(x) given below, determine the intervals on which f(x) is increasing or decreasing. - $f ^ { \prime } ( x ) = ( x + 5 ) ^ { 2 } e ^ { - x }$

Correct Answer
verified

Exam 4: Applications of Derivatives

Correct Answerverified(a) The only critical point in the inter...View AnswerShow Answer

Find the extreme values of the function and where they occur. - y=(x−5) 2/3y = ( x - 5 ) ^ { 2 / 3 }y=(x−5) 2/3

Correct AnswerverifiedShow Answer

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+94f ( x ) = 0.01 x ^ { 5 } - x ^ { 4 } + x ^ { 3 } + 8 x ^ { 2 } - 7 x + 94f(x) =0.01x5−x4+x3+8x2−7x+94

Correct AnswerverifiedShow Answer

Using the derivative of f(x) given below, determine the intervals on which f(x) is increasing or decreasing. - f′(x) =(8−x) (9−x) f ^ { \prime } ( x ) = ( 8 - x ) ( 9 - x ) f′(x) =(8−x) (9−x)

Correct AnswerverifiedShow Answer

Find the function with the given derivative whose graph passes through the point P. - r′(θ) =8−csc⁡2θ,P(π4,0) \mathrm { r } ^ { \prime } ( \theta ) = 8 - \csc ^ { 2 } \theta , \mathrm { P } \left( \frac { \pi } { 4 } , 0 \right) r′(θ) =8−csc2θ,P(4π​,0)

Correct AnswerverifiedShow Answer

Identify the function's local and absolute extreme values, if any, saying where they occur. - h(x) =x−2x2+3x+6h ( x ) = \frac { x - 2 } { x ^ { 2 } + 3 x + 6 }h(x) =x2+3x+6x−2​

Correct AnswerverifiedShow Answer

Find the extreme values of the function and where they occur. - y=x3−12x+2y = x ^ { 3 } - 12 x + 2y=x3−12x+2

Correct AnswerverifiedShow Answer

Find the largest open interval where the function is changing as requested. -Increasing y = 7x - 5

Correct AnswerverifiedShow Answer

Find the derivative at each critical point and determine the local extreme values. - y={4−x,x<04+3x−x2,x≥0y = \left\{ \begin{array} { l l } 4 - x , & x < 0 \\4 + 3 x - x ^ { 2 } , & x \geq 0\end{array} \right.y={4−x,4+3x−x2,​x<0x≥0​

Correct AnswerverifiedShow Answer

Find the extreme values of the function and where they occur. - y=x2exy = x ^ { 2 } e ^ { x }y=x2ex

Correct AnswerverifiedShow Answer

Solve the problem. -On our moon, the acceleration of gravity is 1.6 m/sec21.6 \mathrm {~m} / \mathrm { sec } ^ { 2 }1.6 m/sec2 . If a rock is dropped into a crevasse, how fast will it be going just before it hits bottom 45 seconds later?

Correct AnswerverifiedShow Answer

Find the function with the given derivative whose graph passes through the point P. - r′(t) =sec⁡2t−4,P(0,0) r ^ { \prime } ( t ) = \sec ^ { 2 } t - 4 , P ( 0,0 ) r′(t) =sec2t−4,P(0,0)

Correct AnswerverifiedShow Answer

Find the absolute extreme values of the function on the interval. - f(x) =ln⁡(−x) ,−7≤x≤−1f ( x ) = \ln ( - x ) , - 7 \leq x \leq - 1f(x) =ln(−x) ,−7≤x≤−1

Correct AnswerverifiedShow Answer

Correct AnswerverifiedShow Answer

Find the absolute extreme values of the function on the interval. - f(x) =csc⁡x,−π2≤x≤3π2f ( x ) = \csc x , - \frac { \pi } { 2 } \leq x \leq \frac { 3 \pi } { 2 }f(x) =cscx,−2π​≤x≤23π​

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Provide an appropriate response. -Let f have a derivative on an interval I. f' has successive distinct zeros at x = 1 and x = 5. Prove that there can be at most one zero of f on the interval (1, 5).

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Find the function with the given derivative whose graph passes through the point P. - r′(θ) =4+sec⁡2θ,P(π,0) \mathrm { r } ^ { \prime } ( \theta ) = 4 + \sec ^ { 2 } \theta , \mathrm { P } ( \pi , 0 ) r′(θ) =4+sec2θ,P(π,0)

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Find all possible functions with the given derivative. - y′=32ty ^ { \prime } = \frac { 3 } { 2 \sqrt { t } }y′=2t​3​

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Find the absolute extreme values of the function on the interval. - F(x) =−3x2,0.5≤x≤3F ( x ) = - \frac { 3 } { x ^ { 2 } } , 0.5 \leq x \leq 3F(x) =−x23​,0.5≤x≤3

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Using the derivative of f(x) given below, determine the intervals on which f(x) is increasing or decreasing. - f′(x) =(x+5) 2e−xf ^ { \prime } ( x ) = ( x + 5 ) ^ { 2 } e ^ { - x }f′(x) =(x+5) 2e−x

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(a) The only critical point in the inter...
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Find the extreme values of the function and where they occur. - $y = ( x - 5 ) ^ { 2 / 3 }$

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Use the maximum/minimum finder on a graphing calculator to determine the approximate location of all local extrema. - $f ( x ) = 0.01 x ^ { 5 } - x ^ { 4 } + x ^ { 3 } + 8 x ^ { 2 } - 7 x + 94$

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Using the derivative of f(x) given below, determine the intervals on which f(x) is increasing or decreasing. - $f ^ { \prime } ( x ) = ( 8 - x ) ( 9 - x )$

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Find the function with the given derivative whose graph passes through the point P. - $\mathrm { r } ^ { \prime } ( \theta ) = 8 - \csc ^ { 2 } \theta , \mathrm { P } \left( \frac { \pi } { 4 } , 0 \right)$

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Identify the function's local and absolute extreme values, if any, saying where they occur. - $h ( x ) = \frac { x - 2 } { x ^ { 2 } + 3 x + 6 }$

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Find the extreme values of the function and where they occur. - $y = x ^ { 3 } - 12 x + 2$

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Find the derivative at each critical point and determine the local extreme values. - $y = \left\{ \begin{array} { l l } 4 - x , & x < 0 \\4 + 3 x - x ^ { 2 } , & x \geq 0\end{array} \right.$

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

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Solve the problem. -On our moon, the acceleration of gravity is $1.6 \mathrm {~m} / \mathrm { sec } ^ { 2 }$ . If a rock is dropped into a crevasse, how fast will it be going just before it hits bottom 45 seconds later?

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Find the function with the given derivative whose graph passes through the point P. - $r ^ { \prime } ( t ) = \sec ^ { 2 } t - 4 , P ( 0,0 )$

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Find the absolute extreme values of the function on the interval. - $f ( x ) = \ln ( - x ) , - 7 \leq x \leq - 1$

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Find the absolute extreme values of the function on the interval. - $f ( x ) = \csc x , - \frac { \pi } { 2 } \leq x \leq \frac { 3 \pi } { 2 }$

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Assume there are 2 (or more) zeros and ...
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Find the function with the given derivative whose graph passes through the point P. - $\mathrm { r } ^ { \prime } ( \theta ) = 4 + \sec ^ { 2 } \theta , \mathrm { P } ( \pi , 0 )$

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

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Find the absolute extreme values of the function on the interval. - $F ( x ) = - \frac { 3 } { x ^ { 2 } } , 0.5 \leq x \leq 3$

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Using the derivative of f(x) given below, determine the intervals on which f(x) is increasing or decreasing. - $f ^ { \prime } ( x ) = ( x + 5 ) ^ { 2 } e ^ { - x }$

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