Explanation Given: The function is: f ( x ) = − 2 x x 2 + 1 . Formula used: Write the formula for derivative of power rule. If f ( x ) = x n for any number n , then the derivative. f ′ ( x ) = n x n − 1 If f ( x ) = k for any real number k , then the derivative is defined by the formula. f ′ ( x ) = 0 If the function f ( x ) = u ( x ) v ( x ) , if all indicated derivatives exist, and v ( x ) ≠ 0 , then the derivative of the function f ( x ) by division rule is, f ′ ( x ) = v ( x ) ⋅ u ′ ( x ) − u ( x ) ⋅ v ′ ( x ) ( v ( x ) ) 2 Here, v ( x ) is the function of x and v ′ ( x ) is the derivative of the function v ( x ) . To find local maximum and minimum find the values of x from the equation f ′ ( x ) = 0 . Then find f ″ ( x ) and put the values of x . If f ″ ( x ) equals some positive quantity then it implies Local minimum is located at point x and if f ″ ( x ) equals some negative quantity then Local maximum is located at that point. Calculation: Consider the given function: f ( x ) = x x 2 + 4 …… (1) Differentiate f ( x ) with respect to x . d d x ( f ( x ) ) = d d x ( − 2 x x 2 + 1 ) f ′ ( x ) = ( x 2 + 1 ) d d x ( − 2 x ) − ( − 2 x ) ( d d x ( x 2 + 1 ) ) ( x 2 + 1 ) 2 f ′ (

MATH W/APPLICAT.W/NOTES GDE +ACCESS CODE

11th Edition

ISBN: 9781323751671

Author: Lial

Publisher: PEARSON C

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Chapter 12.3, Problem 14E

To determine

To calculate: The locations of absolute extrema of the function given below on the given interval:

f(x)=−2xx2+1 ;[0,3]

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Chapter 12.3, Problem 13E

Chapter 12.3, Problem 15E

Chapter 12 Solutions

MATH W/APPLICAT.W/NOTES GDE +ACCESS CODE

Ch. 12.1 - Checkpoint 1 For what values of x is the function...Ch. 12.1 - Checkpoint 2 Find all intervals on which is...Ch. 12.1 - Checkpoint 3 Identity the x-values of all points...Ch. 12.1 - Checkpoint 4 Find the critical numbers for each of...Ch. 12.1 - Prob. 5CP Ch. 12.1 - Prob. 6CP Ch. 12.1 - Checkpoint 7 Find the locations of the local...Ch. 12.1 - Prob. 8CP Ch. 12.1 - Checkpoint 9 If a sales function is given by...Ch. 12.1 - Prob. 1E

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The locations of absolute extrema of the function given below on the given interval: f ( x ) = − 2 x x 2 + 1 ; [ 0 , 3 ]

Solution Summary: The author explains that the function f(x) has an absolute maxima and a minima.

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To calculate: The locations of absolute extrema of the function given below on the given interval:

f(x)=−2xx2+1 ;[0,3]

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Chapter 12 Solutions

The locations of absolute extrema of the function given below on the given interval: f ( x ) = − 2 x x 2 + 1 ; [ 0 , 3 ]

Solution Summary: The author explains that the function f(x) has an absolute maxima and a minima.

Concept explainers

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To calculate: The locations of absolute extrema of the function given below on the given interval: f(x)=−2xx2+1 ;[0,3]

Want to see the full answer?

Chapter 12 Solutions

To calculate: The locations of absolute extrema of the function given below on the given interval:

f(x)=−2xx2+1 ;[0,3]