Explanation Given: The function is given as: f ( x ) = − 2 x 3 − x 2 + 4 x − 1 , in the interval [ − 3 , 1 ] . Formula used: The n th derivative formula is, d d x ( x n ) = n ⋅ x n − 1 Calculation: By the use of the extreme-value theorem, consider only the critical points of f ( x ) and the endpoints which are − 3 and 1. Find the derivative and determine the critical points. Equate the equation to zero to find the critical points. The derivative of the function is given as: f ' ( x ) = − 6 x 2 − 2 x + 4 Equate the above equation to zero. − 6 x 2 − 2 x + 4 = 0 3 x 2 + x − 2 = 0 3 x 2

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ISBN: 9780135904282

Author: Lial, HUNGERFO

Publisher: PEARSON

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Chapter 12, Problem 36RE

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To calculate: The location and values of all absolute maxima and absolute minima for given function f(x)=−2x3−x2+4x−1 in provided interval [−3,1].

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Chapter 12, Problem 35RE

Chapter 12, Problem 37RE

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Chapter 12 Solutions

MYMATHLAB ACCESS CODE - MATHEMATICS WIT

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 location and values of all absolute maxima and absolute minima for given function f ( x ) = − 2 x 3 − x 2 + 4 x − 1 in provided interval [ − 3 , 1 ] .

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To calculate: The location and values of all absolute maxima and absolute minima for given function f(x)=−2x3−x2+4x−1 in provided interval [−3,1].

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Chapter 12 Solutions