Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 4: Applications of Derivatives - Section 4.1 - Extreme Values of Functions - Exercises 4.1 - Page 191: 38


Absolute minimum is $-1$ at $x=-1,$ absolute maximum is $32$ at $x=8$

Work Step by Step

To find absolute extrema of a continuous function f on a closed interval: 1. Evaluate $f$ at all critical points and endpoints. 2. Take the largest and smallest of these values. --- f is continuous on $[-1,8]$. Critical points: $f'(x)=0$ $\displaystyle \frac{5}{3}x^{2/3}=0$ $x=0 , \quad f(0)=0.$ Endpoints: $f(-1)=[(-1)^{5}]^{1/3}=(-1)^{1/3}=-1,$ $f(8)=(8^{1/3})^{5}=2^{5}=32$
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