Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 3 - Applications of Differentiation - 3.1 Exercises: 24

Answer

Over the specified interval, the function has an absolute maximum equal to $2$ and an absolute minimum equal to $-2$.

Work Step by Step

$g'(x)=\dfrac{1}{3\sqrt[3]{x^2}}$ $g'(x)$ is never equal to $0$ but is undefined for $x=0.$ Hence, the endpoints of the interval, along with the critical number $x=0,$ are possible points for absolute extrema. $g(-8)=-2.$ $g(0)=0.$ $g(8)=2.$ Over the specified interval, the function has an absolute maximum equal to $2$ and an absolute minimum equal to $-2$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.