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: 23

Answer

Over the specified interval, the function has an absolute maximum equal to $5$ and an absolute minimum equal to $0$

Work Step by Step

$y'=\dfrac{2}{\sqrt[3]{x}}-2=\dfrac{2-2\sqrt[3]{x}}{\sqrt[3]{x}}.$ $y'=0\to2-2\sqrt[3]{x}=0\to x=1.$ $y'$is undefined $\to \sqrt[3]{x}=0\to x=0.$ Since both values are in the specified interval, both, along with the interval endpoints, are possible candidates for absolute extrema. $f(-1)=5.$ $f(0)=0.$ $f(1)=1.$ Over the specified interval, the function has an absolute maximum equal to $5$ and an absolute minimum equal to $0$.
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.