University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 4 - Section 4.2 - The Mean Value Theorem - Exercises - Page 222: 2



Work Step by Step

Here, the function $f(x)=x^{\frac{2}{3}}$ is continuous on $[0,1]$ and differentiable on $(0,1)$ when $f'(x)=\dfrac{2}{3}x^{\frac{-1}{3}}$ . The Mean value Theorem states that there is a point $c \in (0,1)$ such that $f'(c)=\dfrac{f(1)-f(0)}{1-0}=\dfrac{1-0}{1-0}=1$ Then $f'(c)=\dfrac{2}{3}c^{\frac{-1}{3}}=1$ This implies $c=\dfrac{8}{27}$
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.