Calculus 10th Edition

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

Chapter 3 - Differentiation - 3.2 Exercises: 5

Answer

$f'(x)=0$ for $x=\frac{1}{2}.$

Work Step by Step

$f(x)=x^2-x-2=(x-2)(x+1)\to$ The intercepts are $x=2$ and $x=-1.$ Applying Rolle's Theorem over the interval $[-1, 2]$ guarantees the existence of at least one value $c$ such that $-1\lt c\lt 2$ and $f'(c)=0.$ $f'(x)=2x-1\to f'(x)=0\to 2x-1=0\to c=\frac{1}{2}.$
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.