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 - 3.2 Exercises: 44

Answer

The Mean Value Theorem can be applied; $c=-\dfrac{1}{4}.$

Work Step by Step

$f(x)$ is continuous for all values of $x\le2$ and differentiable for at all values of $x\lt2/.$ $a=-7; b=2\to f(a)=3$ and $f(b)=0.$ Since $f(x)$ is continuous over $[a, b]$ and differentiable over $(a, b)$, applying Mean Value Theorem over the interval $[a , b]$ guarantees the existence of at least one value c such that $a\lt c\lt b$ and $f′(c)=\dfrac{f(b)-f(a)}{b-a}.$ $f'(c)=\dfrac{0-3}{2-(-7)}=-\dfrac{1}{3}.$ $f'(x)=-\dfrac{1}{2\sqrt{2-x}}\to f'(c)=-\dfrac{1}{3}\to -\dfrac{1}{2\sqrt{2-c}}=-\dfrac{1}{3}\to$ $\sqrt{2-c}=\dfrac{3}{2}\to c=-\dfrac{1}{4}.$
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.