Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 4 - Applications of the Derivative - 4.3 The Mean Value Theorem and Monotonicity - Exercises - Page 189: 2


c= 16.

Work Step by Step

$f(9)=\sqrt 9= 3$ $f(25)= \sqrt {25}=5$ $\frac{f(25)-f(9)}{25-9}=\frac{2}{16}=\frac{1}{8}$ MVT states that there is a point c$\in$(9,25) such that $f'(c)=\frac{1}{8}$. But $f'(x)= \frac{1}{2\sqrt x}$ which implies c=16. Thus at c=16$\,\in$(9,25), we have $f'(c)=\frac{1}{8}.$
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