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: 1

Answer

c= 4

Work Step by Step

$f'(c) = -\frac{1}{c^{2}}$ MVT states that, if the interval is [a,b] $f'(c)=\frac{f(b)-f(a)}{b-a}$. Then in this case $f'(c)= \frac{f(8)-f(2)}{8-2}=\frac{\frac{1}{8}-\frac{1}{2}}{6}= -\frac{1}{16}$ ⇒ $-\frac{1}{c^{2}}= -\frac{1}{16}$ ⇒ $c^{2}= 16$ or c= 4 $\epsilon$ (2,8) satisfying mean value theorem.
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