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: 4

Answer

$c= 3\sqrt{2}-2 $

Work Step by Step

Given $$y=\frac{x}{x+2}, \quad[1,4]$$ Since $$f'(x)=\frac{\frac{d}{dx}\left(x\right)\left(x+2\right)-\frac{d}{dx}\left(x+2\right)x}{\left(x+2\right)^2}= \frac{2}{\left(x+2\right)^2} $$ Then by MVT, there exists a constant $c\in (1,4)$ such that \begin{align*} f'(c)&= \frac{f(b)-f(a) }{b-a}\\ \frac{2}{(c+2)^2}&=\frac{2/3-1/3}{3}\\ \frac{2}{(c+2)^2}&=\frac{1}{9}\\ (c+2)^2&= 18\\ c+2&=\pm \sqrt{18}\\ c&= \pm 3\sqrt{2}-2 \end{align*} Choose $c= 3\sqrt{2}-2\in (1,4)$
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