University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 4 - Section 4.2 - The Mean Value Theorem - Exercises - Page 222: 3



Work Step by Step

Here, the function $f(x)=x+\dfrac{1}{x}$ is continuous on $[\dfrac{1}{2},2]$ and differentiable on $(\dfrac{1}{2},2)$ when $f'(x)=1-\dfrac{1}{x^2}$ . The Mean value Theorem states that there is a point $c \in (\dfrac{1}{2},2)$ such that $f'(c)=\dfrac{f(2)-f(\dfrac{1}{2})}{2-\dfrac{1}{2}}=\dfrac{(2+\dfrac{1}{2})-(\dfrac{1}{2}+2)}{2-\dfrac{1}{2}}=0$ Then $f'(c)=1-\dfrac{1}{c^2}=0$ This implies $c=1$
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