## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 4: Applications of Derivatives - Section 4.2 - The Mean Value Theorem - Exercises 4.2 - Page 197: 4

#### Answer

$c=\dfrac{3}{2}$

#### Work Step by Step

Since, the given function $f(x)=(\sqrt {x-1})$ is continuous on $[1,3]$ and differentiable on $(1,3)$ and $f'(x)=\dfrac{1}{2\sqrt{x-1}}$ . The Mean value Theorem states that there is a point $c$ and $c \in (1,3)$. This means that $f'(c)=\dfrac{f(3)-f(1)}{(3-1)}=\dfrac{\sqrt {3-1}-\sqrt {1-1}}{(3-1)}=\dfrac{\sqrt 2}{2}$ This implies that $f'(c)=\dfrac{1}{2\sqrt{c-1}}=\dfrac{\sqrt 2}{2}$ Thus, $c=\dfrac{3}{2}$

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