# Chapter 3 - Applications of Differentiation - 3.2 The Mean Value Theorem - 3.2 Exercises - Page 220: 8

$c=1$

#### Work Step by Step

Here, $f(x)=x+\frac1x$ in interval $[\frac12,2]$. 1. $f(x)$ is continuous over the given interval $[\frac12,2]$ & not defined for $x=0$. 2. Similarly, $f(x)$ is differentiable on the open interval $(\frac12,2)$. 3. $f(\frac12)=\frac12+\frac21=\frac52$. Similarly,$f(2)=2+\frac12=\frac52$. Thus, $f(\frac12)=f(2)$. Thus, all three criteria of Rolle's theorem satisfied. This implies that there exists a number $c$ in the open interval $(\frac12,2)$ such that $f'(c)=0$. Let's differentiate $f(x)$, $$f'(x)=1-\frac{1}{x^2}\implies1-\frac{1}{c^2}=0$$ $$c=-1,1$$ Thus, $c=1$ satisfies the required condition.

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