Answer
$c=\sqrt{3}$.
Work Step by Step
The given function $f=\frac{1}{x}$ is defined and continuous on $(-\infty,0) \cup (0,+\infty)$.
Since $[1,3]$ is included into the interval $(0,+\infty)$ it follows that $f$ is continuous on$[1,3]$.
The first derivative of $f$ is:
$$f'(x)=\left(\frac{1}{x}\right)'=\frac{(1)'x-1(x)'}{x^{2}}=\frac{-1}{x^{2}}$$
The derivative of $f$ is defined also on $(-\infty,0) \cup (0,+\infty)$, so since $[1,3]$ is included into the interval $(0,+\infty)$ it follows that $f$ is differentiable on $(1,3)$.
Therefore, the mean value theorem is satisfied.
There is $c \in (1,3)$ such that:
$$f'(c)=\frac{f(3)-f(1)}{3-1}$$
$$\frac{-1}{c^{2}}=\frac{\frac{1}{3}-1}{3-1}$$
$$\frac{-1}{c^{2}}=\frac{\frac{-2}{3}}{2}$$
$$\frac{-1}{c^{2}}=\frac{-2}{6}$$
$$\frac{-1}{c^{2}}=\frac{-1}{3}$$
$$c^{2}=3$$
$$c=\sqrt{3},~~\text{or}~~c=-\sqrt{3}$$
Since $-\sqrt{3}$ does not lie in the interval $(1,3)$ so the value of $c$ is $\sqrt{3}$.
