Answer
$f$ satisfies the hypotheses of the Mean Value Theorem on the interval $[1,3]$
On the interval $(1, 3)$, $f'(c) = -\frac{1}{3}$ when $c = \sqrt{3}$
Work Step by Step
$f(x) = \frac{1}{x}$
1. The function is continuous on the interval $(0, \infty)$. Therefore, the function is continuous on the closed interval $[1,3]$
2. $f'(x) = -\frac{1}{x^2}$ which exists for all numbers in the interval $(0, \infty)$. Therefore, $f'(x)$ exists for all numbers on the interval $(1, 3)$.
Thus $f$ is differentiable on the open interval $(1, 3)$
$f$ satisfies the hypotheses of the Mean Value Theorem on the interval $[1,3]$
Therefore, according to the Mean Value Theorem Theorem, there is a number $c$ in the interval $(1,3)$ such that $f'(c) = \frac{f(3)-f(1)}{3-1}$
$\frac{f(3)-f(1)}{3-1} = \frac{\frac{1}{3}-\frac{1}{1}}{2} = -\frac{1}{3}$
We can find $c$:
$f'(x) = -\frac{1}{x^2} = -\frac{1}{3}$
$x^2=3$
$x = -\sqrt{3}~~$ or $~~x = \sqrt{3}$
On the interval $(1, 3)$, $f'(c) = -\frac{1}{3}$ when $c = \sqrt{3}$
