## Calculus: Early Transcendentals 8th Edition

$f$ satisfies the hypotheses of the Mean Value Theorem on the interval $[1,3]$ On the interval $(1, 3)$, $f'(c) = -\frac{1}{3}$ when $c = \sqrt{3}$
$f(x) = \frac{1}{x}$ 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 $c = \sqrt{3}$