Answer
$f$ satisfies the three hypotheses of Rolle's Theorem on the interval $[\frac{1}{2}, 2]$
$f'(c) = 0$ when $c = 1$
Work Step by Step
1. The function $g(x) = x$ is continuous on the closed interval $[\frac{1}{2},2]$
The function $h(x) = \frac{1}{x}$ is continuous on the closed interval $[\frac{1}{2},2]$
Then $f = x + \frac{1}{x}$ is continuous on the closed interval $[\frac{1}{2},2]$
2. $f'(x) = 1-\frac{1}{x^2}$ which exists for all numbers in the interval $(\frac{1}{2}, 1)$
Then $f$ is differentiable on the open interval $(\frac{1}{2}, 1)$
3. $f(\frac{1}{2}) = \frac{1}{2}+2 = \frac{5}{2}$
$f(2) = 2+\frac{1}{2} = \frac{5}{2}$
Thus $f(\frac{1}{2}) = f(2)$
$f$ satisfies the three hypotheses of Rolle's Theorem on the interval $[\frac{1}{2}, 2]$
Therefore, according to Rolle's Theorem, there is a number $c$ in the interval $(\frac{1}{2},2)$ such that $f'(c) = 0$
We can find $c$:
$f'(x) = 1-\frac{1}{x^2} = 0$
$1 = \frac{1}{x^2}$
$x^2 = 1$
$x = -1$ or $x=1$
On the interval $(\frac{1}{2}, 2)$, $f'(c) = 0$ when $c = 1$
