Answer
$c=\sqrt{ab}$
Work Step by Step
Zero is not in $[a,b]$, so f is continuous and differentiable over the interval.
$f(a)=\displaystyle \frac{1}{a},\qquad f(b)=\frac{1}{b},\qquad $
$f'(x)=-\displaystyle \frac{1}{x^{2}},\qquad $
The Mean Value Theorem tells us that there exists a $c\in(a,b)$ such that
$f'(c)=\displaystyle \frac{f(b)-f(a)}{b-a}$
$-\displaystyle \frac{1}{c^{2}}=\frac{\frac{1}{b}-\frac{1}{a}}{b-a}$
$-\displaystyle \frac{1}{c^{2}}=\frac{\frac{a-b}{ba}}{\frac{b-a}{1}}$
$-\displaystyle \frac{1}{c^{2}}=\frac{-\frac{b-a}{ba}}{\frac{b-a}{1}}$
$\displaystyle \frac{1}{c^{2}}=\frac{1}{ab}\quad \Rightarrow\quad c=\sqrt{ab}$
