Chapter 4: Applications of Derivatives - Section 4.2 - The Mean Value Theorem - Exercises 4.2 - Page 199: 55

See explanations.

Step 1. Given a function $f(x)=\frac{1}{x}$, we can find its derivative as $f'(x)=-\frac{1}{x^2}$ Step 2. In the positive interval of $[a,b]$, the Mean Value Theorem gives $f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{1/b-1/a}{b-a}=\frac{a-b}{ab(b-a)}=-\frac{1}{ab}$ Step 3. Combining the above results, we get $-\frac{1}{c^2}=-\frac{1}{ab}$ which gives $c^2=ab$ and $c=\sqrt {ab}$