Chapter 4 - Applications of the Derivative - 4.2 Extreme Values - Exercises - Page 181: 57

$f(\frac{1}{2})=f(2)=\frac{5}{2}$ and c=1

$f(\frac{1}{2})= \frac{1}{2}+\frac{2}{1}=\frac{5}{2}$ $f(2)=2+\frac{1}{2}= \frac{5}{2}$ We have $f(\frac{1}{2})=f(2)=\frac{5}{2}$ f'(x)= $1-\frac{1}{x^{2}}$ Now let c be a point in $(\frac{1}{2},2)$ and let f'(c) be 0. That is, $1-\frac{1}{c^{2}}=0$ ⇒ 1=$\frac{1}{c^{2}}$ or $c^{2}=1$ ⇒ c= 1.