Chapter 4 - Section 4.2 - The Mean Value Theorem - Exercises - Page 222: 3

$c=1$

Here, the function $f(x)=x+\dfrac{1}{x}$ is continuous on $[\dfrac{1}{2},2]$ and differentiable on $(\dfrac{1}{2},2)$ when $f'(x)=1-\dfrac{1}{x^2}$ . The Mean value Theorem states that there is a point $c \in (\dfrac{1}{2},2)$ such that $f'(c)=\dfrac{f(2)-f(\dfrac{1}{2})}{2-\dfrac{1}{2}}=\dfrac{(2+\dfrac{1}{2})-(\dfrac{1}{2}+2)}{2-\dfrac{1}{2}}=0$ Then $f'(c)=1-\dfrac{1}{c^2}=0$ This implies $c=1$