Answer
$c=1$
Work Step by Step
The Mean value Theorem states that let there is a point $c$ and $c \in (\dfrac{1}{2},2)$
This means that $f'(c)=\dfrac{f(2)-f(\dfrac{1}{2})}{2-\dfrac{1}{2}}=\dfrac{(2+\dfrac{1}{2})-(\dfrac{1}{2}+2)}{2-\dfrac{1}{2}}=0$
Since, the given function $f(x)=(x+\dfrac{1}{x})$ is continuous on $[\dfrac{1}{2},2]$ and differentiable on $(\dfrac{1}{2},2)$ and $f'(x)=1-\dfrac{1}{x^2}$ .
This implies that $f'(c)=1-\dfrac{1}{c^2}=0$
Thus, $c=1$
