#### Answer

$c=1$

#### Work Step by Step

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 $c=1$