Answer
$c=\dfrac{3}{2}$
Work Step by Step
Here, the function $f(x)=\sqrt {x-1}$ is continuous on $[1,3]$ and differentiable on $(1,3)$ when $f'(x)=\dfrac{1}{2\sqrt{x-1}}$ .
The Mean value Theorem states that there is a point $c \in (1,3)$
such that $f'(c)=\dfrac{f(3)-f(1)}{3-1}=\dfrac{\sqrt {3-1}-\sqrt {1-1}}{3-1}=\dfrac{\sqrt 2}{2}$
Then $f'(c)=\dfrac{1}{2\sqrt{c-1}}=\dfrac{\sqrt 2}{2}$
This implies $c=\dfrac{3}{2}$
