Answer
$c=\displaystyle \frac{a+b}{2}$
Work Step by Step
$f9x)=x^{2}$
is continuous and differentiable over the interval $[a,b]$.
$f(a)= a^{2},\qquad f(b)=b^{2},\qquad $
$f'(x)=2x,\qquad $
The Mean Value Theorem tells us that there exists a $c\in(a,b)$ such that
$f'(c)=\displaystyle \frac{f(b)-f(a)}{b-a}$
$2c=\displaystyle \frac{b^{2}-a^{2}}{b-a}$
$2c=\displaystyle \frac{(b-a)(b+a)}{(b-a)}$
$c=\displaystyle \frac{a+b}{2}$