Answer
$$c =\frac{a+b}{2}$$
Work Step by Step
Consider the quadratic equation
$$f(x)=p x^{2}+q x+r$$
Then by using the MVT,
\begin{aligned}
f^{\prime}(c) &=\frac{f(b)-f(a)}{b-a} \\
&=\frac{\left(p b^{2}+q b+r\right)-\left(p a^{2}+q a+r\right)}{b-a} \\
&=\frac{p\left(b^{2}-a^{2}\right)+q(b-a)}{b-a} \\
&=\frac{p(b-a)(b+a)+q(b-a)}{b-a} \\
&=p(b+a)+q
\end{aligned}
On the other hand, we have
$$f'(x) =2px+q $$
Then
\begin{align*}
f^{\prime}(c)&=p(b+a)+q=2 p c+q\\
b+a&=2 c\\
c&=\frac{a+b}{2}
\end{align*}
