Question

Calculus help! Using the mean value theorem!

Answer 1

$f$ is differentiable across its domain, so the MVT says there is some value of $c$ in the open interval $(a,b)$ such that

$f'(c)=\dfrac{f(b)-f(a)}{b-a}$

You have

$f(x)=Ax^2+Bx+C\implies f'(x)=2Ax+B$

so the equation above becomes

$2Ac+B=\dfrac{(Ab^2+Bb+C)-(Aa^2+Ba+C)}{b-a}$

Solve for $c$.

$2Ac+B=\dfrac{A(b^2-a^2)+B(b-a)}{b-a}$
$2Ac+B=A(b+a)+B$
$2Ac=A(b+a)$
$c=\dfrac{b+a}2$

so $c$ is indeed the average of the endpoints, i.e. the midpoint.