## Calculus (3rd Edition)

$c=2$
MVT states that $f'(c)=\frac{f(b)-f(a)}{b-a}$ Then, $2c-4=\frac{[(3-1)(3-3)]-[(1-1)(1-3)]}{3-1}$ $\implies 2c-4=0$ $\implies 2c=4$ $\implies c=2$ $c=2$ lies in the interval $(1,3)$ and is a point that satisfies the conclusion of the MVT in the given function.