Answer
Rolle's theorem can be applied; $c=4.$
Work Step by Step
Since $f(x)$ is a polynomial, it is continuous for all values of $x$ and differentiable at every value of $x$.
$f(2)=f(6)=-7.$
Since $f(x)$ is continuous over $[2 , 6]$ and differentiable over $(2, 6)$, applying Rolle's Theorem over the interval $[2, 6]$ guarantees the existence of at least one value $c$ such that $2\lt c\lt6 $ and $f'(c)=0.$
$f'(x)=2x-8\to f'(x)=0\to 2x-8=0\to c=4.$
