Answer
Rolle's Theorem can be applied, $c=2.$
Work Step by Step
Since $f(x)$ is a polynomial, it is continuous for all values of $x$ and differentiable at every value of $x$.
Since $f(x)$ is continuous over $[ , ]$ and differentiable over $(, )$, applying Rolle's Theorem over the interval $[, ]$ guarantees the existence of at least one value $c$ such that $\lt c\lt $ and $f'(c)=0.$
Product Rule $(f’(x)=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$
$u(x)=(x-4) ;u’(x)= 1$
$v(x)=(x+2)^2 ;v’(x)=2(x+2) $
$f'(x)=(1)(x+2)^2+(x-4)(2x+4)$
$=3x^2-12=3(x-2)(x+2).$
$f'(x)=0\to 3(x-2)(x+2)=0\to c=2$ or $c=-2$( which is rejected since $c\gt-2.$
