Answer
$f'(x)=0$ for $x=-3.$
Work Step by Step
$f(x)=x^2+6x=x(x+6)\to$ The intercepts are $x=0$ and $x=-6.$
Applying Rolle's Theorem over the interval $[-6, 0]$ guarantees the existence of at least one value $c$ such that $-6\lt c\lt 0$ and $f'(c)=0.$
$f'(x)=2x+6\to f'(x)=0\to 2x+6=0\to c=-3.$
