Answer
$f'(x)=0$ for $x=-\dfrac{8}{3}.$
Work Step by Step
$f(x)=x\sqrt{x+4}\to$ The intercepts are $x=0$ and $x=-4.$
Applying Rolle's Theorem over the interval $[-4, 0]$ guarantees the existence of at least one value $c$ such that $-4\lt c\lt 0$ and $f'(c)=0.$
Product Rule $(f’(x)=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$
$u(x)=x ;u’(x)=1 $
$v(x)= \sqrt{x+4};v’(x)=\dfrac{1}{2\sqrt{x+4}} $
$f'(x)=(1)(\sqrt{x+4})+(x)(\dfrac{1}{2\sqrt{x+4}})=\dfrac{3x+8}{2\sqrt{x+4}}$
$f'(x)=0\to\dfrac{3x+8}{2\sqrt{x+4}}=0\to3x+8=0\to c=-\dfrac{8}{3}.$
