Answer
$f'(x)=0$ for $x=-\dfrac{2}{3}.$
Work Step by Step
$f(x)=-3x\sqrt{x+1}\to$ The intercepts are $x=-1$ and $x=0.$
Applying Rolle's Theorem over the interval $[-1, 0]$ guarantees the existence of at least one value $c$ such that $-1\lt c\lt 0$ and $f'(c)=0.$
Product Rule $(f’(x)=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$
$u(x)=-3x ;u’(x)= -3$
$v(x)=\sqrt{x+1} ;v’(x)= \dfrac{1}{2\sqrt{x+1}}$
$f'(x)=(-3)(\sqrt{x+1})+(-3x)(\dfrac{1}{2\sqrt{x+1}})=\dfrac{-9x-6}{2\sqrt{x+1}}.$
$f'(x)=0\to \dfrac{-9x-6}{2\sqrt{x+1}}=0\to-9x-6=0\to c=-\dfrac{2}{3}.$
