Answer
$f'(x)=0$ for $x=\frac{1}{2}.$
Work Step by Step
$f(x)=x^2-x-2=(x-2)(x+1)\to$ The intercepts are $x=2$ and $x=-1.$
Applying Rolle's Theorem over the interval $[-1, 2]$ guarantees the existence of at least one value $c$ such that $-1\lt c\lt 2$ and $f'(c)=0.$
$f'(x)=2x-1\to f'(x)=0\to 2x-1=0\to c=\frac{1}{2}.$
