Answer
Over the specified interval, the function has an absolute maximum equal to $2$ and an absolute minimum equal to $-\frac{5}{2}.$
Work Step by Step
$f'(x)=3x^2-3x=3x(x-1).$
$f'(x)$ is defined for all x in the interval.
$f'(x)=0\to x=0$ or $x=1.$ Since both are in the specified interval, they, along with the endpoints of the interval, are possible candidates for absolute extrema.
$f(-1)=-\frac{5}{2}.$
$f(0)=0.$
$f(1)=-\frac{1}{2}.$
$f(2)=2.$
Over the specified interval, the function has an absolute maximum equal to $2$ and an absolute minimum equal to $-\frac{5}{2}.$
