## Calculus 10th Edition

Over the specified interval, the function has an absolute maximum equal to $2$ and an absolute minimum equal to $-\frac{5}{2}.$
$f'(x)=3x^2-3x=3x(x-1).$ $f'(x)$ 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 $-\frac{5}{2}.$