## Calculus 10th Edition

Over the specified interval, the function has an absolute maximum equal to $36$ and an absolute minimum equal to $-4.$
$f'(x)=6x^2-6=6(x-1)(x+1).$ $f'(x)$ is defined for all x in the interval. $f'(x)=0\to x=1$ or $x=-1$ but only $x=1$ is in the specified interval. The function could attain an absolute extremum at $x=1$ or at the interval endpoints. $f(0)=0.$ $f(1)=-4.$ $f(3)=36.$ Over the specified interval, the function has an absolute maximum equal to $36$ and an absolute minimum equal to $-4.$