Answer
Over the specified interval, the function has an absolute maximum equal to $36$ and an absolute minimum equal to $-4.$
Work Step by Step
$f'(x)=6x^2-6=6(x-1)(x+1).$
$f'(x)$ 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 $-4.$
