Answer
The absolute maximum of f(x) on [-1,6] is 33 at x=4. The absolute minimum of f(x) on [1,6] is 1 at x=0 and x=6.
Work Step by Step
$f'(x)=-3x^{2}+12x$
$f'(x)=0 \rightarrow -3x^{2}+12x=0 \rightarrow x=4, x=0 $
$f(4)=33$
$f(0)=1$
$f(-1)=8$
$f(6)=1$
The absolute maximum of f(x) on [-1,6] is 33 at x=4. The absolute minimum of f(x) on [1,6] is 1 at x=0 and x=6.
