Answer
The absolute maximum of f(x) on [-1,6] is -3 at x=0. The absolute minimum of f(x) on [1,6] is -16 at x=0 and x=-1.
Work Step by Step
$f'(x)=12x^{2}-9x$
$f'(x)=0 \rightarrow -3x^{2}+12x=0 \rightarrow x=\frac{3}{4}, x=0 $
$f(\frac{3}{4})=\frac{-51}{8}$
$f(0)=-3$
$f(-1)=-16$
$f(2)=-7$
The absolute maximum of f(x) on [-1,6] is -3 at x=0. The absolute minimum of f(x) on [1,6] is -16 at x=0 and x=-1
