## Calculus with Applications (10th Edition)

$f'(x)=-3x^{2}+12x$ $f'(x)=0 \rightarrow -3x^{2}+12x=0 \rightarrow x=4, x=0$ $f(4)=33$ $f(0)=1$ $f(-1)=8$ $f(6)=1$ 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.