Answer
The absolute maximum of f(x) on [-4,2] is 39 at x=-3. The absolute minimum of f(x) on [-4,2] is -319/27 at x=5/3
Work Step by Step
$f'(x)=3x^{2}+4x-15$
$f'(x)=0 \rightarrow 3x^{2}+4x-15=0 \rightarrow x=\frac{5}{3}, x=-3 $
$f(\frac{5}{3})=\frac{-319}{27}$
$f(-3)=39$
$f(-4)=31$
$f(2)=-11$
The absolute maximum of f(x) on [-4,2] is 39 at x=-3. The absolute minimum of f(x) on [-4,2] is -319/27 at x=5/3.
