## Calculus with Applications (10th Edition)

$f'(x)=3x^{2}+4x-15$ $f'(x)=0 \rightarrow 3x^{2}+4x-15=0 \rightarrow x=\frac{5}{3}, x=-3$ $f(\frac{5}{3})=\frac{-319}{27}$ $f(-3)=39$ $f(-4)=31$ $f(2)=-11$ 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.