## Calculus with Applications (10th Edition)

The absolute maximum of f(x) on [-3,1] is 29 at x=-3. The absolute minimum of f(x) on [-3,1] is -3 at $x=\pm1$.
$f'(x)=-6x^{2}-4x+2$ $f'(x)=0 \rightarrow -6x^{2}-4x+2=0 \rightarrow x=\frac{1}{3}, x=-1$ $f(\frac{1}{3})=\frac{-17}{27}$ $f(-1)=-3$ $f(-3)=29$ $f(1)=-3$ The absolute maximum of f(x) on [-3,1] is 29 at x=-3. The absolute minimum of f(x) on [-3,1] is -3 at $x=\pm1$.