Answer
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 $x=\pm1$.
Work Step by Step
$f'(x)=-6x^{2}-4x+2$
$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 $x=\pm1$.
