Answer
Absolute maximum: $(-4,32)$
Absolute minimum: $(-6,0)$ and $(0,0)$
Work Step by Step
$ f(x)=x^{3}+6x^{2} \quad$ ... defined everywhere
$ f^{\prime}(x)=3x^{2}+12x\quad$ ... defined everywhere.
$f^{\prime}(x)=0 \quad$ for
$3x^{2}+12x=0$
$3x(x+4)=0$
$x=-4,0\in[-6,1]\quad $(critical numbers)
$\left[\begin{array}{lccccccc}
x, \text{ intervals} & -6 & (-6,-4) & -4 & (-4,0) & 0 & (0,1) & 1\\
t=\text{ test number} & & -5 & & -1 & & 1/2 & \\
f^{\prime}(t) & & +75+60 & & -3-12 & & \frac{3}{4}+6 & \\
f(x) & 0 & \nearrow & +32 & \searrow & 0 & \nearrow & 7\\
& min & & max & & min & &
\end{array}\right]$
