Answer
(a)
Critical numbers: $x=-2,0$
(b)
Increasing on: $(-\infty, -2), (0, \infty)$
Decreasing on: $(-2,0)$
(c)
Relative maximum: $(-2,0)$
Relative minimum: $(0, -4)$
(d)
See image
Work Step by Step
(a)
$f(x)=(x+2)^{2}(x-1)$
$f^{\prime}(x)=2(x+2)(x-1)+(x+2)^{2}(1)$
$=(x+2)[2(x-1)+(x+2)]$
$=(x+2)[2x-2+x+2]$
$=3x(x+2)$
$3x(x+2)=0$
Critical numbers: $x=-2,0$
(b)
$\left[\begin{array}{llll}
Interval & (-\infty, -2) & (-2,0) & (0, \infty)\\
\text{test point} & -3 & -1 & 1\\
f^{\prime}(\text{test point}) & 9 & -3 & 9\\
\text{sign} & + & - & +\\
& \nearrow & \searrow & \nearrow
\end{array}\right] $
Increasing on: $(-\infty, -2), (0, \infty)$
Decreasing on: $(-2,0)$
(c)
From the table in part b,
f has a relative maximum at $x=-2,\qquad f(-2)=0$
f has a relative minimum at $x=0,\qquad f(0)=-4$
Relative maximum: $(-2,0)$
Relative minimum: $(0, -4)$
