Answer
(a)
Critical numbers: $x=-2,1$
(b)
Increasing on: $(-\infty, -2), (1, \infty)$
Decreasing on: $(-2,1)$
(c)
Relative maximum: $(-2,20)$
Relative minimum: $(1, -7)$
(d)
See image
Work Step by Step
(a) $f(x)=2x^{3}+3x^{2}-12x$
$f^{\prime}(x)=6x^{2}+6x-12=6(x+2)(x-1)$
$6(x+2)(x-1)=0$
Critical numbers: $x=-2,1$
(b)
$\left[\begin{array}{llll}
Interval & (-\infty, -2) & (-2,1) & (1, \infty)\\
\text{test point} & -3 & 0 & 2\\
f^{\prime}(\text{test point}) & 24 & -12 & 24\\
\text{sign} & + & - & +\\
& \nearrow & \searrow & \nearrow
\end{array}\right]$
Increasing on: $(-\infty, -2), (1, \infty)$
Decreasing on: $(-2,1)$
(c)
From the table in part b,
f has a relative maximum at $x=-2,\qquad f(-2)=20$
f has a relative minimum at $x=-1,\qquad f(1)=-7$
