Answer
(a)
Critical numbers: $x=0,4$
(b)
Increasing on: $(-\infty, 0), (4, \infty)$
Decreasing on: $(0,4)$
(c)
Relative maximum: $(0,15)$
Relative minimum: $(4, -17)$
(d)
See image
Work Step by Step
(a)
$f(x)=x^{3}-6x^{2}+15$
$f^{\prime}(x)=3x^{2}-12x=3x(x-4)$
$3x(x-4)=0$
Critical numbers: $x=0,4$
(b)
$\left[\begin{array}{llll}
Interval & (-\infty, 0) & (0,4) & (4, \infty)\\
\text{test point} & -1 & 1 & 5\\
f^{\prime}(\text{test point}) & 15 & -9 & 15\\
\text{sign} & + & - & +\\
& \nearrow & \searrow & \nearrow
\end{array}\right]$
Increasing on: $(-\infty, 0), (4, \infty)$
Decreasing on: $(0,4)$
(c)
From the table in part b,
f has a relative maximum at $x=0,\qquad f(0)=15$
f has a relative minimum at $x=4,\qquad f(1)=-17$
Relative maximum: $(0,15)$
Relative minimum: $(4, -17)$
