Answer
(a) Critical number: $x=-3$
(b) Decreasing on: $(-\infty, -3)$
Increasing on: $(-3, \infty)$
(c) Relative minimum: $(-3,1)$
(d) See image
Work Step by Step
(a)
$f(x)=x^{2}+6x+10$
$f^{\prime}(x)=2x+6$
$2x+6=0$
$2x=-6$
Critical number: $x=-3$
(b)
$\left[\begin{array}{lll}
Interval & (-\infty, -3) & (3, \infty)\\
\text{test point} & -4 & 4\\
f^{\prime}(\text{test point}) & -2 & 14\\
\text{sign} & - & +\\
& \searrow & \nearrow
\end{array}\right]$
Decreasing on: $(-\infty, -3)$
Increasing on: $(-3, \infty)$
(c)
From the table in part b, f has a relative minimum at $x=-3$
$f(-3)=(-3)^{2}+6(-3)+10=1$
Relative minimum: $(-3,1)$
