Answer
$$
f(x)=(x+3)^{4}
$$
Critical number: $-3$.
the second derivative test gives no information about extrema, by using the first derivative test, we find that:
$-3$ leads to a relative minimum.
Work Step by Step
$$
f(x)=(x+3)^{4}
$$
First, find the points where the derivative is $0$
Here
$$
\begin{aligned}
f^{\prime}(x) &=(4)(1)(x+3)^{3}\\
&=4(x+3)^{3}\\
\end{aligned}
$$
Solve the equation $f^{\prime}(x)=0 $ to get
$$
\begin{aligned}
f^{\prime}(x) =4(x+3)^{3} &=0\\
(x+3) &=0\\
\Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
x =-3
\end{aligned}
$$
Critical number: $-3$.
Now use the second derivative test. The second derivative is
$$
f^{\prime\prime}(x)=4(3)(x+3)^{2}=12(x+3)^{2}.
$$
Evaluate $f^{\prime\prime}(x)$ at $-3$, getting
$$
f^{\prime\prime}(-3)=12((-3)+3)^{2}=0 .
$$
then the test gives no information about extrema, so we can use the first derivative test.
Check the sign of $f^{\prime}(x)$ in the intervals
$$
(-\infty, -3 ), \quad (-3,\infty).
$$
(1)
Test a number in the interval $(-\infty, -3 )$ say $ -4$
$$
\begin{aligned}
f^{\prime}(-4) =4((-4)+3)^{3} \\
&=4(-1)^{3} \\
&=-4 \\
&\lt 0
\end{aligned}
$$
to see that $ f^{\prime}(x)$ is negative in that interval, so $f(x)$ is decreasing on $(-\infty, -3 )$
(2)
Test a number in the interval $(-3, \infty )$ say $0$
$$
\begin{aligned}
f^{\prime}(0) =4((0)+3)^{3} \\
&=4(3)^{3} \\
&=108 \\
&\gt 0
\end{aligned}
$$
to see that $ f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(-3, \infty ).$
so that by the first derivative test, $-3$ leads to a relative minimum.
