Answer
$$
f(x)=x^{3}
$$
Critical number: $0$.
the second derivative test gives no information about extrema, by using the first derivative test, we find that:
Neither a relative maximum nor relative minimum occurs at $x=0$
Work Step by Step
$$
f(x)=x^{3}
$$
First, find the points where the derivative is $0$
Here
$$
\begin{aligned}
f^{\prime}(x) &=3x^{2}\\
\end{aligned}
$$
Solve the equation $f^{\prime}(x)=0 $ to get
$$
\begin{aligned}
f^{\prime}(x)& =3x^{2} =0\\
\Rightarrow\quad\quad\quad\quad\quad\\
x &=0\\
\end{aligned}
$$
Critical number: $0$.
Now use the second derivative test. The second derivative is
$$
f^{\prime\prime}(x)=3(2)x=6x.
$$
Evaluate $f^{\prime\prime}(x)$ at $0$, getting
$$
f^{\prime\prime}(0)=6(0)=0 .
$$
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
$$
(-\infty, 0 ), \quad (0,\infty).
$$
(1)
Test a number in the interval $(-\infty, 0 )$ say $ -1$:
$$
\begin{aligned}
f^{\prime}(-1) =3(-1)^{2} \\
&=3(1) \\
&=3 \\
&\gt 0
\end{aligned}
$$
to see that $ f^{\prime}(x)$ is positive in that interval, so $f(x)$ is increasing on $(-\infty, 0 )$
(2)
Test a number in the interval $(0, \infty )$ say $1$:
$$
\begin{aligned}
f^{\prime}(1) =3(1)^{2} \\
&=3(1) \\
&=3 \\
&\gt 0
\end{aligned}
$$
to see that $ f^{\prime}(x)$ is positive in that interval, so $f(x)$ is
also increasing on $(0, \infty ).$
So that by the first derivative test, since $f(x)$ is increasing for all $x$, then neither a relative maximum nor relative minimum occurs at $x=0.$
