#### Answer

$$
f(x)=3x^{3}-3x^{2}+1
$$
Critical numbers: $0, \frac{2}{3}$.
By Part 2 of the second derivative test, $0$ leads to a relative maximum.
Also, by Part 1 of the second derivative test, $\frac{2}{3}$ leads to a relative minimum.

#### Work Step by Step

$$
f(x)=3x^{3}-3x^{2}+1
$$
First, find the points where the derivative is $0$
Here
$$
\begin{aligned}
f^{\prime}(x) &=3(3)x^{2}-3(2)x+0\\
&=9x^{2}-6x\\
& =3x(3x-2)
\end{aligned}
$$
Solve the equation $f^{\prime}(x)=0 $ to get
$$
\begin{aligned}
f^{\prime}(x) =3x(3x-2) &=0\\
\Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
x =0 \quad &\text {and} \quad x =\frac{2}{3} \\
\end{aligned}
$$
Critical numbers: $0, \frac{2}{3}$.
Now use the second derivative test. The second derivative is
$$
f^{\prime\prime}(x)=9(2)x-6=18x-6.
$$
Evaluate $f^{\prime\prime}(x)$ first at $0$, getting
$$
f^{\prime\prime}(0)=18(0)-6=-6 \lt 0.
$$
so that by Part 2 of the second derivative test, $0$ leads to a relative maximum.
Also, when $x=\frac{2}{3}$,
$$
f^{\prime\prime}(\frac{2}{3})=18(\frac{2}{3})-6=6 \gt 0.
$$
so that by Part 1 of the second derivative test, $\frac{2}{3}$ leads to a relative minimum.