Answer
$$
f(x)=2x^{3}-4x^{2}+2
$$
Critical numbers: $0, \frac{4}{3}$.
By Part 2 of the second derivative test, $0$ leads to a relative maximum.
Also, by Part 1 of the second derivative test, $\frac{4}{3}$ leads to a relative minimum.
Work Step by Step
$$
f(x)=2x^{3}-4x^{2}+2
$$
First, find the points where the derivative is $0$
Here
$$
\begin{aligned}
f^{\prime}(x) &=2(3)x^{2}-4(2)x+0\\
&=6x^{2}-8x\\
& =2x(3x-4)
\end{aligned}
$$
Solve the equation $f^{\prime}(x)=0 $ to get
$$
\begin{aligned}
f^{\prime}(x) =2x(3x-4) &=0\\
\Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
x =0 \quad &\text {and} \quad x =\frac{4}{3} \\
\end{aligned}
$$
Critical numbers: $0, \frac{4}{3}$.
Now use the second derivative test. The second derivative is
$$
f^{\prime\prime}(x)=6(2)x-8=12x-8.
$$
Evaluate $f^{\prime\prime}(x)$ first at $0$, getting
$$
f^{\prime\prime}(0)=12(0)-8=-8 \lt 0.
$$
so that by Part 2 of the second derivative test, $0$ leads to a relative maximum.
Also, when $x=\frac{4}{3}$,
$$
f^{\prime\prime}(\frac{4}{3})=12(\frac{4}{3})-8=16-8=8 \gt 0.
$$
so that by Part 1 of the second derivative test, $\frac{4}{3}$ leads to a relative minimum.
