Example 4. Investigate the function for extremum

$$
f(x)=(x-2) \sqrt[3]{x^{2}}
$$
Solution. We find the first derivative:

$$
f^{\prime}(x)=\sqrt[3]{x^{2}}+\frac{2(x-2)}{3 \sqrt[3]{x}}=\frac{5 x-4}{3 \sqrt[3]{x}}
$$

The derivative is zero at the point $x_{1}=\frac{4}{5}$, and it does not exist at the point $x_{2}=0$ (it goes to infinity). Therefore, the critical points are $x_{1}=\frac{4}{5}$ and $x_{2}=0$.

We investigate the nature of the critical points. Since $f^{\prime}(x)0)$ and $f^{\prime}(x)>0$ for $x>\frac{4}{5}$, $x_{1}=\frac{4}{5}$ is a point of minimum, and

$$
\min f(x)=f\left(\frac{4}{5}\right)=\left(\frac{4}{5}-2\right) \sqrt[3]{\frac{16}{25}}=-\frac{6}{5} \sqrt[3]{\frac{16}{25}}
$$

Furthermore, since $f^{\prime}(x)>0$ for $x0$ (but $x<\frac{4}{5}$), the function has a maximum at $x_{2}=0$, and

$$
\max f(x)=f(0)=0
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_7ef2114a321cf223ba99g-122.jpg?height=406&width=506&top_left_y=1248&top_left_x=53)

Fig. 3.9
Noting that the function is continuous at the point $x_{2}=0$ and the tangent to the graph coincides with the $O$ axis, we construct the graph of the function (Fig. 3.9).