Example 2. Investigate the function for maximum and minimum

$$
f(x)=\frac{1}{3} x^{3}-\frac{1}{2} x^{2}-6 x+2 \frac{2}{3}
$$
Solution. We find the first derivative:

$$
f^{\prime}(x)=x^{2}-x-6
$$

The derivative is continuous everywhere. Let's find the points where it is zero: $x^{2}-x-6=0$.

The critical points of the function argument are $x_{1}=-2, x_{2}=3$.

We will investigate the critical points using the first derivative.

Since

$$
f^{\prime}(x)=x^{2}-x-6=(x+2)(x-3)
$$

then for $x< -2$, $f^{\prime}(x)>0$ (since $x+2<0, x-3<0$), and for $-2 < x < 3$, $f^{\prime}(x)<0$ (since $x+2>0, x-3<0$), and for $x>3$, $f^{\prime}(x)>0$ (since $x+2>0, x-3>0$). This means that the derivative of the function changes sign from negative to positive as it passes through the point $x_{2}=3$. This indicates that the function has a minimum at this point, specifically

![](https://cdn.mathpix.com/cropped/2024_05_22_7ef2114a321cf223ba99g-120.jpg?height=686&width=487&top_left_y=904&top_left_x=57)

Fig. 3.7

$$
\min f(x)=f(3)=\frac{1}{3} \cdot 3^{3}-\frac{1}{2} \cdot 3^{2}-6 \cdot 3+\frac{8}{3}=-10 \frac{5}{6} \text {. }
$$

Based on the conducted research, we construct the graph of the function (Fig. 3.7).

Remark. When using the first rule, it is always important to keep in mind the local nature of the extremum. When investigating the sign of the first derivative, it is necessary to consider points to the left and right of the critical point and sufficiently close to it. If, in this example, for $x_{1}=-2$, we take points sufficiently far away (for example, $x>3$), it will turn out that $f^{\prime}(x)>0$ for $x>3$, and $f^{\prime}(x)>0$ for $x>-2$ (for example, at $x=4$). From this, one might incorrectly conclude that there is no extremum at the point $x_{1}=-2$.