Example 3. Investigate the function for extremum

$$
f(x)=\frac{1}{4} x^{4}-2 x^{3}+\frac{11}{2} x^{2}-6 x+\frac{9}{4}
$$
Solution. We will use the second rule of the sufficient condition for an extremum. We find the first and second derivatives:

$$
f^{\prime}(x)=x^{3}-6 x^{2}+11 x-6, f^{\prime \prime}(x)=3 x^{2}-12 x+11
$$

We look for points of extremum among the critical points, i.e., points for which $f^{\prime}(x)=0$. The first derivative is zero when

$$
x^{3}-6 x^{2}+11 x-6=0
$$

Using the theorem of Bezout, we can establish that $x=1$ is a root of the last equation. Noting this and factoring the left side of the equation, we get

$$
\begin{gathered}
x^{3}-6 x^{2}+11 x-6=(x-1)\left(x^{2}-5 x+6\right)= \\
=(x-1)(x-2)(x-3)=0
\end{gathered}
$$

from which

$$
x_{1}=1, x_{2}=2, x_{3}=3
$$

We investigate the sign of the second derivative at these values of $x$:

$$
\begin{gathered}
f^{\prime \prime}(1)=3 \cdot 1^{2}-12 \cdot 1+11=2>0 \\
f^{\prime \prime}(2)=3 \cdot 2^{2}-12 \cdot 2+11=-10
\end{gathered}
$$

Therefore, $x=2$ is a point of maximum, and $x=1$ and $x=3$ are points of minimum. We find the values of the extrema:

$$
\begin{aligned}
& \min f(x)=f(1)=\frac{1}{4} \cdot 1^{4}-2 \cdot 1^{3}+\frac{11}{2} \cdot 1^{2}-6 \cdot 1+\frac{9}{4}=0 \\
& \max f(x)=f(2)=\frac{1}{4} \cdot 2^{4}-2 \cdot 2^{3}+\frac{11}{2} \cdot 2^{2}-6 \cdot 2+\frac{9}{4}=\frac{1}{4} \\
& \min f(x)=f(3)=\frac{1}{4} \cdot 3^{4}-2 \cdot 3^{3}+\frac{11}{2} \cdot 3^{2}-6 \cdot 3+\frac{9}{4}=0
\end{aligned}
$$

The graph of the function is shown in Fig. 3.8.

![](https://cdn.mathpix.com/cropped/2024_05_22_7ef2114a321cf223ba99g-122.jpg?height=415&width=493&top_left_y=47&top_left_x=42)

Fig. 3.8