Example 4. Study the maximum and minimum values of the function $y=x^{2}-6 x+4 \ln x$ on $(0,2.5)$.
Since $y(+0)=-\infty, \therefore y$ has no minimum value on $(0,2.5)$. Also, $y^{\prime}=2 x-6+\frac{4}{x}=\frac{2\left(x^{2}-3 x+2\right)}{x}$
$$
=\frac{2(x-1)(x-2)}{x},
$$

the critical points are $x_{1}=1, x_{2}=2$. Comparing $y(1)=-5, y(2)=4 \ln 2 - 2 \doteq -5.2, y(2.5-0)$
$=y(2.5)=4 \ln \frac{5}{2}-\frac{35}{4}=$
-5.1, we know that $y$ has only a maximum value $y(1)=-5$
(Figure 6).