Example 2. Study the maximum and minimum values of the function $\mathrm{y}=\frac{1}{\mathrm{x}(1-\mathrm{x})}$ on $(0,1)$.


Given $x \in (0,1)$, therefore $x>0, 1-x>0$.
When $x \rightarrow +0$, $1-x \rightarrow 1, x(1-x) \rightarrow +0$,
When $x \rightarrow 1-0$, $1-x \rightarrow +0, x(1-x) \rightarrow +0$,
Therefore, $y(+0)=y(1-0)=+\infty$. From (v),

it is known that $y$ has only a minimum value on $(0,1)$. Also, $y' = \frac{2x-1}{x^2(1-x)^2}$, so $x=\frac{1}{2}$ is a critical point. The minimum value of $y$ is $y\left(\frac{1}{2}\right)=4$ (Figure 3).