Example 3. Study the maximum and minimum values of the function $\mathrm{y}=\frac{4 \mathrm{x}^{3}-3 \mathrm{x}^{4}}{12}$ on the interval $(-1$, 2).
Solving, we know $y(-1+0)=y(-1)=-\frac{7}{12}$,
$y(2-0)=y(2)=-\frac{4}{3}$.

Also, $y^{\prime}=x^{2}-x^{3}=x^{2}(1-x)$, the critical points are $x_{1}=0, x_{2}=1$. By comparing $y(0)=0$, $y(1)=\frac{1}{12}$ with the endpoint (limit) values, we can see that $y$ has only a maximum value of $y(1)=\frac{1}{12}$ on $(-1,2)$, but no minimum value (Figure 5).