3.4.2. Find the greatest and the least values of the function $f(x)=2 x^{3}+3 x^{2}-36 x+1$ on the interval $[1 ; 11]$.
Solution. The given function is continuous on $[1 ; 11]$. We find $f^{\prime}(x)=6 x^{2}+6 x-36=0$. Setting $y^{\prime}=0$, we have $x^{2}+x-6=0$,
from which we obtain two critical points: $x_{1}=-3$ and $x_{2}=2$. To find the minimum and maximum values of the function on the given interval, it is sufficient to compute its values at the endpoints of the interval and at the point $x=2$, since the point $x=-3$ does not belong to the interval $[1 ; 11]$. We get:

$$
\begin{gathered}
f(1)=2 \cdot 1+3 \cdot 1-36+1=-30 \\
f(2)=2 \cdot 2^{3}+3 \cdot 2^{2}-36 \cdot 2+1=-43 \\
f(11)=2 \cdot 11^{3}+3 \cdot 11^{2}-36 \cdot 11+1=2630 .
\end{gathered}
$$

Therefore, the minimum value of the function, equal to (-43), is achieved at the critical point $x=2$, and the maximum value is $2630-$ at the right endpoint of the interval, at the point $x=11$.