Example 5. Find the maximum and minimum values of the function $f(x)=2 x^{3}-6 x+5$ on the interval $\left[-\frac{5}{2}, \frac{3}{2}\right]$.
Solution. We find the extrema of the function:

\[
\begin{gathered}
f^{\prime}(x)=6 x^{2}-6=6\left(x^{2}-1\right) \\
f^{\prime}(x)=0,6\left(x^{2}-1\right)=0, x_{1}=-1, x_{2}=1 \\
f^{\prime \prime}(x)=12 x, f^{\prime \prime}(-1)0
\end{gathered}
\]

Therefore, at the point \( x_{1}=-1 \) the function has a maximum, and

\[
\max f(x)=f(-1)=2(-1)^{3}-6(-1)+5=9
\]

at the point \( x_{2}=1 \) a minimum, and

\[
\min f(x)=f(1)=2 \cdot 1^{3}-6 \cdot 1+5=1
\]

We calculate the values of the function at the endpoints of the interval:

\[
\begin{gathered}
f\left(-\frac{5}{2}\right)=2\left(-\frac{5}{2}\right)^{3}-6\left(-\frac{5}{2}\right)+5=-11 \frac{1}{4} \\
f\left(\frac{3}{2}\right)=2\left(\frac{3}{2}\right)^{3}-6\left(\frac{3}{2}\right)+5=2 \frac{3}{4}
\end{gathered}
\]

![](https://cdn.mathpix.com/cropped/2024_05_22_7ef2114a321cf223ba99g-123.jpg?height=600&width=281&top_left_y=904&top_left_x=49)

Fig. 3.10

Thus, the greatest value of the function on the given interval is \( f(-1)=9 \), and the least value is \( f\left(-\frac{5}{2}\right)=-11 \frac{1}{4} \) (Fig. 3.10).