Example 1. Let $F=x^{3}-3 x^{2}-9 x+5$, find the extremum of $F$.
Here $a=1, b=-3$.
Let $x=y-(b / 3a)=y+1$, then
$$
F=f(y)=y^{3}-12 y-6 .
$$

Here $a=1, m=12$. Thus, by Theorem 1, we have
$$
\begin{array}{l}
F_{\text {min }}=f(\sqrt{m} / \sqrt{3a})=f(2)=-22 ; \\
F_{\text {max }}=f(-\sqrt{m / 3a})=f(-2)=10 .
\end{array}
$$