For $x \in R, f(x)=\min \{4 x+2$, $\left.-12 x^{2}+12,-2 x+4\right\}$, find the expression of $f(x)$ and $f(x)_{\max } \cdot$
Solution:
As shown in the figure, in
the same Cartesian
coordinate system,
draw the graphs of $y_{1}=$
$$
\begin{array}{l}
4 x+2, y_{2} \\
=-12 x^{2} \\
+12, y_{3}= \\
-2 x+4
\end{array}
$$

According to the definition of $f(x)$, we can obtain the piecewise expression of $f(x)$
$$
\left\{\begin{array}{ll} 
& f(x)= \\
-12 x^{2}+12, & x \leqslant \frac{-1-\sqrt{31}}{6} \text { or } x \geqslant \frac{1+\sqrt{97}}{12}, \\
4 x+2, & \frac{-1-\sqrt{31}}{6}<x \leqslant \frac{1}{3}, \\
-2 x+4, & \frac{1}{3}<x<\frac{1+\sqrt{97}}{12} .
\end{array}\right.
$$

From the graph, it can be seen that when $x=\frac{1}{3}$,
$$
f(x)_{\max }=\frac{10}{3} \text {. }
$$

The extreme values are always related to symmetry, order, balance, and special values or points, and this is also true for composite extreme values. Therefore, we can use these associated conditions to first determine the upper or lower bounds of the given problem, and then further find the composite extreme values. The upper bound obtained at this time is the maximum value, and the lower bound is the minimum value.