13・37 For each real number $x$, let $f(x)$ be the minimum of the three functions $4x+1$, $x+2$, and $-2x+4$. Then the maximum value of $f(x)$ is
(A) $\frac{1}{3}$.
(B) $\frac{1}{2}$.
(C) $\frac{2}{3}$.
(D) $\frac{5}{2}$.
(E) $\frac{8}{3}$.
(31st American High School Mathematics Examination, 1980)
[Solution] In the figure, the graphs of $y=4 x+1, y=x+2$ and $y=-2 x+4$ are given, where the solid line represents the graph of function $f$.

Obviously, the intersection point of the lines $y=x+2$ and $y=-2 x+4$ is the maximum value of the function, that is,
$$
x=\frac{2}{3}, \quad f\left(\frac{2}{3}\right)=\frac{8}{3} .
$$

Therefore, the answer is $(E)$.