For any real number $x$, let $f(x)$ be the minimum of the values $4 x+1, x+2,-2 x+4$. What is the largest value of $f(x)$?
Solution. Let's plot the given functions in a Cartesian coordinate system and calculate the coordinates of the intersection points pairwise:

![](https://cdn.mathpix.com/cropped/2024_05_02_b065edf438b07be68381g-1.jpg?height=791&width=588&top_left_y=266&top_left_x=758)

From the system of equations, the coordinates of the intersection point are: \( M_{1}\left(\frac{1}{3} ; \frac{7}{3}\right) \). From the system of equations

\[
\left\{\begin{array}{l}
y=4 x+1 \\
y=-2 x+4
\end{array}\right.
\]

the coordinates of the intersection point are: \( M_{2}\left(\frac{1}{2} ; 3\right) \). From the system of equations

\[
\left\{\begin{array}{l}
y=x+2 \\
y=-2 x+4
\end{array}\right.
\]

the coordinates of the intersection point are: \( M_{3}\left(\frac{2}{3} ; \frac{8}{3}\right) \). From the positions of the intersection points and the lines, we can read that up to the intersection point \( M_{1} \), the value of the function \( 4 x+1 \) is the smallest, between the intersection points \( M_{1} \) and \( M_{3} \), the value of the function \( x+2 \) is the smallest, and after the intersection point \( M_{3} \), the value of the function \( -2 x+4 \) is the smallest.

Since the functions \( 4 x+1 \) and \( x+2 \) are strictly increasing, and the function \( -2 x+4 \) is strictly decreasing, the highest point on the graph of \( f(x) \) is \( M_{3}\left(\frac{2}{3} ; \frac{8}{3}\right) \), so the maximum value of the function is \( \frac{8}{3} \).