For each real number $x$, let $f(x)$ be the minimum of the numbers $4x+1$, $x+2$, and $-2x+4$. Then the maximum value of $f(x)$ is 
$\text{(A)} \ \frac 13 \qquad \text{(B)} \ \frac 12 \qquad \text{(C)} \ \frac 23 \qquad \text{(D)} \ \frac 52 \qquad \text{(E)} \ \frac 83$
1. We are given the function \( f(x) \) which is defined as the minimum of three linear functions: \( 4x + 1 \), \( x + 2 \), and \( -2x + 4 \). We need to find the maximum value of \( f(x) \).

2. To determine the intervals where each function is the minimum, we first find the points of intersection of the three lines:
   - Set \( 4x + 1 = x + 2 \):
     \[
     4x + 1 = x + 2 \implies 3x = 1 \implies x = \frac{1}{3}
     \]
   - Set \( 4x + 1 = -2x + 4 \):
     \[
     4x + 1 = -2x + 4 \implies 6x = 3 \implies x = \frac{1}{2}
     \]
   - Set \( x + 2 = -2x + 4 \):
     \[
     x + 2 = -2x + 4 \implies 3x = 2 \implies x = \frac{2}{3}
     \]

3. We now have the points of intersection \( x = \frac{1}{3} \), \( x = \frac{1}{2} \), and \( x = \frac{2}{3} \). These points divide the real line into intervals. We will analyze each interval to determine which function is the minimum in that interval.

4. For \( x \leq \frac{1}{3} \):
   - Compare \( 4x + 1 \), \( x + 2 \), and \( -2x + 4 \):
     \[
     4x + 1 \leq x + 2 \quad \text{and} \quad 4x + 1 \leq -2x + 4
     \]
   - At \( x = \frac{1}{3} \):
     \[
     4\left(\frac{1}{3}\right) + 1 = \frac{4}{3} + 1 = \frac{7}{3}
     \]

5. For \( \frac{1}{3} \leq x \leq \frac{2}{3} \):
   - Compare \( 4x + 1 \), \( x + 2 \), and \( -2x + 4 \):
     \[
     x + 2 \leq 4x + 1 \quad \text{and} \quad x + 2 \leq -2x + 4
     \]
   - At \( x = \frac{2}{3} \):
     \[
     \frac{2}{3} + 2 = \frac{2}{3} + \frac{6}{3} = \frac{8}{3}
     \]

6. For \( x \geq \frac{2}{3} \):
   - Compare \( 4x + 1 \), \( x + 2 \), and \( -2x + 4 \):
     \[
     -2x + 4 \leq 4x + 1 \quad \text{and} \quad -2x + 4 \leq x + 2
     \]
   - At \( x = \frac{2}{3} \):
     \[
     -2\left(\frac{2}{3}\right) + 4 = -\frac{4}{3} + 4 = \frac{8}{3}
     \]

7. From the above analysis, the maximum value of \( f(x) \) is \( \frac{8}{3} \).

The final answer is \(\boxed{\frac{8}{3}}\).