The first two given functions intersect at $\left(\frac{1}{3},\frac{7}{3}\right)$, and last two at $\left(\frac{2}{3},\frac{8}{3}\right)$. Therefore $$f(x)=\left\{ \begin{matrix} 4x+1 & xx>\frac{2}{3} \\ -2x+4 & x>\frac{2}{3} \end{matrix}\right.$$
Which attains a maximum at
$\boxed{(E)\ \frac{8}{3}}$

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 $\boxed{\frac{8}{3}}$.