Let $f(x)=|x-2|+|x-4|-|2x-6|$ for $2 \leq x\leq 8$. The sum of the largest and smallest values of $f(x)$ is 
$\textbf {(A)}\ 1 \qquad  \textbf {(B)}\ 2 \qquad  \textbf {(C)}\ 4 \qquad  \textbf {(D)}\ 6 \qquad  \textbf {(E)}\ \text{none of these}$
Note that at $x=2,3,4,$ one of the three absolute values is equal to $0.$
Without using absolute values, we rewrite $f(x)$ as a piecewise function:
\[f(x) = \begin{cases} (x-2)+(4-x)-(6-2x) & \mathrm{if} \ 2\leq x<3 \\ (x-2)+(4-x)-(2x-6) & \mathrm{if} \ 3\leq x<4 \\ (x-2)+(x-4)-(2x-6) & \mathrm{if} \ 4\leq x\leq8 \end{cases},\]
which simplifies to
\[f(x) = \begin{cases} 2x-4 & \mathrm{if} \ 2\leq x<3 \\ -2x+8 & \mathrm{if} \ 3\leq x<4 \\ 0 & \mathrm{if} \ 4\leq x\leq8 \end{cases}.\]
The graph of $y=f(x)$ is shown below.

The largest value of $f(x)$ is $2,$ and the smallest value of $f(x)$ is $0.$ So, their sum is $\boxed{\textbf {(B)}\ 2}.$
~MRENTHUSIASM