Function f(x)=$\left| {x - 2} \right| - 2\left| {x - 4} \right|\,is$ discontinous at:

$f(x)=\mid x-2 \mid -2\mid x-4 \mid$

$\therefore f\left(x \right)=\begin{Bmatrix} -(x-2)+2(x-4),\qquad x<2\\ (x-2)+2(x-4), \qquad 2\le x \le 4\\ (x-2)-2(x-4) \qquad x\gt 4 \\ \end{Bmatrix}$

At $x=2$

L.H.L(Left Hand Limit)

$\displaystyle\lim_{x\to\ 2^{-}}=-(2-2)+2(2-4) = -4$

R.H.LRight Hand Limit/0

$\displaystyle\lim_{x\to\ 2^{+}}=(2-2)+2(2-4)=-4$

$\therefore \displaystyle\lim_{x\to\ 2^{-}}f(x) = \displaystyle\lim_{x\to\ 2^{+}}f(x)$

At $x=4$

L.H.L(Left Hand Limit)

$\displaystyle\lim_{x\to\ 4^{-}}=(4-2)+2(4-4)=2$

R.H.L(Right Hand Limit)

$\displaystyle\lim_{x\to\ 4^{+}}=(4-2)-2(4-4)=2$

$\therefore \displaystyle\lim_{x\to\ 4^{-}}f(x) = \displaystyle\lim_{x\to\ 4^{+}}f(x)$

$\therefore \ f(x)$ is continuous at everywhere.