Let f:[-1,3]

Given function is,

$f(\mathrm{x})= \begin{cases}|x|+[x], & -1 \leq x<1 \\ x+|x|, & 1 \leq x<2 \\ x+[x], & 2 \leq x \leq 3\end{cases}$

$= \begin{cases}-x-1, & -1 \leq x<0 \\ x, & 0 \leq x<1 \\ 2 x, & 1 \leq x<2 \\ x+2, & 2 \leq x<3 \\ 6, & x=3\end{cases}$

$\Rightarrow f(-1)=0, f\left(-1^{+}\right)=0$

$f\left(0^{-}\right)=-1, f(0)=0, f\left(0^{+}\right)=0$

$f\left(1^{-}\right)=1, f(1)=2, f\left(1^{+}\right)=2$

$f\left(2^{-}\right)=4, f(2)=4, f\left(2^{+}\right)=4$

$f\left(3^{-}\right)=5, f(3)=6$

$f(x)$ is discontinuous at $x=\{0,1,3\}$

Hence, $f(x)$ is discontinuous at only three points.