Here both the modulus function is continuous in its domain.

At $x=-1$, 

$\underset{x\to -1^{-}}{\lim }\left|2x^2-3x-7\right|=2 \& \underset{x\to -1^{+}}{\lim }\left[4x^2-1\right]=2$

Hence $f\left(x\right)$ is continuous at $x=-1$

At $x=1$.

$\underset{x\to 1^{-}}{\lim }\left[4x^2-1\right]=2 \& \underset{x\to 1^{+}}{\lim }\left|x+1\right|+\left|x-2\right|=3$

Hence, $f\left(x\right)$ is discontinuous at $x=1$

For $x\in \left(-1,1\right)$ the function $f\left(x\right)$ is discontinuous when $\left[4x^2-1\right]=0/ 1/ 2$

i.e when $x\in \left\{\pm \frac{1}{\sqrt{2}},\pm \frac{1}{2},\pm \frac{\sqrt{3}}{2}\right\}$