\(f(x) = \begin{cases}|2x^2 - 3x - 7|&; x\le - 1\\2&;-1 < x \le\frac{-\sqrt3}2\\1&; \frac{-\sqrt3}2 < x \le \frac{-1}{\sqrt2}\\1 &; \frac{-1}{\sqrt2} < x \le \frac{-1}2\\ - 1&;\frac{-1}2 < x < \frac12\\0&;\frac12 \le x < \frac1{\sqrt2}\\1&;\frac1{\sqrt2} \le x<\frac{\sqrt3}2 \\2&; \frac{\sqrt3}2 \le x< 1\\3&; 1 \le x < 2\\2x -1&;x\ge 2 \end{cases}\)
Hence, f(x) is discontinuous at
\(x = \frac{-\sqrt3}2 , \frac{-1}{\sqrt2}, \frac{-1}2, \frac12, \frac1{\sqrt2}, \frac{\sqrt3}2, 1.\)
Total number of points of discontinuity is 7.