1. We check discontinuities piecewise and also at the boundary points $x=-1$ and $x=1$.

The function is

where $[t]$ is the greatest integer function.

2. For $x\le -1$, consider $f(x)=|2x^2-3x-7|.$ Since $2x^2-3x-7$ is a polynomial and modulus preserves continuity, this part is continuous for all $x\le -1$.

So there is no discontinuity inside $(-\infty,-1)$.

3. For $-1<x<1$, consider $f(x)=[4x^2-1].$ The greatest integer function $[g(x)]$ is discontinuous where $g(x)$ takes an integer value and actually crosses it.

Let $g(x)=4x^2-1.$ For $-1<x<1$, we have $-1\le 4x^2-1<3.$ Possible integer values in this range are $-1,0,1,2.$

Now solve:

• $4x^2-1=-1 \Rightarrow 4x^2=0 \Rightarrow x=0$
• $4x^2-1=0 \Rightarrow 4x^2=1 \Rightarrow x=\pm \frac12$
• $4x^2-1=1 \Rightarrow 4x^2=2 \Rightarrow x=\pm \frac{1}{\sqrt2}$
• $4x^2-1=2 \Rightarrow 4x^2=3 \Rightarrow x=\pm \frac{\sqrt3}{2}$

At these points, check whether discontinuity actually occurs.

• At $x=0$, near $x=0$, we have $4x^2-1>-1$ for $x\ne 0$ and close to $0$, so $[4x^2-1]=[-1+\text{small positive}]=-1,$ and at $x=0$, $[4x^2-1]=[-1]=-1$. Hence continuous at $x=0$.

• At $x=\pm \frac12$, $g(x)=0$. Since crossing an integer $0$ occurs, $[g(x)]$ jumps. So discontinuous at both points.

• At $x=\pm \frac{1}{\sqrt2}$, $g(x)=1$. Again a jump occurs. So discontinuous at both points.

• At $x=\pm \frac{\sqrt3}{2}$, $g(x)=2$. Again a jump occurs. So discontinuous at both points.

Thus in $(-1,1)$, discontinuities occur at $x=\pm \frac12,\ \pm \frac1{\sqrt2},\ \pm \frac{\sqrt3}{2}$ which gives 6 points.

4. For $x\ge 1$, consider $f(x)=|x+1|+|x-2|.$ This is continuous for all $x\ge 1$ because sums of modulus functions are continuous.

So there is no discontinuity inside $(1,\infty)$.

5. Check boundary point $x=-1$.

Value at $x=-1$ comes from first piece: $f(-1)=|2(-1)^2-3(-1)-7|=|2+3-7|=|-2|=2.$

Left-hand limit: $\lim_{x\to -1^-}|2x^2-3x-7|=2.$

Right-hand limit from middle piece: $\lim_{x\to -1^+}[4x^2-1].$ For $x\to -1^+$, $x^2\to 1^-$, so $4x^2-1\to 3^-,$ hence $[4x^2-1]=2$ for all $x$ sufficiently close to $-1$ from the right. So $\lim_{x\to -1^+}f(x)=2.$

Thus both one-sided limits equal $2=f(-1)$, so $f$ is continuous at $x=-1$.

6. Check boundary point $x=1$.

Value at $x=1$ comes from third piece: $f(1)=|1+1|+|1-2|=2+1=3.$

Left-hand limit from middle piece: $\lim_{x\to 1^-}[4x^2-1].$ As $x\to 1^-$, $4x^2-1\to 3^-,$ so $[4x^2-1]=2$ near $1$ from the left. Therefore $\lim_{x\to 1^-}f(x)=2.$

Right-hand limit from third piece: $\lim_{x\to 1^+}(|x+1|+|x-2|)=3.$

Since $\lim_{x\to 1^-}f(x)=2 \ne 3=\lim_{x\to 1^+}f(x),$ $f$ is discontinuous at $x=1$.

7. Total number of discontinuity points:

• From middle piece: $6$
• At $x=-1$: $0$
• At $x=1$: $1$

Hence total number of discontinuities is $6+1=7.$

Therefore, the required integer answer is $\boxed{7}.$