1. We need the number of points in $(-20,20)$ where $f(x)=4|2x+3|+9\left[x+\frac12\right]-12[x+20]$ is not differentiable.

A sum of functions is not differentiable at points where any term is not differentiable, unless a cancellation makes the total differentiable. So we examine each term carefully.

2. Non-differentiability of $4|2x+3|$

The absolute value function is not differentiable when its inside is zero: $2x+3=0 \implies x=-\frac32.$ So this term is not differentiable at $x=-\frac32.$

3. Non-differentiability of $9\left[x+\frac12\right]$

A greatest integer function $[u]$ has jump discontinuities when $u$ is an integer. Thus $x+\frac12 \in \mathbb Z \implies x=n-\frac12, \quad n\in\mathbb Z.$ So possible points are all half-integers of the form $x=k+\frac12, \quad k\in\mathbb Z,$ including negative ones as well.

Now restrict to $(-20,20)$: $-20<x<20,$ so for $x=n-\frac12$, $-20<n-\frac12<20$ $-\frac{39}{2}<n<\frac{41}{2}$ $-19.5<n<20.5.$ Hence $n=-19,-18,\dots,20,$ which gives $20-(-19)+1=40$ points.

So $9\left[x+\frac12\right]$ is not differentiable at 40 points in $(-20,20)$.

4. Non-differentiability of $-12[x+20]$

Similarly, $[x+20]$ has jumps when $x+20\in \mathbb Z \implies x=m-20, \quad m\in\mathbb Z.$ Thus the points are integers: $x\in \mathbb Z.$ Inside $(-20,20)$, these are $-19,-18,\dots,19,$ which are $19-(-19)+1=39$ points.

So $-12[x+20]$ is not differentiable at 39 points.

5. Check overlaps

We now combine the non-differentiability sets:

• from $4|2x+3|$: $\left\{-\frac32\right\}$
• from $9\left[x+\frac12\right]$: all half-integers in $(-20,20)$
• from $-12[x+20]$: all integers in $(-20,20)$

Observe:

• Half-integers and integers are disjoint.
• $-\frac32=-1.5$ is itself a half-integer.

So the point $x=-\frac32$ is already included among the 40 half-integer points. Hence it does not add a new point.

Thus total number of distinct points is $40+39=79.$

6. Can any jumps cancel and make the sum differentiable?

We must verify this. At integer points, only $[x+20]$ jumps; $\left[x+\frac12\right]$ does not jump there, and $|2x+3|$ is smooth there. So the function has a jump, hence not differentiable.

At half-integer points, only $\left[x+\frac12\right]$ jumps; $[x+20]$ does not jump there. At $x=-\frac32$, the absolute value term has a cusp, but the floor term also jumps there; in any case the function is not differentiable.

So no cancellation changes the count.

Therefore, the number of points is $\boxed{79}.$