Let a function $$g : [0, 4] \to R$$ be defined as
$$g(x) =\begin{cases}\max\limits_{0 \le t \le x} \{ t^3 - 6t^2 + 9t - 3 \}, & 0 \le x \le 3 \\4 - x, & 3 < x \le 4\end{cases}$$
then the number of points in the interval $$(0, 4)$$ where $$g(x)$$ is NOT differentiable, is ___.

Let $$h(t) = t^3 - 6t^2 + 9t - 3$$. We find its critical points: $$h'(t) = 3t^2 - 12t + 9 = 3(t-1)(t-3)$$. So $$h$$ has a local maximum at $$t=1$$ with $$h(1) = 1 - 6 + 9 - 3 = 1$$, a local minimum at $$t=3$$ with $$h(3) = 27 - 54 + 27 - 3 = -3$$, and $$h(0) = -3$$.

For $$0 \le x \le 3$$, $$g(x) = \max_{0 \le t \le x} h(t)$$. We track how this running maximum evolves as $$x$$ increases from 0 to 3.

At $$x=0$$: $$g(0) = h(0) = -3$$.

For $$0 < x \le 1$$: $$h$$ is increasing (since $$h'(t) > 0$$ on $$(0,1)$$), so the maximum on $$[0,x]$$ is $$h(x)$$. Thus $$g(x) = h(x)$$ on $$(0,1]$$.

For $$1 < x \le 3$$: $$h$$ decreases from $$h(1)=1$$ to $$h(3)=-3$$. The maximum on $$[0,x]$$ is still $$h(1) = 1$$. Thus $$g(x) = 1$$ on $$(1,3]$$.

For $$3 < x \le 4$$: $$g(x) = 4 - x$$.

Now we check differentiability at each transition point in $$(0,4)$$:

At $$x = 1$$: For $$x \le 1$$, $$g(x) = h(x)$$, so $$g'(1^-) = h'(1) = 0$$. For $$x > 1$$, $$g(x) = 1$$, so $$g'(1^+) = 0$$. Both one-sided derivatives equal 0, so $$g$$ is differentiable at $$x=1$$.

At $$x = 3$$: For $$x \le 3$$, $$g(x) = 1$$, so $$g'(3^-) = 0$$. For $$x > 3$$, $$g(x) = 4-x$$, so $$g'(3^+) = -1$$. Since $$0 \neq -1$$, $$g$$ is NOT differentiable at $$x = 3$$.

We also check $$x = 0$$ (the running maximum): at $$x = 0$$, $$h(0) = -3$$ is a local min, but the maximum function starts equalling $$h(x)$$ immediately and $$h'(0) = 9 > 0$$, so no issue there for $$x \in (0,4)$$.

Therefore, the number of points in $$(0, 4)$$ where $$g$$ is not differentiable is $$\boxed{1}$$.