1. Given function

We have

Let $f(t)=t^3-6t^2+9t-3.$ Then for $0\le x\le 3$, $g(x)=\max_{0\le t\le x} f(t).$

We must find the number of points in $(0,4)$ where $g$ is not differentiable.

2. Study $f(t)$ on $[0,3]$

Differentiate: $f'(t)=3t^2-12t+9=3(t-1)(t-3).$

So:

• $f'(t)>0$ for $0<t<1$,
• $f'(t)<0$ for $1<t<3$,
• $f'(t)=0$ at $t=1,3$.

Hence on $[0,3]$:

• $f$ increases on $[0,1]$,
• $f$ decreases on $[1,3]$.

Now compute key values: $f(0)=-3,$ $f(1)=1-6+9-3=1,$ $f(3)=27-54+27-3=-3.$

Thus the maximum of $f$ up to $x$ behaves as follows:

• for $0\le x\le 1$, since $f$ is increasing, the maximum on $[0,x]$ is attained at $t=x$,
• for $1\le x\le 3$, since $f$ decreases after $1$, the maximum on $[0,x]$ remains the value at $t=1$.

Therefore,

for the first part.

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

So overall,

3. Check differentiability at possible joining points

Inside each open interval:

• $(0,1)$: polynomial, so differentiable.
• $(1,3)$: constant function, so differentiable.
• $(3,4)$: linear function, so differentiable.

Thus only possible non-differentiable points are $x=1$ and $x=3$.

4. At $x=1$

From the left, $g'(x)=3x^2-12x+9=3(x-1)(x-3),$ so $g'_-(1)=3(0)(-2)=0.$

From the right, $g(x)=1$, so $g'_+(1)=0.$

Hence $g$ is differentiable at $x=1$.

5. At $x=3$

From the left, $g(x)=1$, so $g'_-(3)=0.$

From the right, $g(x)=4-x$, so $g'_+(3)=-1.$

Since $g'_-(3)\ne g'_+(3),$ $g$ is not differentiable at $x=3$.

Also, $g$ is continuous there because $g(3)=1, \qquad \lim_{x\to 3^+}(4-x)=1.$

6. Count of non-differentiable points in $(0,4)$

There is only one such point, namely $x=3$.

Therefore, the required number is $\boxed{1}.$