Let $$f(x) = x|x^2 - 1| - 2|x - 3| + x - 3, x \in \mathbb{R}$$. If $$m$$ and $$M$$ are respectively the number of points of local minimum and local maximum of $$f$$ in the interval $$(0, 4)$$, then $$m + M$$ is equal to ______.

Given,

$$f(x)=x|x^2-1|-2|x-3|+x-3,\qquad x\in(0,4)$$

The modulus terms change sign at

$$x=1,\qquad x=3$$

So, consider the function case-wise.

Case 1: $$0<x<1$$

Here,

$$x^2-1<0,\qquad x-3<0$$

Thus,

$$|x^2-1|=1-x^2,\qquad |x-3|=3-x$$

Hence,

$$f(x)=x(1-x^2)-2(3-x)+x-3$$

$$=-x^3+4x-9$$

Therefore,

$$f'(x)=-3x^2+4$$

Now,

$$f'(x)=0\Rightarrow -3x^2+4=0$$

$$x=\frac{2}{\sqrt3}>1$$

So, there is no critical point in $$(0,1)$$.

Also,

$$f'(x)>0\quad \text{for all}\quad x\in(0,1)$$

Hence, $$f$$ is increasing on $$(0,1)$$.

Case 2: $$1<x<3$$

Here,

$$x^2-1>0,\qquad x-3<0$$

Thus,

$$|x^2-1|=x^2-1,\qquad |x-3|=3-x$$

Hence,

$$f(x)=x(x^2-1)-2(3-x)+x-3$$

$$=x^3+2x-9$$

Therefore,

$$f'(x)=3x^2+2>0$$

Hence, $$f$$ is increasing on $$(1,3)$$.

Case 3: $$3<x<4$$

Here,

$$x^2-1>0,\qquad x-3>0$$

Thus,

$$|x^2-1|=x^2-1,\qquad |x-3|=x-3$$

Hence,

$$f(x)=x(x^2-1)-2(x-3)+x-3$$

$$=x^3-2x+3$$

Therefore,

$$f'(x)=3x^2-2>0$$

Hence, $$f$$ is increasing on $$(3,4)$$.

Now, check the corner points.

At

$$x=1$$

$$f'_-(1)=1,\qquad f'_+(1)=5$$

Both are positive, so no local extremum occurs at $$x=1$$.

At

$$x=3$$

$$f'_-(3)=29,\qquad f'_+(3)=25$$

Again, both are positive, so no local extremum occurs at $$x=3$$.

Thus, the function is increasing throughout $$(0,4)$$.

Therefore,

$$m=0,\qquad M=0$$

Hence, $$\boxed{m+M=0}$$.