The number of points, at which the function $$f(x) = |2x + 1| - 3|x + 2| + |x^2 + x - 2|$$, $$x \in R$$ is not differentiable, is ______

We have $$f(x) = |2x + 1| - 3|x + 2| + |x^2 + x - 2|$$. Note that $$x^2 + x - 2 = (x + 2)(x - 1)$$, so $$|x^2 + x - 2| = |x + 2||x - 1|$$.

The critical points where the expressions inside the absolute values change sign are $$x = -\frac{1}{2}$$ (from $$|2x+1|$$), $$x = -2$$ (from $$|x+2|$$), and $$x = 1$$ (from $$|x-1|$$).

We analyze each interval and check differentiability at each critical point.

For $$x < -2$$: $$|2x+1| = -(2x+1)$$, $$|x+2| = -(x+2)$$, $$|x+2||x-1| = (-(x+2))(-(x-1)) = (x+2)(x-1)$$. So $$f(x) = -(2x+1) + 3(x+2) + (x+2)(x-1) = -2x - 1 + 3x + 6 + x^2 + x - 2 = x^2 + 2x + 3$$.

For $$-2 < x < -\frac{1}{2}$$: $$|2x+1| = -(2x+1)$$, $$|x+2| = x+2$$, $$|x+2||x-1| = (x+2)(-(x-1)) = -(x+2)(x-1)$$. So $$f(x) = -(2x+1) - 3(x+2) - (x+2)(x-1) = -2x - 1 - 3x - 6 - x^2 - x + 2 = -x^2 - 6x - 5$$.

At $$x = -2$$: Left derivative from $$x^2 + 2x + 3$$ is $$2(-2) + 2 = -2$$. Right derivative from $$-x^2 - 6x - 5$$ is $$-2(-2) - 6 = -2$$. Both equal $$-2$$, so $$f$$ is differentiable at $$x = -2$$.

For $$-\frac{1}{2} < x < 1$$: $$|2x+1| = 2x+1$$, $$|x+2| = x+2$$, $$|x+2||x-1| = (x+2)(1-x)$$. So $$f(x) = (2x+1) - 3(x+2) + (x+2)(1-x) = 2x + 1 - 3x - 6 - x^2 - x + 2 = -x^2 - 2x - 3$$.

At $$x = -\frac{1}{2}$$: Left derivative from $$-x^2 - 6x - 5$$ is $$-2(-\frac{1}{2}) - 6 = -5$$. Right derivative from $$-x^2 - 2x - 3$$ is $$-2(-\frac{1}{2}) - 2 = -1$$. Since $$-5 \neq -1$$, $$f$$ is not differentiable at $$x = -\frac{1}{2}$$.

For $$x > 1$$: $$|2x+1| = 2x+1$$, $$|x+2| = x+2$$, $$|x+2||x-1| = (x+2)(x-1)$$. So $$f(x) = (2x+1) - 3(x+2) + (x+2)(x-1) = 2x + 1 - 3x - 6 + x^2 + x - 2 = x^2 - 7$$.

At $$x = 1$$: Left derivative from $$-x^2 - 2x - 3$$ is $$-2(1) - 2 = -4$$. Right derivative from $$x^2 - 7$$ is $$2(1) = 2$$. Since $$-4 \neq 2$$, $$f$$ is not differentiable at $$x = 1$$.

Therefore, the number of points where $$f$$ is not differentiable is $$2$$.