Find the number of points of non differentiability in $(x^2-3x+2)(|x^3-6x^2+11x-6|)$ The expression is factorized as $$(x-2)(x-1)(|(x-1)(x-2)(x-3)|)$$
I expected the points to be $1,2,3$, put looking at the graph, it’s only $x=3$
I kinda figured out the reason for this, because the expression would end up simplifying to
$\pm (x-2)^2(x-1) ^2|x-3|$ but I am not finding this very convincing. What can be a concrete reason for this?
 A: $$F(x)=(x-1)(x-2)|(x-1)(x-2)(x-3)|.$$ Only $x=3$ is the point of non-differentiability. At $x=1,2$ is it differentiable. At these points, we have $(x-a)|x-a|$ which is differentiable at $x=a$.
Note that $g(x)=(x-a)|x-a|= (x-a)(a-x), ~if~ x<a$ and $g(x)=(x-a)^2, ~if ~x\ge a$, hence both the left and right derivatives are equal and zero at $x=a$.
A: The function $f(x)=x|x|$ is differentiable everywhere. Therefore, $$F(x)=(x-2)(x-1)(|(x-1)(x-2)(x-3)|)=f(x-1)f(x-2)|x-3|$$ is differentiable everywhere except at $x=3$.
A: Exact simplification looks like
$$(x-2)^2(x-1) ^2|x-3|\text{sign}(x-2)\text{sign}(x-1)$$
now in every point except $x=3$ you have multiplication of differentiable functions. Is this enough convince now?
A: When f(x) is differentiable at "a" and if f(a)=0 then |f(x)| is differentiable at "a" iff f'(a)=0.  (Refer Real Analysis by "Bartle and Sherbert". At 1 and 2 your f(x) satisfies this hypothesis whereas at 3 it does not. Hence you get that it's differentiable at 1 and 2 and not at 3.
