Where is the absolute value function differentiable? I have that $f(x) = |x^2 -4x|$
What I've done is trying to define $f(x)$ with the zero values being 0 and 4. But not really sure if that's how I'm supposed to go about solving this.

Note that 
$$\lim_{x\to 0^+}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0^+}\frac{|x^2-4x|}{x}=\lim_{x\to 0^+}\frac{-(x^2-4x)}{x}=-\lim_{x\to 0^+}(x-4)=4$$
and
$$\lim_{x\to 0^-}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0^-}\frac{|x^2-4x|}{x}=\lim_{x\to 0^-}\frac{(x^2-4x)}{x}=\lim_{x\to 0^+}(x-4)=-4$$
What may we conclude about the differentiability of $f$ at $0$?
What about the other point $x=4$?
P.S. Recall that $|g(x)|=g(x)$ if $g(x)\geq 0$ and $|g(x)|=-g(x)$ otherwise.
