$|P(x)|$ differentiable at a root $x_0$ Let $p(x)$ be a polynomial and suppose that $x_0 \in \bf R$ is a real root i.e. $p(x_0) = 0$. When will $|p(x)|$ be differentiable at $x_0$?
My Thoughts
For polynomials such as $f(x) = x$, we run into trouble at roots of odd multiplicity i.e. $|x|$ is not differentiable at $x = 0$. So my thoughts here are that it would be sufficient for the root to be of even multiplicity, in which case $|p(x)|$ would behave identically to $p(x)$ around $x_0$. Is this correct?
 A: Since $\,P(x_0)=0\,$ , we can write $\,P(x)=(x-x_0)^mg(x)\,\,,\,g(x_0)\neq 0\,$ , so
$$|P(x_0)|':=\lim_{h\to 0}\frac{|P(x_0+h)|}{h}=\lim_{h\to 0}\frac{|h|^m\,|g(x_0+h)|}{h}$$
so if $\,m>1\,$ then
$$|P(x_0)|'=\lim\frac{|h|^m\,|g(x_0+h)|}{h}=\lim_{h\to 0}\frac{\pm h^m\,|g(x_0+h)|}{h}=0$$
If $\,m=1\,$ then clearly, and as noted already in the comments above, the limit doesn't exist.
A: It doesn't actually matter that $p$ is a polynomial -- it is enough that it is a differentiable function. Then assuming $p(x_0)=0$, then $|p(x)|$ is differentiable at $x_0$ if and only if $p'(x_0)=0$.
The easy direction is the one where we know that $p'(x_0)=0$. Then the difference quotients for $|p(x)|$ around $x_0$ are just $\pm$ the difference quotients of $p(x)$ and if one set of quotients tend towards zero, then obviously the other set does too.
On the other hand, if $p'(x_0)=a>0$, then there are points immediately to the right of $x_0$ with difference quotients close to $a$ and points immediately to the left of $x_0$ with difference quotients close to $-a$. Since $a\ne -a$, the difference quotients cannot tend to a limit, so $|p(x)|$ is not differentiable.
For $p'(x_0)<0$ it's the same as in the previous case, with the roles of $a$ and $-a$ swapped.
(Beware that some more explicit reasoning about the various limits will be probably be necessary to make this into an acceptable homework answer).
