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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?

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If and only if $x_0$ is NOT a simple root. – Did Aug 19 '12 at 13:15
You should look at $f(x)=x^3$. – Chris Eagle Aug 19 '12 at 13:16
Just observe that for $p(x) = x^3$, $|p(x)|$ is also differentiable at $x = 0$. The geometric reason that $p(x) = x$ fails to satisfy this property is that taking absolute value flips over the negative part of the graph of $y = p(x)$, resulting in a wedge-like peak. So as long as this wedge is avoided, taking absolute value does not spoil differentiability. This is the case for zeros of multiplicity $\geq 2$. – Sangchul Lee Aug 19 '12 at 13:23
@sos440 , you should write down as answer. +1 – DonAntonio Aug 19 '12 at 13:27
@DonAntonio, I think my answer is still insufficient to be a nice answer. It is just a phenomenal explanation, lacking details, hence it does not deserve of an answer. That's why I just wrote it as a comment. – Sangchul Lee Aug 19 '12 at 13:34
up vote 1 down vote accepted

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.

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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).

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Your initial statement can be strengthened marginally: If $p(x_0) = 0$, then $|p|$ is differentiable at $x_0$ iff $p$ is differentiable at $x_0$ and $p'(x_0) = 0$. – copper.hat Aug 19 '12 at 16:28

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