# Do absolute values of differentiable functions with roots always exhibit cusps?

I am interested in the following question because I am hoping to be surprised by an exotic counterexample.

Conjecture: Let $x=a$ be a root of a differentiable function $f$ such that $f(x)>0$ for $k<x<a$ and $f(x)<0$ for $a<x<q$, where $k, q \in \mathbb{R} \cup \{\pm \infty\}$. Then $x=a$ is a cusp of $|f(x)|$.

I am essentially conjecturing that if you have a differentiable function $f$ which is non-constant around one of it's roots, then the absolute value function of $f$ will be non-differentiable at the root of $f$ specified. Would anyone be able to think of a proof sketch for this claim, or suggest a counterexample?

-
$f(x) = - x^3 .$ Note that I am not sure what you mean by a cusp, but $|x|^3$ is a $C^1$ function. –  Will Jagy Sep 2 '12 at 5:03
To add to Will's (correct) comment, note that your conjecture will be violated by any function which approaches $0$ "sufficiently slowly". –  Alex Becker Sep 2 '12 at 5:11

As the comments to your question show, you don't have to go to very exotic functions to get a counterexample to your conjecture; it suffices for $f$ to have derivative $0$ at its root (and to satisfy the remainder of your requirements). The simplest kind of counterexample would be $x\mapsto x\,|x|$, that is $$f(x)=\begin{cases}x^2&\text{if }x\geq0\\-x^2&\text{if }x<0.\end{cases}$$ Now $|f(x)|=x^2$ is as nice as one could hope a function to be. Of course the original function is now only once differentiable at $0$, but you did not ask for more. If you want an example where both the original function and its absolute value are $\mathcal C^\infty$ (infinitely often differentiable everywhere) then contemplate $$g(x)=\begin{cases}xe^{-1/x^2}&\text{if }x\neq0\\0&\text{if }x=0.\end{cases}$$
A function that satisfies your condition will be differentiable in absolute value if and only if the left and right semi-derivatives are zero, i.e iff $$\lim_{t \to 0^-} \frac{f(a+t) - f(a)}t = \lim_{t \to 0^+} \frac{f(a+t) - f(a)}t = 0,$$ because by continuity you have $f(a) = 0$ and you know that $|f(a+t)| = f(a+t)$ for $t < 0$, $|f(a+t)| = -f(a+t)$ for $t > 0$, so putting all this together means that for $|f(x)|$ to be differentiable at $x=a$, you need $$\lim_{t \to 0^+} \frac{-|f(a+t)|}t = \lim_{t \to 0^+} \frac{f(a+t)}t = \lim_{t \to 0} \frac{f(a+t)}t = \lim_{t \to 0^-} \frac{f(a+t)}t = \lim_{t \to 0^-} \frac{|f(a+t)|}t.$$ This gets you to the desired conclusion, with some understanding.