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
and $f(x)<0$ for $a , 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?