Suppose that we have a function $f: \mathbb{R} \rightarrow \mathbb{R}$ that is defined on a closed, bounded, and continuous set. Suppose further that there are one or more inflection points in this set.
Is it possible for $f$ to be locally invertible at any of the inflection points? My intuition tells me no, but I don't know how to show this in a rigorous manner if it is indeed true.
Sure, all functions $$f(x) = x^{2n+1}$$ for $n\ge 1$ are such functions.
Their inflection points are at $x=0$ and they are locally invertible at $x=0.$