# Is twice-differentiability at a point required for there to exist a point of inflection?

I'm going off of a similar post and I don't quite understand what mathematicians believe - can $f(x)$ have a point of inflection at $c$ if $f''(c)$ does not exist? The post I linked seems to hint that this is not possible but does not address if there is a change in concavity. If the concavity changes, then shouldn't there exist a point of inflection at the point even if its second derivative does not exist?

Definitions differ, but if you define a point of inflection as one where $f$ changes concavity (i.e. concave in $(x-\epsilon, x]$ and convex in $[x,x+\epsilon)$ or vice versa), then yes, it does not require $f''(x)$ to exist, or even $f'(x)$ to exist.
You don't need existence of $f''$ at a particular point in order to have a point of inflection at that point. A point of inflection is a point at which concavity changes. To say that $f$ is concave downward on a particular interval means that all of the chords of the graph within that interval are on or below the graph. A chord of the graph is just a straight line segment whose endpoints are on the graph. Similarly, upward concavity on an interval means all of the chords within that interval are on or above the graph.
The function $x\mapsto\sqrt[\Large 3] x$ has a point of inflection at $x=0$ although the second derivative does not exist at that point. So does this function: $$f(x) =\begin{cases} \phantom{-}x^2 & \text{if } x\ge0, \\ -x^2 & \text{if } x<0. \end{cases}$$