I am wondering about the following question: A strictly convex (concave) differentiable function $f:\mathcal{R}\to\mathcal{R}$ has the geometrical property that its graph lies completely above (below) the tangential line at any point (except at the point of contact). Now, this is a special property of such functions but is a relaxed version of this also valid for all (not necessarily continuously) differentiable functions? More specifically, let's call $x\_0$ a *convex* point if there is a neighborhood of $x\_0$ such that
$$f(x) > f(x\_0) + f'(x\_0)(x- x\_0)$$
for all $x$ in that neighborhood. Similarly, let's call $x\_0$ concave if the reverse inequality holds for some neighborhood:
$$f(x) < f(x\_0) + f'(x\_0)(x- x\_0)$$
Then, is it true that any non-linear differentiable function has at least one concave or convex point?


Full disclosure: this question was first stated as true on a different forum without any additional information or explanations. It is possible this is well-known and I have never come across this result, or that this is completely wrong. Regardless, I am hoping to get some additional information as I think this is quite a neat result either way.



Convex/concave points of a differentiable function
Let $f$ be non linear in the interval $[-1,1]$. We can suppose $f(-1)=f(1)=0$ and $f(x)>0$ for some $x\in(-1,1)$. Now let $N$ be so big that the closed ball of center $(0,-N)$ and with radius $\sqrt{N^2+1}$ doesn't contain the whole graph of $f$. Let $(x\_0,f(x\_0))$ be one of the points of the graph furthest from $(0,-N)$. Then $f$ is concave at $x\_0$.


Indeed, the whole graph of $f$ is contained in the ball of center $(0,-N)$ and boundary passing through $(x\_0,f(x\_0))$. This implies that the derivative of $f$ at $x\_0$ has to be the slope of the tangent to that ball at $(x\_0,f(x\_0))$, and that tangent only intersects the graph of $f$ at $(x\_0,f(x\_0))$ (in fact it only intersects the ball at that one point).


