Given $f:\mathbb{R}\to\mathbb{R}$, $f(x)=x^3$, how could I show (using the definition of convexity and concavity of functions) that it has a unique inflection point?
Seeing as $f$ is continuous, I didn't use the general definition of convexity/concavity, but the Jensen one, i.e. (for convexity)
$$ f \left( \frac{x+y}{2} \right) \leq \frac{f(x)+f(y)}{2}. $$
First I wrote out what the equivalent condition would be for $f(x)=x^3$. With some simple algebraic manipulations I get that the equivalent condition for convexity (the above formulation) would be that
$$ (x-y)^2(x+y) \geq 0, $$
meaning $x+y\geq 0$, for all $x,y$ on the part of the domain on which $f$ is convex.
I could do the same for concavity and obtain the reverse inequality, however I'm not sure that helps me. Any hints?
EDIT: Since the condition for convexity is $x+y\geq 0$ for all $x,y$ on the part of the domain on which $f$ is convex, could I use the fact that then $x+x$ must be greater than $0$ (because obviously $x$ is in the same part of the domain as $x$), and with that I get $x\geq 0$.
Conversely, for concavity I get $x\leq 0$.
So the only point in which the function is both (or neither) is $x=0$?
