I want to know what properties does the non-zero points in the domain of $f(x)=x^3$ have?
For example, $0$ is an inflection point, since the curvature changes sign at $0$ ($f''$ changes sign near $0$). In fact it is an stationary point as well $(f'(0)=0)$. So, it is also a saddle point.
What about if $x\neq 0$. Are these points inflection point or something else? I think such points ($x_0\neq 0$) are not inflection points nor saddle points, since then $f''(x_0)=6x_0\neq 0$ if $x_0\neq 0$.
But from the graph, it seems such points are neither maximum nor minimum points. Can someone please explain in detail what should these non-zero points be called?
Thanks a lot.
First Derivative Test:
A necessary condition for a function $f: \mathbf{R} \rightarrow \mathbf{R}$ to have a critical point at $x_0 \in \mathbf{R}$ is that $f'(x_0) = 0$.
Using this test, it is easy to deduce that the only critical point for $$ f(x) = x^3 $$ is at the origin $x = 0$.
It is easy to verify that $x = 0$ is a saddle point for $f$ since $$ f''(0) = 0, \ \ f'''(0) \neq 0 $$
The question is about the non-zero points $x \neq 0$.
Obviously they cannot be critical points since $f'$ vanishes only at $x = 0$.
Geometrically, the function $f$ is concave upwards in the region $x > 0$.
The function $f$ is concave downwards in the region $x < 0$.
(The function $f$ changes from concave downwards to concave upwards at $x = 0$.)