In Wikipedia and many other places, it is stated that $f''(a)=0$ is a necessary condition for a function to have an inflection point at $x=a$.
I was wondering, if a function had a vertical tangent, would an inflection point also exist there?
For example, let $f(x)=x^{1/3}(x-1)$,
$f'(x)=\frac 4 3 x^{1/3} - \frac 1 3 x^{-2/3} $ which has a root only at $x=1/4$ and for $x=0$, it has a vertical tangent.
$f''(x) = \frac 4 9 x^{-2/3} + \frac 2 9 x^{-5/3}$ which has a root only at $x=-1/2$.
By nth derivative test or by observing that $f''(x)$ changes sign in the neighborhood of $-1/2$ we conclude that $x=-1/2$ is an Inflection point.
But on drawing the graph of $f(x)$, we observe that even $x=0$ is an inflection point, where the nature of the function changes from concave down to concave up.
So is my assertion true that it is not necessary for the double derivative of a function to be $0$, to have an inflection point, provided a vertical tangent exists at that point?