Special case of Point of Inflection or an incorrect interpretation?

Question

It's said that if $f''(a) = 0\ \text{ and } \,f'''(a) \neq 0\, (i.e., f'''(a)\,$ is actually some well-defined number and that number is not $0$), then there is an inflection point at $x=a$.

Let's say,
$f(x) = 4x^{5}$

$f '(x) = 20x^{4}$

$f ' '(x) = 80x^{3}$

$f ' ' '(x) = 240x^{2}$

Now, it can be said that $x=0 $ is an inflection point of $f(x)$ as $f ' '(0)=0$ and $f ' '(x)$ changes sign on either side of $x=0$. Here, if we proceed to $f ' ' '(x)$ without checking the sign of $f ' '(x)$, then we can see that $f ' ' '(0)=0$ which contradicts the rule above that states if $f ' '(a)=0$ and $f ' ' '(a)≠0$ then $x=a$ is said to be an inflection point of $f(x)$. In this example, $x=0$ is an inflection point while $f ' ' '(0)=0$. How is this possible? Is this a special case that disobeys the rule?

Answer 1

Your logic is wrong.

The statement says that if $f''(a) = 0$ and $f'''(a) \neq 0$ then $f$ has an inflection at argument $a$. That is a true statement.

It does not say that if $f$ has an inflection at argument $a$, then necessarily it must be true that $f''(a) = 0$ and $f'''(a) \neq 0$. Your example shows that this is not true.

So the two statements "$f''(a) = 0$ and $f'''(a) \neq 0$" and "$f$ has an inflection at argument $a$" are not equivalent. From the first follows the second, but not vice versa.

Generally the theory says that if you have, $f'(a)=f''(a)=\ldots = f^{(n)}(a)=0$, but $f^{(n+1)}(a) \neq 0$, then $f$ has a local extremum if $n$ is odd and an inflection point if $n$ is even. In your choice of $f$, $n=4$ for $x=0$, so there is an inflection point at argument $0$.

Answer 2

Note that what you have is a sufficient condition for inflection, i.e., if certain conditions are met $\implies$ we have inflection point. They are by no means necessary.

In fact, by looking at monomials $x^n$ with $n=0,1,2,\dots$ you can always construct cases with/without inflection points (depending on whether $n$ is odd or even), which cannot be distinguished based only on the first few derivatives. One can take it even further, by considering smooth functions $$ f_{\pm}(x) = \begin{cases} \exp(-1/x^2), &\qquad x\geq 0,\\ \pm\exp(-1/x^2), &\qquad x<0.\end{cases} $$ For $f_{+}$, $a=0$ is not an inflection point, but all its derivatives at $a=0$ vanish. For $f_{-}$, $a=0$ is an inflection point, while all its derivatives at $a=0$ vanish. So one can always construct functions, where we cannot make a determination based on infinitely many derivatives (at a point) either.