Can a function have a inflection point at somewhere non-differentiable?

Question

Imagine this function :

$$ f(x)=\begin{cases} x^2,\quad x<0 \\ \sqrt{x},\quad x\ge0. \end{cases} $$

In my lecture, my professor told me $f(x)$ is not differentiable at $0$ but $f(x)$ has an inflection point at $x=0$.

I get the idea that when $x<0$, $f(x)$ is concave up and when $x>0$, $f(x)$ concaves down; but I believe that if $f(x)$ is not differentiable at $0$, then $0$ fails the definition of being an inflection point.

For context, this is how "inflection point" and concavity" is defined in my course:

Let $f$ be a differentiable function defined on open interval $I$.
We say that $f$ concave-up on $I$ when $f'$ is increasing on $I$.
We say that $f$ is concave-down on $I$ when $f'$ is decreasing on $I$.
Let $c$ be an interior point to $I$.
We say that has an inflection point at $c$ when
"$f$ changes concavity at $c$".
We say "$f$ changes concavity at $c$" when $\exists\delta > 0$ such that
$f$ is concave-up on $[c-δ,c]$ and concave-down on $[c,c+ δ]$
or vice versa.

So according to the definition, an inflection point must be differentiable because

1. it is defined within a context of a differentiable interval;
2. it is defined as a point when $f'$ is increasing on a small CLOSED interval with $c$ being the right endpoint and decreasing on a small CLOSED interval with $c$ being the left endpoint (or vice versa).

Both of these conditions require $f'$ to be defined at $c$ which means $f$ is differentiable at $c$.

However, for my $f(x)$ it is not differentiable at $c=0$.

The definition of an inflection point on Wikipedia also requires $f'$ to be defined at that point:

For example, the graph of the differentiable function has an inflection point at $(x, f(x))$ if and only if its first derivative $f'$ has an isolated extremum at $x$. (this is not the same as saying that $f$ has an extremum). That is, in some neighborhood, $x$ is the one and only point at which $f'$ has a (local) minimum or maximum.

So what do u guys think? Further more, consider this function: $$f(x)=\sqrt[3]{x}.$$

It's also non-differentiable at $0$ but concave up when negative and concave up when negative. Would you consider $0$ an inflection point?

Are there any alternative, more rigorous definitions of concavity that can handle these cases?

Answer 1

There's a flaw in the definition.

Let $f$ be a differentiable function defined on open interval $I$.
We say that $f$ concave-up on $I$ when ...

This gives us definitions of concavity on an open interval.

We say "$f$ changes concavity at $c$" when $∃δ>0$ such that
$f$ is concave-up on $[c-δ,c]$ and concave-down on $[c,c+ δ]$
or vice versa.

This requires concavity to be defined on a closed interval. But we only have a definition of concavity on open intervals.

A logical extension of the definition of concavity on open intervals to concavity on closed intervals would be that a function is concave up (for example) on a closed interval if it is concave-up on an open interval containing the closed interval. But that definition would prevent any function from having an inflection point.

In fact, as written, the definition makes no sense, because if a function is concave-up on an interval we would like to say it is concave-up on ever point of that interval. But then we have a definition that says $f$ must be both concave-up and concave-down at $c$ in order to have an inflection point at $c$.

When you look at it closely, it doesn't appear that we can even use this definition to conclude that $f=x^3$ has an inflection point at $c=0$.

All of these problems with the definitions are instantly cleared up if you change the closed intervals to open intervals. It seems likely that someone just made a mistake typesetting the intervals $[c-δ,c]$ and $[c,c+ δ]$; they should have used round brackets.

Answer 2

The french version of Wikipedia https://fr.wikipedia.org/wiki/Point_d%27inflexion is a bit less restrictive concerning the regularity assumptions on $f.$

In the sequel, "locally at $x_0$" means "on some interval centered at $x_0$".

Assume that $f$ is locally defined and continuous at $x_0$, that $f$ has a tangent at $x_0$, eventually a vertical one (so that $f$ is not necessarily differentiable at $x_0$), that $f$ is twice left-differentiable and twice right-differentiable locally at $x_0$, and finally that $x_0$ is locally the only possible point satisfying $f''(x)=0.$

Then the three following conditions are equivalent, and thus any of them may be taken as a definition for $x_0$ to being a flex point:

• the concavity of $f$ changes at $x_0$

• the graph of $f$ cuts the tangent at $x_0$

• the sign of $f''$ changes at $x_0$ ($f''(x_0)$ may exist or not).

In your case, $f$ presumably satisfy all the required conditions (i didn't check)

Answer 3

The main point is that one can define convexity/concavity without reference to the derivative at all, although your text does not do so. A function $f$ is convex [resp., concave] on an interval if for all $a$ and $b$ in the interval the line segment between $(a,f(a))$ and $(b,f(b))$ lies above [resp., below] the graph of $f$.

A point $c$ is an inflection point (whether $f$ is differentiable at $c$ or not) if the function is convex on some interval $(a,c)$ and concave on some interval $(c,b)$ or vice versa.

It is a fun exercise to show that if $f$ is convex (or concave) on some open interval, then it is necessarily continuous on that interval.