If a continuous, diff'l function $f: \mathbb{R} \rightarrow \mathbb{R}$ has at most one inflection pt is it 1-1?

Question

One of my calculus students asked me this. More precisely:

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous & differentiable function with the property: $f'(s) = 0$ $\implies$ $s$ is unique and $s$ is an inflection point. Is $f$ necessarily one-to-one?

I think the answer is yes, although I haven't gotten a lot of sit-down time to work out the details. I am thinking this:

Assume $f$ is not one-to-one. Then there is $a < b$ such that $f(a) = f(b)$. By the Mean Value Theorem, there is $a \leq c \leq b$ such that $f'(c) = 0$. Thus $f$ has a critical point. By uniqueness of $s$, we must have $c = s$, and hence $c$ is an inflection point. [The part I am shaky on:] Since $c$ is an inflection point, $f'$ is, say, decreasing on $(-\infty, c)$ and increasing on $(c, \infty)$. That is, $f$ is increasing on $\mathbb{R}$, whence $f$ is one-to-one. This contradicts that $f(a) = f(b)$.

I feel like I am missing details, or that there may be a hole in this argument - especially in the second half. Any help is appreciated.

Answer 1

We know that $f'$ is non zero and has the same sign on each of $(-\infty, s)$ and $(s, \infty)$.

Since $f$ has an inflection point at $s$, $f'$ has an isolated extremum at $s$ and hence $f'$ has the same sign everywhere. By taking $-f$ as necessary, we can assume that $f'(x) \ge 0$.

Hence $f$ is strictly increasing on $(-\infty, s) \cup (s, \infty)$ and hence everywhere.

Addendum:

As Eric noted in the comment below, $f$ is not assumed to be $C^1$, just differentiable, so my first point needs a little elaboration (it would be straightforward if $f$ was $C^1$). The relevant result here is Darboux's theorem which is an intermediate value theorem for differentiable functions. In particular, it states that $[f'(a),f'(b)] \subset f'([a,b])$.

So, for example, if $x,y < s$, then $f'(x),f'(y)$ must have the same sign, otherwise there would be some point $\zeta \in [x,y]$ such that $f'(\zeta) = 0$ which would contradict $s$ being the only such point.

Answer 2

Yes. Since $s$ is an inflection point, $f'$ does not change sign: that is, there is a neighborhood of $s$ on which $f'$ is always either nonpositive or nonnegative. Suppose that $f'$ is nonnegative in a neighborhood of $s$ (the other case is similar). Then $f'$ must be nonnegative everywhere, since it satisfies the intermediate value theorem and so if it were negative somewhere then it would have to pass through $0$ in between that point and our neighborhood of $s$. So $f$ is increasing, and is in fact strictly increasing since there is no interval on which $f'$ is $0$. In particular, $f$ is injective.