Showing a function is injective using that $f'(x)\ne0$

Question

Given a differentiable function $f\colon \mathbb R\to\mathbb R$ which we must prove to be injective, does it suffice to show $f'(x)≠0$ for all $x$ (for which the function is defined)? It makes sense, of course, due to the mean value theorem, but maybe there's some subtleties I am missing since this method doesn't seem to be used very often.

Answer 1

Yes, this suffices.

To see this suppose $f$ is differntiable on the domain of consideration but not injective. Then $f(a)=f(b)$ for some $a,b$ with $a\neq b$ and by the mean value theorem the derivative has a zero between $a$ and $b$.

Most of the time a function is not describable by a differentiable function so this method is used less often then by directly showing the claim. In algebra and general set theory for example this method is basically never applicable.

Answer 2

Yes, it is sufficient to show that $f'(x) \neq 0$ for all $x$.

One way to show the sufficiency is using Darboux's theorem. From this theorem, it follows that if we have $a,b$ such that $f'(a) > 0$ and $f'(b) < 0$, then there is a $c$ such that $f'(c) = 0$.

Thus, if we have $f'(x) \neq 0$ for all $x$, we may conclude that either $f'(x) \geq 0$ or $f'(x) \leq 0$ for all $x$. Thus, $f'(x) > 0$ for all $x$, or $f'(x) < 0$ for all $x$, since we never have $f'(x) = 0$. The conclusion follows.

Answer 3

There's a little bit of subtlety. The mean value property will work if you can conclude that $f'$ is either always positive or always negative. But how do you conclude this from $f' \neq 0$?

If the derivative is assumed continuous, that will suffice, because if you know that $f'$ is positive at some point, and never zero, then it must remain positive, by the intermediate value theorem applied to $f'$.

But even if you don't assume $f'$ is continuous, you can still conclude $f'$ remains strictly positive or strictly negative, because a derivative always has the intermediate value property, namely:

If $f'(x) = a$ and $f'(y) = b$, and $a<c<b$, then there is some point between $x$ and $y$ where $f'(x)=c$.

That is true even if $f'$ is not continuous. To see how one can hold but not the other, consider the function $$x \mapsto \begin{cases} \sin (1/x) & \text{if } x \neq 0 \\ 0 & \text{if } x =0, \end{cases}$$ which has the intermediate value property but is not continuous at zero*. This allows you to conclude that if $f'$ is never zero, it must remain strictly positive or strictly negative.

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*And in fact it is the derivative of a function, namely $x^2 \cos(1/x) - 2 \int_0^x \xi \cos(1/\xi) \, d\xi$.

Answer 4

This is an interesting question, and I have a few thoughts. You are right that $f'(x)\neq 0$ implies $f$ is injective. Here are some reasons I think it may not usually be used:

• Injective functions designed to be easily used for set theory problems may not be differentiable.
• As far as the use of injective functions, functions $\mathbb R\to\mathbb R$ are used less than functions between more complicated spaces, where the mean value theorem doesn't apply.
• In introductory courses when injectivity is first described, differentiation has not yet been rigorously introduced.
• This condition is not necessary for injectivity.

Answer 5

Yes. This is simply the inverse function theorem. If $f$ is continuously differentiable with nonzero derivative at some point $a$, then $f$ is invertible in a neighborhood of $a$.

Answer 6

Rolle's theorem state that if $f$ is differentiable and continuous on $[a, b]$, with $a \neq b$, you have:

$f(a) = f(b) \implies \exists c\in (a,b): f'(c) = 0$

Using the contrapositive:

$\forall c \in (a,b): \neg \left[f'(c) =0\right] \implies f(a) \neq f(b)$

And starting from $a \neq b$ you got $f(a) \neq f(b)$ that is just the desired result.

Also, this implies that $f$ will be stricty monotonous function(using fermat's theorem).

In the trivial case $a = b$ is the same point, so there is nothing to do.