If Derivative is Not Zero Anywhere Then Function is Injective.

[caffeinemachine]

Hello MHB.

I am sorry that I haven't been able to take part in discussions lately because I have been really busy.

I am having trouble with a question.

In a past year paper of an exam I am preparing for it read:

Let $f: (a,b)\to \mathbb R$ be a differentiable function with $f'(x)\neq 0$ for all $x\in(a,b)$. Then is $f$ necessarily injective?

I know that a function can be differentiable at all points and have a discontinuous derivative.
This makes me think that $f$ is not necessarily injective. But I am not able to construct a counterexample.

Can anybody help?

 

[Ackbach]

I know that a function can be differentiable at all points and have a discontinuous derivative.

It can? Can you come up with an example of a function that does this? For me, I think of $f(x)=|x|$. It is not differentiable at $0$; its derivative is discontinuous at the origin.

 

[Opalg]

Hello MHB.

I am sorry that I haven't been able to take part in discussions lately because I have been really busy.

I am having trouble with a question.

In a past year paper of an exam I am preparing for it read:

Let $f: (a,b)\to \mathbb R$ be a differentiable function with $f'(x)\neq 0$ for all $x\in(a,b)$. Then is $f$ necessarily injective?

I know that a function can be differentiable at all points and have a discontinuous derivative.
This makes me think that $f$ is not necessarily injective. But I am not able to construct a counterexample.

Can anybody help?

Rolle's theorem.

 

[caffeinemachine]

It can? Can you come up with an example of a function that does this? For me, I think of $f(x)=|x|$. It is not differentiable at $0$; its derivative is discontinuous at the origin.

calculus - Discontinuous derivative. - Mathematics Stack Exchange
See Mark McClure's answer. He provides such an example.
Also see Haskell Curry's answer. He doesn't provide an example but his post is useful.

 

[caffeinemachine]

Rolle's theorem.

Thanks.

 

[Ackbach]

Another result of interest, which I found here: the Darboux theorem. If a function is differentiable, then its derivative must satisfy the Intermediate Value property.

 

[Deveno]

Let us suppose by way of contradiction a counter-example exists.

Thus we have two points $c < d \in (a,b)$ such that:

$f(c) = f(d)$, but $c \neq d$.

By supposition, $c$ and $d$ are, of course, interior points of $(a,b)$, and thus since $f$ is differentiable on $(a,b)$, $f$ is continuous on $[c,d]$ and differentiable on $(c,d)$.

Hence we may apply the mean value theorem to deduce there exists a point $x_1 \in (c,d)$ such that:

$f'(x_1) = \dfrac{f(d) - f(c)}{d - c} = 0$

violating the condition $f'(x) \neq 0$ for all $x \in (a,b)$.

Thus no such pair exists, which thus means if for $c,d \in (a,b), f(c) = f(d)$, we must have $c = d$, that is, $f$ is injective.

(Note this proof takes advantage of the trichotomy rule, a consequence of the order properties of $\Bbb R$).