Welcome to our community

Be a part of something great, join today!

[SOLVED] If Derivative is Not Zero Anywhere Then Function is Injective.

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
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

Indicium Physicus
Staff member
Jan 26, 2012
4,191
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

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
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

Well-known member
MHB Math Scholar
Mar 10, 2012
834

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
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

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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 $\Bbb R$).