# Determining Injectivity by Calculation of the First Derivative

Firstly, I was taught that the injectivity of a function can be determined by calculating its derivative- that if $$f'(x) > 0$$, it is injective. Can you give me the idea behind this method?

Secondly, it appears that this method doesn't apply to all functions. For instance, consider $$f(x) = \tan x , \qquad f'(x) = \sec^2 x > 0$$ But $$\tan x$$ is not injective!

So, it would be great if you could also specify the constraints for the use of this method.

The function $$f$$ is required to be continuous and differentiable (on some open interval). The general idea is that, if $$f'(x) > 0$$ the function is strictly increasing, so no two points can be equal.

A rigorous proof is by contradiction:

Suppose $$f(a)=f(b)$$ for some $$a \ne b$$.

By Rolle's theorem, $$f'(c) = 0$$ for some $$a, which contradicts $$f'(x) >0$$.

Also notice that, on $$(-\pi/2, \pi/2$$), $$\tan x$$ is injective (and continuous and differentiable with $$f'(x)=\sec^2 x > 0$$, as you have noticed.)

If the function is strictly decreasing ($$f'(x)<0$$) a similar proof also shows that the function is injective.

The idea is that if $$f'(x) >0$$ for all $$x\in I$$, where $$I$$ is an interval (possibly the whole real line) then the function is strictly increasing (this is proven using mean-value theorem, which i hope you find graphically intuitive). Then it is easy to show that strictly increasing functions are injective (again i hope you can see this easily graphically). The issue with tangent function is that it's domain is not a single interval. If you restrict to one interval (such as $$(-\pi/2, \pi/2)$$) then of course tangent is an injective function there (which is why we can define arctan).

Note that it is in the application of the mean-value theorem that you need the domain you're looking at to be an interval.

There are, really, two (sufficient) conditions,which make the result a direct consequence of Rolle's theorem:

1. The domain of the function is connected,
2. The function is differentiable on its domain and the derivative has a constant sign.

The $$\tan$$ function doesn't satisfy the first condition, so you can only conclude from the sign of the derivative that it is increasing (hence injective) on each interval of its domain.

• What do you mean by the domain being 'connected'? Commented Oct 5, 2020 at 13:43
• It's a topological notion, which intuitively means ‘in one piece’. In the case of $\mathbf R$, connected subsets are just intervals. Commented Oct 5, 2020 at 13:51
• Alright, Thanks:) Commented Oct 5, 2020 at 13:57