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.
 A: 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 $a<c<b$, 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.
A: 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.
A: There are, really, two (sufficient) conditions,which make the result a direct consequence of Rolle's theorem:

*

*The domain of the function is connected,

*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.
