Generally, the method used is - for the function, $f$, to be one-one we prove that for all $x, y$ within domain of the function, $f$, $f(x)=f(y)$ implies that $x=y$.
Another method is by using calculus.
If $f '(x)>0$ or $f '(x)<0$ for all $x$ in domain of the function, then the function is one-one. But if $f '(x)=0$ at some points (let the set of such points be $A$) then at those points we check $f ''(x)$. If $f ''(x)$ is not equal to zero at all points in set $A$, then the function is not one-one. If $f ''(x)=0$ at some set, $B$ of points (set $B$ is a subset of set $A$) , then we check $f '''(x)$ at all points in set $B$. If $f '''(x)$ is not equal to zero at all those points in set $B$ then the function is one-one.
My problem here is -
- Is the above method (using calculus) correct to the extent that I have written it? If it is correct, then please complete it (since I have not considered what happens if $f '''(x)=0$ ). If it is incorrect, please give the reasons as to why it is incorrect and provide (if there exists) another method to prove that the function is one-one (apart from the methods mentioned above).
- Do we have to continue going like this for eternity? Checking $f ''''(x)$, then $f '''''(x)$, then $f ''''''(x)$......(you get the idea) if they successively come out to be zero? Or do we just prove that every derivative after a certain derivative is 0, and leave it at that?
- While writing the method, I have written it, keeping only functions having both domain and range as subsets of real numbers, in mind. In case the method for other kinds of functions is different (which, in my opinion, should not be the case), please elaborate on that front too.
Please try and use examples where possible
Let $$f: (0,1) \cup (2,3) \rightarrow \mathbb{R}$$
$$f(x) = \begin{cases} x, & x \in (0,1) \\ x-2 & , x \in(2,3)\end{cases}$$
The function is not injective as $f(\frac12)=f(\frac52).$
I don't think derivative alone is sufficient to conclude that it is injective.