How do I check if a function has an inverse function?

Question

From what I've learnt, a function $f$ has an inverse function $f^{-1}$ if the function $f$ is injective (one-to-one, horizontal rule applies).

How can I check if a function has an inverse in the first place? Given two examples:

1:

$f(x) = \frac{\arcsin x-1}{\arcsin x+2}$

its inverse is:

$f^{-1}(x)= \sin \frac{-2x-1}{x-1}$

2:

$g(x) = \frac{\ln x}{x}$

its inverse is:

$g^{-1}(x)= e^{\frac{-x}{x-1}}$

But how do I properly check if a function is an injective fuction? Preferably without using derivatives.

Thanks.

Answer 1

If you want to check that a function is injective, of course you need the formal definition

A function $f:A\to B$ is injective if and only if $f(x)=f(y)$ implies $x=y$ for all $x,y\in A$.

Let's take your second example, $f:(0,\infty)\to\mathbb R$, where $f(x)=\ln x/x$. Supopse $f(x)=f(y)$, that is, $$ \frac{\ln x}{x}=\frac{\ln y}y. $$ This implies $$y\ln x = x\ln y \iff xe^y=ye^x.$$ But if you look at the graph of the implicit equation $xe^y=ye^x$, you see that for any $x$ on $(0,\infty)$, there is in fact two $y$s in the same range which satisfy the equation. That is, $f(x)=f(y)$ does not imply $x=y$, so this function is not injective.

By the way, it is wrong to say that $f$ has an inverse if and only if it is injective, if we accept ``inverse'' to mean a function $g$ so that $g\circ f\equiv f\circ g\equiv 1$ where $1$ is the identity mapping. We need instead $f$ to be bijective; i.e. both injective and surjective.