The function $f:(0,1)\to(0,1)$ is defined by the rule $f(x)= \frac{2x}{1+x^2}$. Show that f is a one-to-one and onto function. The function $f:(0,1)\to(0,1)$ is defined by the rule $f(x)= \frac{2x}{1+x^2}$. Show that f is a one-to-one and onto function. Find the rule $f^{-1}(x)$ of the inverse function $f^{-1}$
I tried to prove this is an injective function by using $f(m) = f(n)$, but the result is false. Help guys!!!
 A: $\lim_{x\to0}f(x)=0$ and $\lim_{x\to1}f(x)=1$, use the Intermediate Value Property of continuous functions to see that  $f$ takes all values in $(0,1)$. For one-one see that you have two conditions $$(x-y)(xy-1)=0$$ now the case $xy=1$ is never satisfied by two numbers in $(0,1)$ can you observe this?. After you show one-one and onto, it remains trivial to show the inverse.
A: It depends on which analytic methods you are able to use. The most basic, and calculation heavy, would be to give the inverse function.
It is rather standard to compute the inverse.
We have to solve $x=\dfrac{2y}{1+y^2}$ for $y$.
After a few steps we get
$xy^2-2y+x=0$
For $x\neq 0$, where we note that $x\in (0,1)$ in the first place, we can divide by $x$ and get
$y^2-\dfrac2xy+1=0$.
Now we can use the quadratic formula to solve for $y$.
$y_{1,2}=\dfrac1x\pm\sqrt{\dfrac1{x^2}-1}$
As $\dfrac1{x^2}-1\geq 0$ holds for $x\in (0,1)$.
We have to find the correct choice for $y$.
This is $y=\dfrac{1}{x}-\sqrt{\dfrac{1}{x^2}-1}$ as this expression takes values in $(0,1)$.
So our inverse function is given by
$f^{-1}: (0,1)\to (0,1), x\mapsto \frac{1}{x}-\sqrt{\frac1{x^2}-1}$
We would have to calculate now $f(f^{-1}(x))$ and $f^{-1}(f(x))$ and show that is is $x$ respectivly.
I leave that to you.
The calculation might seem complicated, but it is straight forward.
Nevertheless time consuming and there are somethings to look out for. So it is not the nicest solution, but everything involved is taught in school, so it is rather elementary.
Also keep in mind that for the proof of surjectivity, when you go by definition, you have to give an inverse function anyways, or better said, when you try to find a preimage, you will calculate the inverse function anyways. So when you go straight by definition, you kinda have to do this calculation anyways, and can avoid to show that the function is injective, as you get that 'for free' when you find the left and right inverse.
A: Equation
$$f(x)=\frac{2x}{1+x^2}=\frac{2y}{1+y^2}=f(y)$$
is same as $xy^2-(1+x^2)y+x=0$  and have solutions only $(1,1)$ and $(0,0)$.
For any $y \in $ $0<\frac{2x}{1+x^2}=y<1$ we have $yx^2-2x+y=0$, $x_{1,2}=\frac{1}{y}(1\pm\sqrt{1-y^2})$ on $(0,1)$ and we have $x_1\cdot x_2=1$, which means, that only one from them is on $(0,1)$.
