Show bijectivity of $f:(-1,1)\rightarrow \mathbb{R}, f(x)=\frac{x}{1-|x|}$

Question

Show bijectivity of $f:(-1,1)\rightarrow \mathbb{R}, f(x)=\frac{x}{1-|x|}$

So in order to show injectivity $f(a)=f(b) \Rightarrow a=b$

so $\frac{a}{1-|a|}=\frac{b}{1-|b|}$. But how do I prove that?

For the surjectivity I have to show that $f(X)=Y$ when $X$ is the domain and $Y$ the Image.

so I have to prove that $f$ image is $f(x)\in \mathbb{R}$. But how do I do that...?

Answer 1

Prove that $f$ is strictly increasing. Then, $x<y\implies f(x)<f(y)$. And it is surjective bby the intermediate value theorem and because $\lim_{x\to\pm1}f(x)=\pm\infty$.

If you want to avoid limits, you can solve the equation $f(x)=y$. Suppose that $y\geqslant0$. Then $x$ will have to be non-negative. So$$f(x)=y\iff\frac x{1-x}=y\iff-1+\frac1{1-x}=y\iff x=\frac y{1+y}.$$And, if $y\leqslant 0$, a similar argument shows that can simply take $x=\frac y{1-y}$.

Answer 2

For injectivity note that for $a\in (-1,1)$ and $b\in (-1,1)$ we have $1-|a|>0$ and $1-|b|>0$, so $a$ and $b$ have the same sign.

Thus $$\frac{a}{1-|a|}=\frac{b}{1-|b|}\implies a=b$$

For surjectivity let $$x=\frac{y}{1+|y|}$$ then you have $f(x)=y$

The trick with the absolute value is that $x$ and $f(x)$ always have the same sign.

Answer 3

I rewrite your function for determined domain bellow:

$$f(x)=\begin{cases} \frac{x}{1-x}, & \mbox{if } 0\le x <1 \\ \frac{x}{1+x}, & \mbox{if } -1< x <0 \end{cases}$$

by taking first derivative of this function we reach to this:

$$f^{'}(x)=\begin{cases} \frac{1}{(1-x)^2}, & \mbox{if } 0< x <1 \\ 0, & \mbox{if } x=0 \\ \frac{1}{(1+x)^2}, & \mbox{if } -1< x <0 \end{cases}$$

so for all time we have $f^{'}(x)\ge0$ so for defined domain this function is bijective.