Question

I have the graph \[f(x)=\frac{ x }{ 1-x }, x \in]0,1[\]. And I need to show that it is a bijection of the interval \[]0,infinity\] on the interval \[]0,infinity[\]

Answer 1

$$f(x)=\frac{x}{1-x}=\frac{1}{1-x}-1$$we want to show it's a bijection between \((0,\infty)\) and \((0,\infty)\).

consider that the singularity at \(x=1\) suggests we investigate \((0,1)\) and \((1,\infty)\) separately. note that: $$f(0)=0\\\lim_{x\uparrow1} f(x)=\infty\\\lim_{x\downarrow1}f(x)=-\infty\\\lim_{x\to\infty} f(x)=-1$$
and clearly \(f'(x)=1/(1-x)^2>0\) on both \((0,1),(1,\infty)\) so our function is strictly increasing

Answer 2

so clearly it is a bijection between \((0,1)\) and \((0,\infty)\), and so it cannot be a bijection between \((0,\infty)\supset (0,1)\) and \((0,\infty)\)