Let $f:(0,1)\rightarrow \mathbb{R}$ such that $f(x) = \frac{2x-1}{x-x^2}$. Prove that $f$ a bijection.
I proved that it's injective like this:
Let $x_1, x_2 \in (0,1)$ and $f(x_1)=f(x_2)$, then
$$\frac{2x_1-1}{x_1-x_1^2} = \frac{2x_2-1}{x_2-x_2^2} \implies (x_1-x_2)(2x_1x_2-x_1-x_2+1)=0$$ So $x_1=x_2$ or $(2x_1x_2-x_1-x_2+1)=0$.
The second case takes us to $x_2 = \frac{x_1-1}{2x_1-1}$, if $x_1>0.5$ then $x_2<0$ which is a contradiction.
if $x_2<0.5$ then $1-x_1>1-2x_1$ and $x_2>1$ which is also a contradiction.
if $x_1=0.5$ then we find $0=0.5$ which is a contradiction, so the first case must be true. Which means $f$ is injective.
But how can I prove that $f$ is surjective? (I can't find an inverse function for $f$).