Proof of onto under interval $(0,1)$ How do I prove that the function $$f:(0,1)\rightarrow \mathbb R$$ defined by:
$$f(x) = \frac{-2x+1}{(2x-1)^2-1}$$
is onto?
 A: Just check that$$\lim_{x\to0^+}f(x)=-\infty\text{ and that }\lim_{x\to1^-}f(x)=+\infty.$$Now, the fact that the restricton of $f$ to $(0,1)$ is onto is a consequenc of the intermediate value theorem.
A: Note that 


*

*$\lim_{x\to 0^+} f(x)=-\infty$

*$\lim_{x\to 1^-} f(x)=\infty$


therefore since $f(x)$ is continuous for $x\in(0,1)$ by IVT we have that $f(x)$ is onto.
Moreover note that for $x\in(0,1)$ we have that $f(x)$ 
$$f'(x)=\frac{2x^2-2x+1}{x(x-1)^2x^2}\ge 0$$
therefore it is strictly increasing and also one to one on that interval.
Therefore $f(x):(0,1)\to \mathbb{R}$ is also bijective and invertible.
A: Alternatively, set $t = 2x-1$ and for each $r \in \mathbb{R}$ we show we can find a number $t$ and then an $x \in (0,1)$ such that$f(x) = r$. We have: $\dfrac{t}{1-t^2}=r$. So this gives $t = r - rt^2\implies rt^2+t -r = 0$. If $r = 0 \implies t = 0\implies x = \dfrac{1}{2}$. If $r \neq 0$, $t = \dfrac{-1\pm \sqrt{1+4r^2}}{2r}$. If $r > 0$, take $t = \dfrac{-1+\sqrt{1+4r^2}}{2r}>0$, and $t = \dfrac{2r}{1+\sqrt{1+4r^2}}< \dfrac{2r}{\sqrt{1+4r^2}}<1\implies t \in (0,1)\implies x = \dfrac{t+1}{2} > \dfrac{1}{2} > 0$, and $x < \dfrac{1+1}{2} = 1\implies x \in (0,1)$. If $r < 0$, choose $t = \dfrac{-1+\sqrt{1+4r^2}}{2r}\implies x = \dfrac{-1+2r+\sqrt{1+4r^2}}{4r}$. Observe that with $r < 0\implies \sqrt{1+4r^2} < 1-2r$ ( clear ),and this yields $-1+2r+\sqrt{1+4r^2} < 0\implies x > 0$. To complete the proof, you show $x < 1$. But $x = \dfrac{1-2r-\sqrt{1+4r^2}}{-4r}< 1 \iff 1-2r-\sqrt{1+4r^2} < -4r\iff 1+2r < \sqrt{1+4r^2}\iff 4r < 0$, and this is true since $r < 0$ ( by assumption ) . 
