How to prove that $f(x) = \frac{-2x+1}{(2x-1)^2-1}$ is one-to-one on $(0,1)$? How do I prove that the function
$$f(x) = \frac{-2x+1}{(2x-1)^2-1}$$
is one-to-one on the interval $(0,1)$? I have simplified the function $f(x)=f(y)$ and have a multivariable quadratic equation equal to zero. How do I conclude that the function is one to one? 
 A: Alt. hint: $\,h(x)=1-2x\,$ is a bijection, and $\,g(x) = \dfrac{x}{x^2-1}\,$ is not injective on $\,\Bbb R\,$ (though it is injective on $\,(-1,1)\,$ and $\Bbb R \setminus [-1,1]$ respectively). It follows that $\,f = g \circ h\,$ is not injective on $\,\Bbb R\,$.

[ EDIT ]   Addressing OP's clarification that the question refers to is one-to-one from the interval (0,1), note that $\,g(x)\,$ is indeed injective on $\,(-1,1)\,$ since $\,a,b \in (-1,1) \implies ab+1 \ne 0\,$ and:

$$
0 = g(a) - g(b) = \frac{a}{a^2-1} - \frac{b}{b^2-1}= \frac{ab^2-a-a^2b+b}{(a^2-1)(b^2-1)} \\[10px] = \frac{(ab+1)(b-a)}{(a^2-1)(b^2-1)} \quad\implies\quad a = b
$$
Given that $\,h\big((0,1)\big)=(-1,1)\,$ it follows that $\,f = g \circ h\,$ is injective on $\,(0,1)\,$.
A: Note that $f(x)$ has $2$ vertical asymptotes at 
$$(2x-1)^2=1 \implies 2x-1=\pm 1 \implies x=0,1$$
and it easy to check that


*

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

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

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

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

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

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


and therefore, since the function is continuous in its domain, by intermediate value theorem it can't be one to one.
A: $f(x) = \frac{-2x+1}{(2x-1)^2-1}$
$f'(x) = \frac {(-2x+1)'[(2x-1)^2 -1]- (-2x+1)[(2x-1)^2 - 1]'}{[(2x-1)^2 - 1]^2}$
$=\frac {-2[(2x-1)^2 - 1]-(-2x + 1)2(2x-1)2}{[(2x-1)^2 - 1]^2}=\frac {2[1-(2x-1)^2] +4(2x -1)^2}{{[(2x-1)^2 - 1]^2}}$
Now if $0 < x < 1$ then $-1 < 2x -1 < 1$ and $0\le (2x -1)^2 < 1$ and $0< 1-(2x-1)^2 \le 1$, while $[(2x-1)^2 - 1]^2 >0$ and $(2x -1)^2 > 0$.
So for $0 < x < 1$, $f'(x) > 0$ and $f$ is therefore strictly increasing and $1-1$.
A: You can't prove that, because it is not true. For instance,$$f\left(\frac{1-\sqrt5}4\right)=f\left(\frac{1+\sqrt5}4\right)=1.$$
A: One other pair of $x, y$ that give the same image are of the form: $$2xy-x-y+1$$ , with $(2x-1)^2-1 \neq0$  and  $(2y-1)^2-1 \neq0$
