<s>[INST] 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? 
 [/INST] $f(x) = \frac{-2x+1}{(2x-1)^2-1}$
$f'(x) = \frac {(-2x+1)'[(2x-1)^2 -1]- (-2x+1)[(2x-1)^2 - 1]'}{[(2x-1)^2 - 1]^2}$
$=\frac {-2[(2x-1)^2 - 1]-(-2x + 1)2(2x-1)2}{[(2x-1)^2 - 1]^2}=\frac {2[1-(2x-1)^2] +4(2x -1)^2}{{[(2x-1)^2 - 1]^2}}$
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 $1-1$.
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