Proof of 1-1 under an interval (0,1)

Question

How do I prove that the function

$$f(x) = \frac{-2x+1}{(2x-1)^2-1}$$

is one-to-one from 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?

Answer 1

Note first that $(2x-1)^2 - 1 = 0 \Longrightarrow (2x-2)2x = 0 \Longrightarrow (x = 1 \vee x = 0$ , or, in interval (0,1) this function is well-defined. Now. as both $-2x+1$ and $(2x-1)^2-1$ are differentiable, and the latter has no zeros where we are interested, the quotient is differentiable. Now, we count : $f' = \frac{-2 (2x -1)^2 - 1) - (-2x + 1) (4x-4)}{((2x-1)^2 - 1)^2} = \frac{-2(4x^2-4x) - (-8x^2+12x -4)}{(4x^2-4x)^2} = \frac{4 -4x} {(4x^2-4x)^2}$ Note that this is strictly greater then zero on $(0,1)$. So, the function has a constant rise, hence is one-to-one

Answer 2

Let $$f(x) = f(y)\\ \implies \frac{-2x+1}{(2x-1)^2 -1} = \frac{-2y+1}{(2y-1)^2 -1} \\ \text{Let } t=(2x-1), u=(2y-1) \\ \implies t(u^2-1) = u(t^2-1) \\ \implies (ut+1)(u-t) = 0 $$

Now, $0 < x < 1 \implies -1 < 2x-1 < 1$. Hence, $-1 < u,t < 1$. So, $ut+1>0$ Therefore, $u-t = 0 \implies x=y$. Hence, $f$ is injective on $(0,1)$. Proved.

Answer 3

You can rewrite the function as $$ f(x)=\frac{-2x+1}{4(x^2-x)} $$ Suppose $x,y\in(0,1)$ and $f(x)=f(y)$. Then $$ (-2x+1)(y^2-y)=(-2y+1)(x^2-x) $$ that can be simplified into $$ 2xy(x-y)-(x^2-y^2)+x-y=0 $$ or, equivalently, $$ (x-y)(2xy-x-y+1)=0 $$ If $x\ne y$, we conclude $$ 2xy-x-y+1=0 $$ or $$ y=\frac{1-x}{1-2x}=\frac{1}{2}\left(1+\frac{1}{1-2x}\right) $$ Note that $1-x>0$, so in order to have $y>0$ we need $0<x<1/2$. This implies $0<1-2x<1$, so $$ \frac{1}{1-2x}>1 $$ and therefore $$ y=\frac{1}{2}\left(1+\frac{1}{1-2x}\right)>1 $$ which is a contradiction.

Answer 4

$y=2x-1$, $x \in (0,1)$ is

strictly increasing (why?).

We have $y \in (-1,1)$.

$F(y)= - \dfrac{y}{(y-1)(y+1)} = $

$-(1/2)[\dfrac{1}{y-1} +\dfrac{1}{y+1}]$

$F'(y) =$

$(1/2)(y-1)^{-2} +(1/2)(y+1)^{-2} >0.$

Hence $F$ is strictly increasing.

Combining:

$F(y)= F(y(x))$.

Since $y(x)$ is strictly increasing, and $F(y)$ is strictly increasing $F(y(x))$ is strictly increasing, i.e.

$x_1 < x_2$ then $y_1<y_2$ and finally

$F(y_1) <F(y_2)$.