Show that f is 1-1

[kmath]

I need to show that [latex] f: (0,1) \rightarrow R[/latex] is 1-1, where [latex] f(x) [/latex] is given by [latex] (2x-1)/(x^2-x) [/latex].

 

My attempt: suppose:[latex] f(x) = f(y)[/latex]

 

Then: [latex] (2x-1)/(x^2-x) [/latex] = [latex] (2y-1)/(y^2-y) [/latex].

 

[latex]y(y-1)(2x-1) = (x(x-1))(2y-1)[/latex]

 

Reducing this leads to:

 

[latex]2xy^2 - y^2 + y = 2x^2 - x^2 + x[/latex]

 

 

Of course, I need to show that [latex]x = y[/latex] but I'm not sure how to reduce this equality any further. Any ideas?

 

Thanks

Edited May 1, 2013 by kmath

[krash661]

hmm, it's been a while

 

sorry i have to refresh my memory to help.

 

same value maybe ?

Edited May 1, 2013 by krash661

[shah_nosrat]

Reducing this leads to:

 

[latex]2xy^2 - y^2 + y = 2x^2 - x^2 + x[/latex]

 

 

Of course, I need to show that [latex]x = y[/latex] but I'm not sure how to reduce this equality any further. Any ideas?

 

Thanks

There's a mistake in your equality; it's supposed to be [latex]2xy^2 - y^2 + y = 2yx^2 - x^2 + x [/latex], factoring both terms on each side of the equality we have: [latex] y(2xy -y + 1) = x(2xy -x + 1) [/latex], which we can argue for x to equal to y we should have [latex]2xy -y + 1 = 2xy - x + 1[/latex], which reduces to [latex]x=y[/latex] as required, hence, your function is injective.