# Please find this inverse function

• Oct 22nd 2008, 11:46 AM
petike
Hi all,
could anybody find inverse function to this function:

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

Please write here the whole solution.

• Oct 22nd 2008, 03:42 PM
masters
Quote:

Originally Posted by petike
Hi all,
could anybody find inverse function to this function:

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

Please write here the whole solution.

A few have looked at this, but no takers. Let's see if I can get this thing started. First restate the function with y, then interchange the x and y.

$\displaystyle y = \frac{2x}{1 - x^2}$

$\displaystyle x = \frac{2y}{1 - y^2}$

$\displaystyle x(1-y^2)=2y$

$\displaystyle x-xy^2=2y$

$\displaystyle -xy^2+x-2y=0$

$\displaystyle xy^2+2y=x$

$\displaystyle y^2+\frac{2}{x}y=1$

Complete the square on the left

$\displaystyle y^2+\frac{2}{x}y+\left(\frac{1}{x}\right)^2=1+\fra c{1}{x^2}$

$\displaystyle \left(y+\frac{1}{x}\right)^2=\frac{x^2+1}{x^2}$

$\displaystyle y+\frac{1}{x}=\frac{\pm \sqrt{x^2+1}}{x}$

$\displaystyle y=\frac{-1}{x}\frac{\pm \sqrt{x^2+1}}{x}$

$\displaystyle y=\frac{-1\pm \sqrt{x^2+1}}{x}$

What do you think?
• Oct 23rd 2008, 01:57 AM
petike
Soluted
Oh,
thaaank you very much.
You have really helped me.

You've got my thanks.