Hi all,
could anybody find inverse function to this function:
$\displaystyle
f(x) = \frac{2x}{1 - x^2}
$
Please write here the whole solution.
Thanks in advance.
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?