# [SOLVED] inverse of a function

• Feb 11th 2010, 10:07 AM
satis
[SOLVED] inverse of a function
I'm trying to take the inverse of the function $\frac{2x-3}{x+4}$
I think I've got it following this procedure
$x=\frac{2y-3}{y+4}$
$x(y+4) = 2y-3$
$xy+4x = 2y-3$
$xy-2y = -3-4x$
$y(x-2) = -3-4x$
[tex]y = \frac{-3-4x}{x-2}[tex]
$y = -\frac{4x+3}{x-2}$

I worked that a few times and am pretty confident in my answer. The proof is in plugging that back into the original equation and seeing if I get x, and this is where I'm fumbling about
$\frac{2(-\frac{4x+3}{x-2})-3}{-(\frac{4x+3}{x-2})+4}$
$\frac{-(\frac{6+8x-3x+6}{x-2})}{-(\frac{3+4x+4x-8}{x-2})}$
$\frac{-(\frac{5x+12}{x-2})}{-(\frac{8x-5}{x-2})}$
$-\frac{5x+12}{x-2} * -\frac{x-2}{8x-5}$
$\frac{5x+12}{8x+5}$
which, unless I'm blind, does not equal x. I hope the above isn't too confusing.

Assuming my initial answer is correct, what am I doing wrong when I'm trying to verify it?
• Feb 11th 2010, 10:52 AM
Nyrox
$\frac{2(-\frac{4x+3}{x-2})-3}{-(\frac{4x+3}{x-2})+4}$
$\not = \frac{-(\frac{6+8x-3x+6}{x-2})}{-(\frac{3+4x+4x-8}{x-2})}$
You factor by $-1$ but keep the $-3$ on the numerator (should be $3$). I strongly recommedn writing the original answer as
$y = \frac{4x+3}{2-x}$