# Thread: [SOLVED] inverse of a function

1. ## [SOLVED] inverse of a function

I'm trying to take the inverse of the function $\displaystyle \frac{2x-3}{x+4}$
I think I've got it following this procedure
$\displaystyle x=\frac{2y-3}{y+4}$
$\displaystyle x(y+4) = 2y-3$
$\displaystyle xy+4x = 2y-3$
$\displaystyle xy-2y = -3-4x$
$\displaystyle y(x-2) = -3-4x$
[tex]y = \frac{-3-4x}{x-2}[tex]
$\displaystyle y = -\frac{4x+3}{x-2}$

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
$\displaystyle \frac{2(-\frac{4x+3}{x-2})-3}{-(\frac{4x+3}{x-2})+4}$
$\displaystyle \frac{-(\frac{6+8x-3x+6}{x-2})}{-(\frac{3+4x+4x-8}{x-2})}$
$\displaystyle \frac{-(\frac{5x+12}{x-2})}{-(\frac{8x-5}{x-2})}$
$\displaystyle -\frac{5x+12}{x-2} * -\frac{x-2}{8x-5}$
$\displaystyle \frac{5x+12}{8x+5}$
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?

$\displaystyle \frac{2(-\frac{4x+3}{x-2})-3}{-(\frac{4x+3}{x-2})+4}$

$\displaystyle \not = \frac{-(\frac{6+8x-3x+6}{x-2})}{-(\frac{3+4x+4x-8}{x-2})}$

You factor by $\displaystyle -1$ but keep the $\displaystyle -3$ on the numerator (should be $\displaystyle 3$). I strongly recommedn writing the original answer as

$\displaystyle y = \frac{4x+3}{2-x}$

Keeping unecessary minus signs is a recipe for disaster

3. Thanks a lot for the assistance. I agree, all those minus signs in there were causing me some serious difficulties. I'll try to avoid them as much as possible in the future.