# Finding Inverse of Functions

• Sep 30th 2011, 05:46 PM
Ingersoll
Finding Inverse of Functions
Here's what I have:

f(x)= $\frac{x+3}{x+4}$; x not equal to -4

$f^-1(x)=\frac{3-4x}{x-1}$; x not equal to -1

$f(x): y=\frac{x+3}{x+4}$

$x=\frac{y+3}{y+4}$

$x(y+4)=y+3$

$xy +4x=y+3$

$xy=y-1$

And that's pretty much it for me. I wish I could copy and paste the graph but looking at it it seems that I might just be able to read the graph or take the zeros of all the x's and simply construct the lines of the graph based on that, thereby avoiding the algebra. I have a feeling that's just wishful thinking though. Any and all help is greatly appreciated. Thanks. (Bow)
• Sep 30th 2011, 06:28 PM
Soroban
Re: Finding Inverse of Functions
Hello, Ingersoll!

I'll try to clarify what you wrote . . .

Quote:

$\text{Find the inverse of: }\:f(x) \:=\: \frac{x+3}{x+4}\quad x \ne -4$

$\text{Answer: }\:f^{\text{-}1}(x)\:=\:\frac{3-4x}{x-1}\quad x \ne 1$

$y \;=\;\frac{x+3}{x+4}$

$x \;=\;\frac{y+3}{y+4}$

$x(y+4)\;=\;y+3$

$xy +4x\;=\;y+3$

$xy\;=\;y-1$ . What?

You had: -- $xy +4x \;=\;y + 3$

Rearrange: . $xy - y \;=\;3 - 4x$

Factor:a--- $(x-1)y \;=\;3-4x$

Therefore: - - . . . $y \;=\;\frac{3-4x}{x-1}$