# Inverse help!

• April 8th 2008, 09:50 AM
antz215
Inverse help!
If f= x+1/x+3 , find g, it's inverse, if it exists.

• April 8th 2008, 10:00 AM
topher0805
Finding the Inverse of a Function

This site gives a good explanation.

$y=\frac {x+1}{x+3}$

Solve for x:

$y(x+3)=x+1$

$
yx+3y=x+1$

$3y-1=x-yx$

$3y-1=x(1-y)$

$x=\frac {3y-1}{1-y}$

Now switch the x and y:

$y = \frac {3x-1}{1-x}$

And y is your inverse. :)
• April 8th 2008, 10:05 AM
earboth
Quote:

Originally Posted by antz215
If f= x+1/x+3 , find g, it's inverse, if it exists.

$y = f(x) = \frac{x+1}{x+3}~,~x\neq -3$

Now change the variables:

$g: x = \frac{y+1}{y+3}~\iff~x(y+3)=y+1$

$g: xy+3x = y+1~\iff~ 3x-1=y-xy~\iff~3x-1=y(1-x)$

$g(x)=y=\frac{3x-1}{1-x}~,~x\neq 1$
• April 8th 2008, 10:27 AM
antz215
Thank you both very much!!
• April 8th 2008, 11:42 AM
abender
for these inverse problems, simply switch all x's with y's and all y's with x's. Note that for f(x), just consider it to mean y. Ok, so you switched the x's and y's. Now solve for y. That's it.
-Andy

Edit: And I almost forgot, check your restriction(s). Only God can divide by zero.