# Math Help - Inverse help!

1. ## Inverse help!

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

2. 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:

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

3. 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:

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

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

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

4. Thank you both very much!!

5. 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.