# Finding the inverse of a rational function.

• September 28th 2009, 08:56 PM
Evan.Kimia
Finding the inverse of a rational function.
Im solving to find the inverse of f(x)=(2x+1)/(x-1) and ran into a snag.

heres what i did so far...

1. f(x)=(2x+1)/(x-1)
2. y=(2x+1)/(x-1)
3. x=(2y+1)/(y-1)
4 x(2y+1)=y-1
5. 2xy+1=y-1
6. 2xy+2=y

I cant figure out how to propery factor out the y on the left because im not sure what equation would yield 2xy+2 since 2y(x+2) obviously wont work out. where do you think im going off course?

Thanks.
• September 28th 2009, 10:19 PM
mr fantastic
Quote:

Originally Posted by Evan.Kimia
Im solving to find the inverse of f(x)=(2x+1)/(x-1) and ran into a snag.

heres what i did so far...

1. f(x)=(2x+1)/(x-1)
2. y=(2x+1)/(x-1)
3. x=(2y+1)/(y-1)
4 x(2y+1)=y-1 Mr F says: Wrong. It should be x(y -1) = 2y + 1.

[snip]
where do you think im going off course?

Thanks.

Expand, group terms with y in them, take y out as a common factor, make y the subject.

Learn something form this thread: http://www.mathhelpforum.com/math-he...tml#post373677.
• September 29th 2009, 12:27 AM
A Beautiful Mind
$y = (2x+1)/(x-1)$

You know, you can do this informally in your head. This is the hand crank, mechanical method of doing things. Think about undoing something.

Anyway.

An example to clear up some confusion would be:

$Y = (3x+1)/(x-1)$

$X = (3y+1)/(y-1)$

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

$xy-x = 3y+1$

Pay attention
: (Fubar) Here you'd want to bring the $x$ onto the right side and subtract $3y$ from the right side.

$xy-3y = x + 1$

Take out a $y$.

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

Divide now by $(x-3)$ to isolate the $Y$ variable.

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