# Thread: Finding the inverse of a rational function.

1. ## 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.

2. 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.

3. $\displaystyle y = (2x+1)/(x-1)$

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:

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

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

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

$\displaystyle xy-x = 3y+1$

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

$\displaystyle xy-3y = x + 1$

Take out a $\displaystyle y$.

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

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

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