Finding an Inverse for f(x) = (x - 3)/(x - 1)

[kinerry]

It's been awhile since I worked with inverse functions.

Here's the problem;

. . .f(x) = (x - 3)/(x - 1)

Here's the work I have done;

. . .y = (x - 3)/(x - 1)

. . .x = (y - 3)/(y - 1)

. . .y = (x - 1)/(x - 3)

. . .f^-1(x) = (x - 1)/(x - 3)

First, if my work correct?

Second, how exactly do I determine if it is an inverse of the original function?

I have a feeling I am making an elementary mistake here...

 

[stapel]

Re: Finding an Inverse

f(x) = (x-3)/(x-1)

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

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

I'm with you to this point. You began the basic process by renaming "f(x)" as "y", and then switching the variables. You now need to solve for "y=". But how did you get "y = (x - 1)/(x - 3)"? I'm not seeing all the "solving" steps in between...?

Also, the two functions do not compose to produce just "x". Letting your new function be g(x), we have:

. . .\(\displaystyle \L f(g(x))\, =\, f\left(\frac{x\,-\,1}{x\,-\,3}\right)\)

. . . . . . . . . . . .\(\displaystyle \L =\,\frac{\frac{x\,-\,1}{x\,-\,3}\,-\,3}{\frac{x\,-\,1}{x\,-\,3}\,-\,1}\)

. . . . . . . . . . . .\(\displaystyle \L =\,\frac{\frac{x\,-\,1}{x\,-\,3}\,-\,\frac{3x\,-\,9}{x\,-\,3}}{\frac{x\,-\,1}{x\,-\,3}\,-\,\frac{x\,-\,3}{x\,-\,3}\)

. . . . . . . . . . . .\(\displaystyle \L =\,\frac{\left(\frac{8\,-\,2x}{x\,-\,3}\right)}{\left(\frac{2}{x\,-\,3}\right)}\)

. . . . . . . . . . . .\(\displaystyle \L =\,\frac{8\,-\,2x}{2}\)

. . . . . . . . . . . .\(\displaystyle \L =\,4\,-\,x\)

. . . . . . . . . . . .\(\displaystyle \L \neq\,x\)

Please reply showing your steps for the solving, so that we can try to find the error. Thank you.

Eliz.

 

[kinerry]

I found the answer to be (x-3)/(x-1) and then checked using the method you provided, that is also the method for determining if a function is the inverse if I remember correctly. I transposed the 3 and 1 by mistake, something elementary like I assumed.

 

[stapel]

The working looks like this:

. . . . .\(\displaystyle \L f(x)\,=\,\frac{x\,-\,3}{x\,-\,1}\)

. . . . .\(\displaystyle \L y\,=\,\frac{x\,-\,3}{x\,-\,1}\)

. . . . .\(\displaystyle \L x\,=\,\frac{y\,-\,3}{y\,-\,1}\)

. . . . .\(\displaystyle \L xy\,-\,1x\,=\,y\,-\,3\)

. . . . .\(\displaystyle \L xy\,-\,1y\,=\,1x\,-\,3\)

. . . . .\(\displaystyle \L y(x\,-\,1)\,=\,x\,-\,3\)

. . . . .\(\displaystyle \L y\,=\,\frac{x\,-\,3}{x\,-\,1}\)

. . . . .\(\displaystyle \L f^{-1}(x)\,=\,\frac{x\,-\,3}{x\,-\,1}\)

So the inverse happens to be the same as the original function.

Eliz.