# Thread: Composite Functions

1. ## Composite Functions

I know this is really simple, but can someone help?

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

find (f o f)(x)

Thanks!!

2. Originally Posted by rtwilton
I know this is really simple, but can someone help?

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

find (f o f)(x)

Thanks!!
Is this your function: $\displaystyle f(x)=x+\frac{1}{x}-1$?

Or is it this: $\displaystyle f(x)=\frac{x+1}{x-1}$?

3. The lower function is mine! Sorry, forgot the brackets!

4. Originally Posted by masters
$\displaystyle f(x)=\frac{x+1}{x-1}$?
$\displaystyle [f\circ f](x)=f[f(x)]$

$\displaystyle f\left(\frac{x+1}{x-1}\right)=\frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1}$

Can you simplify?

5. Wow, I knew it was simple, thanks so much, so basically it simplifies down to -1, correct?

6. Originally Posted by rtwilton
Wow, I knew it was simple, thanks so much, so basically it simplifies down to -1, correct?
I get something different. Can you show your work?

7. Hmm I guess I thought that the fractions in the numerator and denominator would just cancel, but as I test my theory I see that does not work. I can hack away at it a little longer. . .

8. Originally Posted by rtwilton
Hmm I guess I thought that the fractions in the numerator and denominator would just cancel, but as I test my theory I see that does not work. I can hack away at it a little longer. . .
You are right. You cannot cancel over addition or subtraction. Here's a start:

$\displaystyle f\left(\frac{x+1}{x-1}\right)=\frac{\dfrac{x+1}{x-1}+1}{\dfrac{x+1}{x-1}-1}=\frac{\dfrac{x+1+x-1}{x-1}}{\dfrac{x+1-(x-1)}{x-1}}=\frac{\dfrac{2x}{x-1}}{\dfrac{2}{x-1}}=$

9. Alright I am going to look at your work a little bit, run some errands, and come back to it. I think (hope) I can get it from here. I'll post and work/solution I have, thanks so much for everything!

10. Hello, rtwilton

$\displaystyle f(x) \:= \:\frac{x+1}{x-1}$

Find: .$\displaystyle (f\circ f)(x)$

$\displaystyle (f\circ f)(x) \;=\;f(f(x)) \;=\;f\left(\frac{x+1}{x-1}\right) \;=\;\frac{\dfrac{x+1}{x-1} + 1}{\dfrac{x+1}{x-1} - 1}$

Multiply by $\displaystyle \frac{x-1}{x-1}\!:\quad\frac{(x-1)\left(\dfrac{x+1}{x-1} + 1\right)} {(x-1)\left(\dfrac{x+1}{x-1} - 1\right)} \;=\;\frac{(x+1) + (x-1)}{(x+1) - (x-1)} \;=\;\frac{2x}{2} \;=\;\boxed{ x}$

Note: This means that $\displaystyle f(x)$ is its own inverse!