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Old 11-19-2008, 10:02 AM
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Default Composite Functions
I know this is really simple, but can someone help?

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

find (f o f)(x)

Thanks!!
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  #2  
Old 11-19-2008, 11:02 AM
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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: f(x)=x+\frac{1}{x}-1?

Or is it this: f(x)=\frac{x+1}{x-1}?
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Old 11-19-2008, 11:05 AM
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The lower function is mine! Sorry, forgot the brackets!
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Old 11-19-2008, 11:21 AM
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f(x)=\frac{x+1}{x-1}?
[f\circ f](x)=f[f(x)]

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

Can you simplify?
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Old 11-19-2008, 11:29 AM
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Wow, I knew it was simple, thanks so much, so basically it simplifies down to -1, correct?
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Old 11-19-2008, 11:43 AM
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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?
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Old 11-19-2008, 11:47 AM
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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. . .
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Old 11-19-2008, 12:07 PM
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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:

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}}=
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Last edited by masters; 11-19-2008 at 01:34 PM.
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Old 11-19-2008, 12:18 PM
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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!
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Old 11-19-2008, 12:33 PM
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Hello, rtwilton

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

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

(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 \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 f(x) is its own inverse!
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