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I have the problem :

Find all functions \(f:\mathbb R \to \mathbb R\) that satisfy

 

 \(f(x) + 3 f\left( \frac {x-1}{x} \right) = 7x\) 

 

for all nonzero \(x\). Note that the domain of \(f\) is all real numbers. Make sure you define your function everywhere!

 

Hint: If you substitute \((x - 1)/x\) in for \(x\), then you get a new equation that also involves \(f\left( \frac{x - 1}{x} \right)\)Then do this substitution again and you may get another useful equation to consider.

 

I think this is related to the fact that \(f(x) = \frac{x - 1}{x}\) is cyclic with order 3, but I don't know how to use this to find the answer.

Guest Nov 21, 2018
 #1
 #3
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That proof is incomplete. The problem needs f(x) to be defined for all x, but in the proof the function is not defined for x=0,1.

Guest Nov 21, 2018
 #4
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Ok, Guest, I'll see if I can help you. smiley

PartialMathematician  Nov 21, 2018
 #2
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StackExchange...huh...I answered some questions there about computer science. I didn't know they also had a math webpage.  

PartialMathematician  Nov 21, 2018
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In the proof, 0 is restricted because the denominator of a fraction cannot be 0 (cannot divide by 0). If we set x to 1, then the part \(3f(\dfrac{x-1}{x})\) cancels out and turns into 0 because \(3f(0) = 0\). Also, if you look around in the proof, you will find \(x\)  and \(1-x\) many times as the denominator, which cannot equal 0. Simply stated, 0 and 1 are restricted, AKA not in the domain of the function. 

 

Hope this helps,

- PartialMathematician

PartialMathematician  Nov 21, 2018
 #6
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Here's my take on this:

 

Alan  Nov 21, 2018

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