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.
+776 StackExchange...huh...I answered some questions there about computer science. I didn't know they also had a math webpage.
+776 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
