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I have the problem :

Find all functions f:RR that satisfy

 

 f(x)+3f(x1x)=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 (x1)/x in for x, then you get a new equation that also involves f(x1x)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)=x1x is cyclic with order 3, but I don't know how to use this to find the answer.

 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.  

 Nov 21, 2018
 #5
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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(x1x) cancels out and turns into 0 because 3f(0)=0. Also, if you look around in the proof, you will find x  and 1x 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

 Nov 21, 2018
 #6
avatar+33654 
+3

Here's my take on this:

 

 Nov 21, 2018

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