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- Feb 14, 2012

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This problem has given me a very hard time because I have exhausted all the methods that I know to figure out a way to find for the values for both a and b but no, there must be a trick to this problem and I admit that it is a question that is out of my reach...

Could you show me how to attack this problem, please?

Thanks in advance.

Problem:

For real numbers $a$ and $b$ define $f(x)=\dfrac{1}{ax+b}$. For which $a$ and $b$ are there three distinct real numbers $x_1, x_2, x_3$ such that $f(x_1)=x_2$, $f(x_2)=x_3$ and $f(x_3)=x_1$?