1. ## Functional Equation

Find all functions $\displaystyle f : \mathbb{R} - \left\{ -\frac{1}{3}, \frac{1}{3} \right\} \rightarrow \mathbb{R}$ that satisfy the identity

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

2. Consider the change of variables: $\displaystyle x : \to {\textstyle{{x - 1} \over {1 + 3x}}}$

Then we get: $\displaystyle f\left( x \right) = {\textstyle{{x - 1} \over {1 + 3x}}} - f\left( {{\textstyle{{x - 1} \over {1 + 3x}}}} \right)$ ( check that the denominator is never 0)

So we've got these 2 equations: $\displaystyle \left[ \begin{array}{l} f\left( x \right) = {\textstyle{{x - 1} \over {1 + 3x}}} - f\left( {{\textstyle{{x - 1} \over {1 + 3x}}}} \right) \\ f\left( {{\textstyle{{x + 1} \over {1 - 3x}}}} \right) = x - f\left( x \right) \\ \end{array} \right.$

Thus: $\displaystyle f\left( {{\textstyle{{x + 1} \over {1 - 3x}}}} \right) - f\left( {{\textstyle{{x - 1} \over {1 + 3x}}}} \right) = x - {\textstyle{{x - 1} \over {1 + 3x}}} = {\textstyle{{3x^2 + 1} \over {3x + 1}}}$

Again take: $\displaystyle x: \to {\textstyle{{x - 1} \over {1 + 3x}}}$

Then: $\displaystyle f\left( x \right) - f\left( {{\textstyle{{x + 1} \over {1 - 3x}}}} \right) = {\textstyle{{3 \cdot \left( {{\textstyle{{x - 1} \over {1 + 3x}}}} \right)^2 + 1} \over {3 \cdot \left( {{\textstyle{{x - 1} \over {1 + 3x}}}} \right) + 1}}}$

Consider then the system of equations: $\displaystyle \left\{ \begin{array}{l} f\left( x \right) - f\left( {{\textstyle{{x + 1} \over {1 - 3x}}}} \right) = {\textstyle{{3 \cdot \left( {{\textstyle{{x - 1} \over {1 + 3x}}}} \right)^2 + 1} \over {3 \cdot \left( {{\textstyle{{x - 1} \over {1 + 3x}}}} \right) + 1}}} \\ f\left( {{\textstyle{{x + 1} \over {1 - 3x}}}} \right) = x - f\left( x \right) \\ \end{array} \right.$

and solve for $\displaystyle f\left( x \right)$

3. Originally Posted by PaulRS
and solve for $\displaystyle f\left( x \right)$
How would I do this?

4. Originally Posted by Winding Function
How would I do this?
As you'd solve in a normal system of equations.

5. Originally Posted by PaulRS
As you'd solve in a normal system of equations.
I'm confused. - What are the "variables" of the equation, and how do I isolate them?

6. Originally Posted by Winding Function

I'm confused. - What are the "variables" of the equation, and how do I isolate them?
Originally Posted by PaulRS

and solve for $\displaystyle f\left( x \right)$

7. Originally Posted by Krizalid
But what do I do with terms like $\displaystyle f\left( \frac{x+1}{1-3x} \right)$?

8. Originally Posted by Winding Function
But what do I do with terms like $\displaystyle f\left( \frac{x+1}{1-3x} \right)$?
What happens if you add them?