# Solving $f(x)$ in a functional equation

Find of general form for $$f(x)$$ given $$f(x)+xf\left(\displaystyle\frac{3}{x}\right)=x.$$

I think we need to substitute $$x$$ as something else, but I'm not sure. Will $$x=\displaystyle\frac{3}{x}$$ help me?

• It should. Why don't you try it? Jul 8 '20 at 19:25
• But why substitute $x=3/x$, how do we determine which substitution is useful or not? Jul 8 '20 at 19:26
• Notice that $3/(3/x) = x$, so this gives you another equation involving the same $f(x)$ and $f(3/x)$. If you can eliminate the $f(3/x)$ from the two equations, you have a chance of being able to solve for $f(x)$. Jul 8 '20 at 19:27
• Ohh I see. Thank you! Jul 8 '20 at 19:30
• Peek at Evan Chen's "Introduction to Functional Equations", a lucid discussion on how to tackle functional equations in general. Jul 9 '20 at 3:52

From $$f(x)+xf(\frac{3}{x})=x\tag{*}$$ we get $$f(\frac{3}{x})+\frac{3}{x}f(x)=\frac{3}{x}$$ or $$xf(\frac{3}{x})+3f(x)=3\tag{**}$$ from (*) anf (**), we have: $$-2f(x)=x-3$$ or $$f(x)=\frac{3-x}{2}$$
$$f\left(\frac{3}{x}\right)+\frac{3}{x}f(x)=\frac{3}{x}$$ Thus, putting this expression of $$f(x/3)$$ in the first equation gives $$f(x)=x-xf(3/x)=x+3f(x)-3$$ We finally have $$f(x)=\frac{3-x}{2}$$ and this function satisfies your equation.