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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?

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    $\begingroup$ It should. Why don't you try it? $\endgroup$ Jul 8 '20 at 19:25
  • $\begingroup$ But why substitute $x=3/x$, how do we determine which substitution is useful or not? $\endgroup$
    – Frost Bite
    Jul 8 '20 at 19:26
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    $\begingroup$ 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 $f(x)$. $\endgroup$ Jul 8 '20 at 19:27
  • $\begingroup$ Ohh I see. Thank you! $\endgroup$
    – Frost Bite
    Jul 8 '20 at 19:30
  • $\begingroup$ Peek at Evan Chen's "Introduction to Functional Equations", a lucid discussion on how to tackle functional equations in general. $\endgroup$
    – vonbrand
    Jul 9 '20 at 3:52
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Yes, it helps, as follows:

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 $$f(x)=\frac{3-x}{2}$$

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$$f\left(\frac{3}{x}\right)+\frac{3}{x}f(x)=\frac{3}{x}$$ 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.

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