let $f(\frac{x}{3})+f(\frac{2}{x})=\frac{4}{x^2}+\frac{x^2}{9}-2$ then find $f(x)$

Question

let $f(\frac{x}{3})+f(\frac{2}{x})=\frac{4}{x^2}+\frac{x^2}{9}-2$ then find $f(x)$

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My Try :

$$f(\frac{x}{3})+f(\frac{2}{x})=(\frac{2}{x})^2-1+(\frac{x}{3})^2-1$$

So we have :

$$f(x)=x^2-1$$

it is right ?Is there another answer?

Answer 1

Suppose $f(x)$ can be given by the power series expansion $f(x) = a_{0} + a_{1} x + a_{2} x^{2} + \cdots$. Now, for the equation $$f\left(\frac{x}{3}\right) + f\left(\frac{2}{x}\right) = \frac{x^{2}}{9} - 2 + \frac{4}{x^{2}}$$ it is seen that: \begin{align} \frac{x^{2}}{9} + \frac{4}{x^{2}} - 2 &= 2 a_{0} + a_{1} \, \left(\frac{x}{3} + \frac{2}{x}\right) + a_{2} \left(\frac{x^{2}}{9} + \frac{4}{x^{2}}\right) + \cdots. \end{align} It is easy to determine that $a_{m+3} = 0$, for $m \geq 0$, $a_{1} = 0$, $a_{2} = 1$, and $a_{0} = -1$ which leads to the result $$f(x) = x^{2} - 1.$$

Answer 2

A plausible $f(x)$ is in fact $f(x)=x^2-1$. A proof of that is as follows.

Suppose $f(x)=x^2+h$ For $x=\sqrt6$ one has $\dfrac 2x=\dfrac x3$ so $$2f(\frac{\sqrt6}{3})=2(\frac{\sqrt6}{3})^2-2=2((\frac{\sqrt6}{3})^2+h)\Rightarrow h=-1$$

Answer 3

A particular solution is $f(x)=x^2-1$. The general solution of the associated homogeneous problem $$f\left({x\over3}\right)+f\left({2\over x}\right)=0$$ is $$f_{\rm hom}(x)=u\left(\log\bigl(\sqrt{3/2}\> x\bigr)\right)\qquad(x>0)\ ,$$ whereby $u$ is an arbitrary odd function. It follows that the general solution of the original problem is given by $$f(x)=x^2-1+ u\left(\log\bigl(\sqrt{3/2}\> x\bigr)\right)\qquad(x>0)\ .$$