Find value of a functional equation

Question

Find $f(x)$ such that $$2 f(n) + \frac{1}{3}f\left(\frac{1}{n}\right) = 12.$$ Can anybody suggest me a way to solve this kind of functional equations?

Answer 1

Write $f\left({\frac1n}\right)$ in terms of $f(n)$: $$f\left({\frac1n}\right)=36-6f(n)$$ This is true for all $n\ne0$, so $$f(n)=36-6f\left({\frac1n}\right)$$ Now let $a=f(n)$ and $b=f\left({\frac1n}\right)$, then solve linear system of equations. You will get that $f(x)$ is constant function $$f(x)=\frac{36}7$$ for all $x\in\mathbb{R}\setminus\{0\}$.

Answer 2

If we set, for any $x>0$, $f(x)=g(\log x)$, then $g$ is a solution of: $$ 2 g(m) + \frac{1}{3}g(-m) = 12, \tag{1}$$ so we may take any $g$ such that $g(0)=\frac{36}{7}$, define it over $\mathbb{R}^+$ as we like then take: $$ g(-m)=36-6g(m) $$ to define $g$ over $\mathbb{R}^-$.