# Rational Functional Equations

Suppose $f(x)$ is a rational function such that $3 f \left( \frac{1}{x} \right) + \frac{2f(x)}{x} = x^2$ for all $x \neq 0$. Find $f(-2)$.

I tried substituting different values of $x$ to get a system of equations to solve for $f(x)$, but this didn't work. How should I take this from here?

• Hint: get two equations by taking $x=-2,-\frac 12$ and solve. – lulu Apr 24 '18 at 1:29

Following lulu's comment, set $$x=-2$$ and $$-\frac{1}{2}$$ to find, respectively,
$$\begin{cases}3f(-\frac{1}{2})+\frac{2f(-2)}{-2}=4 \\ 3f(-2)+\frac{2f(-\frac{1}{2})}{-\frac{1}{2}}=\frac{1}{4}\end{cases}$$so that we have a system of simultaneous equations in $$f(-2)$$ and $$f(-\frac{1}{2})$$. Solving this system yields $$f(2)=\frac{67}{20}.$$
To motivation this substitution, notice that $$f(x)=\frac{1}{x}$$ is an involution. In fact, comparing the original functional equation with the equation formed by setting $$x=\frac{1}{x}$$ allows us to find the general solution for $$f(x)$$.