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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?

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

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