If $f(x)$ is a polynomial satisfying $ f(x)f(\frac 1x) = f(x)+f(\frac 1x)$ and $f(3)=28$, then how could we find $f(4)$? 
If $f(x)$ is a polynomial satisfying $  f(x)f(\frac 1x) = f(x)+f(\frac
 1x)$ and $f(3)=28$, then how could we find $f(4)$ ?


Solving the functional equation for $f\left(\frac{1}{x} \right) = \frac{f(x)}{f(x)-1}$. This means that $f(x)-1$ must be a monomial. Let $f(x) = 1 + c x^d$. Then 
$$
   c \left( \frac{1}{x} \right)^d +1 = \frac{1}{c} \left( \left( \frac{1}{x} \right)^d + c \right)
$$
This, implies $c^2 = 1$. Now use $f(3) = 28$ to determined $c$ and $d$. Since $28 = 1 + 1 \times 3^3$, we conclude $c=1$ and $d=3$.
Thus $f(4) = 1 + 4^3 = 65$.
