If $f(x)$ is a polynomial satisfying $$f(x)=\frac{1}{2} \begin{vmatrix} f(x) & f(\frac{1}{x})-f(x) \\ 1 & f(\frac{1}{x}) \end{vmatrix} $$ and $f(3)=244$ then $f(2)$ is what?
My attempt— Replacing $x$ by $\frac{1}{x}$ we get $$f\left(\frac{1}{x}\right)=\frac12\begin{vmatrix} f\left(\frac{1}{x}\right) & -(f(\frac{1}{x})-f(x)) \\ 1 & f(x) \end{vmatrix}.$$ Which gave me $f\left(\frac{1}{x}\right)-f(x)=\frac{2f(x)-f(x)^2}{f(x)-1}$. Now putting the value of $f\left(\frac{1}{x}\right)$ obtained in terms of $f(x)$ in the original determinant I became helpless when it reduced to $f(x)=f(x)$ and I achieved nothing. Please help me out with this problem.thanks.
Your starting idea is indeed great. Using $f(x)$ and $f(1/x)$ both, and adding them up, we have $$ f(x)+f(1/x) = f(x)f(1/x). $$
This gives $$ f(1/x)-1=\frac{1}{f(x)-1}. \ \ \ \ (*) $$
As $x\rightarrow\infty$, we have $f(0)=1$.
By assuming that $n=\textrm{deg}(f)$, we have by multiplying $x^n$ both sides, $$ \frac{x^n}{f(x)-1} \ \textrm{ is a polynomial.} $$
Thus, $f(x)-1$ must be a divisor of $x^n$. Then only possibilities that we have $$ f(x) = ax^k + 1. $$ By (*), we have $a^2=1$, giving that $a=\pm 1$. With $f(3)=244$, we are left with $a=1$. Then $f(x) = x^k +1$. Again by $f(3)=244$, we obtain $k=5$. Therefore, $$ f(x)=x^5+1, \ \textrm{this gives } f(2) = 33. $$