Determinant involving function of $x$ 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, 
$$
f(x)=x^5+1, \ \textrm{this gives } f(2) = 33.
$$
