Let $\mathbb C(x)$ be the field of rational functions with complex coefficients i.e. it is the fraction field of $\mathbb C[x]$ .
I am trying to explicitly determine the subfield of $\mathbb C(x)$ given by $\{f(x) \in \mathbb C(x) : f(x)=f(1/x) \}$ .
I can show that any element in the subfield of the form $p(x)/q(x)$ with $p(x), q(x)$ are polynomial of degree $1$ is actually a member of $\mathbb C$ . I'm not sure what would happen with other elements.
Please help.
Note that the element $x + 1/x$ is invariant under $x \mapsto 1/x$. Consider the field extension $K := \mathbb{C}(x + 1/x) \subset L := \mathbb{C}(x)$, it is an easy exercise to see that $K \not= L$. Consider the polynomial $$ p(t) = t^2 -at+1 $$ where $a = x + 1/x \in K[t]$, then, since $p(x) = 0$, this is the minimal polynomial of $x$ over $K$ and this extension has degree 2 and is Galois (every degree 2 extension is Galois), then its Galois group has order two and contains $x \mapsto 1/x$ and the identity, the only possibility is that $Gal(L/K) = \{ x \mapsto 1/x, Id\}$. Therefore $K = L^{Gal(L/K)}$, i.e. $K$ is the fixed field of $Gal(L/K)$ in L.
Edit: The Fundamental theorem_of Galois theory was used.