Solution: Let $\alpha$ and $\beta$ be roots of $f(x)$. Since $f(x)$ is separable, there exists $\sigma \in \operatorname{Aut}(K(\alpha, \beta)/K)$ such that $\sigma(\alpha) = \beta$. By the Fundamental Theorem of Galois Theory, $\sigma$ corresponds to an element of the Galois group of $f(x)$, which therefore acts transitively on the roots of $f(x)$.