Find a polynomial $f \in \mathbb{Q}[x]$ with $f(x)=\frac{1}{x}$

Question

I have this small exercise where I am stuck since one hour. Can someone help me?

I have to find a polynomial $f \in \mathbb{Q}[x]$ with $f(x)=\frac{1}{x}$

Answer 1

Let $\mathbb{Q}[x] \ni f(x) = \sum_{i=1}^n a_i x_i$, assume that $f$ has the property $f(x) = x^{-1}$ for every $x$. If that is the case, then that implies that $$f^{(n + 1)}(x) = \left(\frac1x \right)^{(n + 1)},$$ and it is easy to see that $f^{(n+1)}(x) = 0$ and $(x^{-1})^{(n+1)} = (-1)^{n+1}(n+1)!x^{-(n+2)}$. Clearly, the latter expression is nonzero, which implies that if such a function exists, then it is not a polynomial with rational coefficients.