Let $f(x)$ be a polynomial of degree $8$ such that $f(r)=\frac1r$, for $r=1,2,3,\ldots,9$. Find $\frac1{f(10)}$.

Question

Question:

Let $f(x)$ be a polynomial of degree $8$ such that $f(r)=\frac1r$, for $r=1,2,3,\ldots,9$. Find $\frac1{f(10)}$.

My Approach: We know that $f(r)=\frac1r$, which implies that $$rf(r)-1=0$$ Using the information that $f(r)=\frac1r$, for $r=1,2,3,4...8,9$, we get that $1,2,3,4,5...,8,9$ are roots of the equation $rf(r)-1=0$. Which implies that $$rf(r)-1=(r-1)(r-2)(r-3)(r-4)(r-5)(r-6)(r-7)(r-8)(r-9)$$. Putting $r=10$ in the above equation, we get, $$f(10)=\frac{1+9!}{10}$$ and $\frac{1}{f(10)}$ as $$\frac{1}{f(10)}=\frac{10}{1+9!}$$. But the answer is 5. Please help.

Answer 1

You are right that $rf(r)-1$ is a polynomial of degree $9$ with zeros at $1, 2, \ldots, 9$. But that determines the polynomial only up to a constant factor, i.e. $$ rf(r)-1=C(r-1)(r-2)(r-3)(r-4)(r-5)(r-6)(r-7)(r-8)(r-9) $$ for some constant $C$ (which can be determined by setting $r=0$).