f(0)=f(11)=3628800, & f(n) is the same for all integers {n: 1≤n≤10}. What is the polynomial f(x)?

Question

... f(0)=f(11)=3628800
yahoo capitalized f in the question!
its the same function...

Or rather, what is the f(x) of least degree?

To avoid the degenerate, please note
f(n)≠f(0) for all {n: 1≤n≤10} !

Answer 1

Excluding the trivial polynomial f(x) = 10!, all of the solutions f(x) of minimal degree have the form C[k=1, 10]∏(x-k) + D, for appropriate constants C and D. Specifically, any C and D such that 10!*C + D = 10! will work. So in particular, we can choose C = 1 and D=0, to obtain:

f(x) = [k=1, 10]∏(x-k) = x^10 - 55x^9 + 1320x^8 - 18150x^7 + 157773x^6 - 902055x^5 + 3416930x^4 - 8409500x^3 + 12753576x^2 - 10628640x + 3628800

Answer 2

There are infinitely many solutions if you don't restrict
the degree of the polynomial. If f(x) is a polynomial of degree
at most 11, then there is a unique polynomial solving
your problem. It's given by Lagrange interpolation.