Given that f(x) is a polynomial of degree EIGHT such that
f(t) = 1/t for t = 1,2,3,4,5,...,9
Find f(10)
let $\displaystyle g(t)=tf(t) - 1.$ so $\displaystyle g$ is a polynomial of degree 9 with 9 roots t = 1, 2, ... , 9. thus $\displaystyle tf(t)-1=k(t-1)(t-2) \ ... \ (t-9), \ \ \ \ (1)$
for some constant $\displaystyle k.$ now in (1) put t = 0 to to get: $\displaystyle k=\frac{1}{9!}.$ therefore (1) becomes: $\displaystyle tf(t)=\frac{1}{9!}(t-1)(t-2) \ ... \ (t-9)+1. \ \ \ \ \ \ (2)$
finally in (2) put t = 10 to get: $\displaystyle 10f(10)=\frac{1}{9!} \times 9! + 1 = 2,$ which gives us: $\displaystyle f(10)=\frac{1}{5}. \ \ \ \square$
Remark: it's clear that the problem can be easily generalized in different ways and the above method will still work quite nicely!