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Question

Suppose f (x) is a polynomial of degree 5 and with leading coefficient 2009. Suppose further that f (1)=1, f (2)=3,f (3)=5,f (4)=7,f (5)=9. What is the value of f (6).

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Solution

If you look at the function at the given five values, it looks like f(x)=2x−1
But the question also adds the mundane requirement that it is of degree 5 and has 2009 as its leading coefficient. We can do this by manipulating our proposed f in such a way so that it is not affected at 1, 2, 3, 4, 5. This can be easily done by adding a fifth degree polynomial which has roots at those points. Such a polynomial would look like
(x−1)(x−2)(x−3)(x−4)(x−5)
In order to satisfy the leading coefficient criteria, we just multiply it by 2009. So we get the exact polynomial as
f(x)=2x−1+2009(x−1)(x−2)(x−3)(x−4)(x−5)

Next substituting x==6, we get
f(6)=11+2009×5!
=241091

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