0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
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).

Open in App
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

Suggest Corrections
14
Join BYJU'S Learning Program
Related Videos
Degree of a Polynomial
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program