Question

suppose f(x) is polynomial of degree 5 and with leading coefficient 2001 . suppose further that f(1) = 1 , f(2) = 3 , f(3) = 5 ,f(4) = 7 , f(5) = 9 . then find value of f(6):
a)240120
b)11
c)240130
d)240131

Answer 1

Answer:

240131

Step-by-step explanation:

Let f(x) = g(x) + 2x - 1

since f(1) = 1, f(2) = 3, f(3) = 5, f(4) = 7, f(5) = 9, it then follows that g(1) = g(2) = g(3) = g(4) = g(5) = 0.

since f(x) is the 5th degree with leading coefficient 2001, the same is true for g(x).

Therefore g(x) = 2001(x - 1)(x - 2)(x - 3)(x - 4)(x - 5).

So now, we have

f(x) = 2001(x - 1)(x - 2)(x - 3)(x - 4)(x - 5) + 2x - 1

f(6) = 2001( 5 )( 4 )( 3 )( 2 )( 1 ) + 12 - 1 = 240131