Question

let f(x) be a 2020 degree polynomial with leading coefficient 1, given that f(1) =-2,....., f(2020)=-2020 find out the value of f(2021)

Answer 1

Answer:

Step-by-step explanation:

Here's how I would approach the problem..

Edit 1 - Typed the solution and removed the images of the solution as per QUORA'S POLICY..

Given:

f(1) = 1

f(2) = 2

f(3) = 3

f(4) = 16

To find : f(5)

Solution :

Assume another function g(x) such that g(x) = f(x) - x

Hence,

g(1) = 0 = x-1 ..... (1)

g(2) = 0 = x-2 ...... (2)

g(3) = 0 = x-3 ...... (3)

From (1,2,3), we can say that

g(x) = k (x-1)(x-2)(x-3) ......  (4)

f(4) = 16 .... Given

g(4) = 16 - 4 = 12

Therefore, g(4) = k(4-1)(4-2)(4-3)

12= k(4-1)(4-2)(4-3)

K=2

Equation 4 becomes

g(x) = 2 (x-1)(x-2)(x-3)

g(5) = 2(5-1)(5-2)(5-3)

g(5) = 2*4*3*2

g(5) = 48

But we know that

g(x) = f(x) - x

g(5) = f(5) - 5

f(5) = 48 + 5

Thus, f(5) = 53

Feel free to ask doubts and correct me if the need be!! After all, even I'm a student..

Hope this helped!!

13.1k views · View 45 Upvoters

Related Questions

More Answers Below

   If f(1) =1, f(2) = 2, f(3) = 3 and f(4) = 16, what is f(5)?

   A cubic equation f(x) is such that f(1) =1 ,f(2) = 2 ,f(3) =3 and f(4) =5 then f(6) =?

   If f(1)=1 , f(2)=2 , f(3)=3 , f(10)=10, can you find f(11)=?

   Suppose that f(x)

is a polynomial of degree 5

with 2009

as the leading coefficient. If f(1)=1,

f(2)=3,

f(3)=5,

f(4)=7,

f(5)=9,

can you find f(6)?

If f(1) =1, f(2) =2,f(3) =3 ,f(4) =5, what is f(x)?