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# If f is a polynomial of degree 3, such that f(0) = f(2) = f(3) = 1 and f(1) = 0, find f(4).

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If f is a polynomial of degree 3, such that f(0) = f(2) = f(3) = 1 and f(1) = 0, find f(4).

Feb 15, 2021

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We  have  the form,  ax^3  + bx^2  + cx  +  d

Since  f(0) =  = 1,  then  d  =1

And  we have  this system

a(1)^2  + b(1)^2  + c(1)  + 1  =  0

a(2)^3  + b(2)^2  + c(2)  + 1  = 1

a^(3)^3  + b(3)^2  + c(3)  + 1  =1        simplify

a  + b  +  c    = -1

8a + 4b  + 2c   = 0

27a + 9b + 3c  = 0

Multiply the first  equation by - 2  and  add it to  the second

Multiply the first equatio by  -3  and add it to the third.....this  results in  :

6a + 2b   = 2  →    3a + b  = 1   →   -6a  - 2b  = -2   (1)

24a + 6b  = 3  →   8a + 2b  =  1  (2)

Add  (1)  and (2)   and we  have that    2a =  -1        →  a  = -1/2

So  3(-1/2) + b  = 1 →   -3/2  + b   =1  →   b =  5/2

And  -1/2  + 5/2  + c  =-1

2 + c = -1

c =-3

So....the polynomial  is  f(x)  = (-1/2)x^3 + (5/2)x^2  - 3x  + 1

And  f(4)  =  (-1/2)4^3  + (5/2)4^2  - 3(4)  + 1   =  -32 + 40  - 12 +  1  =  -3

See the graph here  :  https://www.desmos.com/calculator/guifyz4nhd

Feb 15, 2021