Asked by bryan
If P(x) is a cubic polynomial with P(1)=1, P(2)=2, P(3)=3, P(4)=5, find P(6).
Answers
Answered by
Bosnian
A cubic polynomial has the form:
P(x) = a x³ + b x² + c x + d
P(1) mean value of a polynomial when x = 1
P(2) mean value of a polynomial when x = 2
P(3) mean value of a polynomial when x = 3
P(4) mean value of a polynomial when x = 4
P(6) mean value of a polynomial when x = 6
In this case:
P(1) = a ∙ 1³ + b ∙ 1² + c ∙ 1 + d = 1
a + b + c + d = 1
P(2) = a ∙ 2³ + b ∙ 2² + c ∙ 2 + d = 2
8 a + 4 b + 2 c + d = 2
P(3) = a ∙ 3³ + b ∙ 3² + c ∙ 3 + d = 3
27 a + 9 b + 3 c + d = 3
P(4) = a ∙ 4³ + b ∙ 4² + c ∙ 4 + d = 5
64 a + 16 b + 4 c + d = 5
Now you must solve system of equations :
a + b + c + d = 1
8 a + 4 b + 2 c + d = 2
27 a + 9 b + 3 c + d = 3
64 a + 16 b + 4 c + d = 5
____________________
The solutions are:
a = 1 / 6 , b = -1 , c = 17 / 6 , d = -1
So your polynomial:
P(x) = ( 1 / 6 ) x³ - x² + ( 17 / 6 ) x - 1
P(6) = ( 1 / 6 ) ∙ 6³ - 6² + ( 17 / 6 ) ∙ 6 - 1 =
( 1 / 6 ) ∙ 216 - 36 + 17 - 1 =
36 - 36 + 17 - 1 = 16
P(6) = 16
P(x) = a x³ + b x² + c x + d
P(1) mean value of a polynomial when x = 1
P(2) mean value of a polynomial when x = 2
P(3) mean value of a polynomial when x = 3
P(4) mean value of a polynomial when x = 4
P(6) mean value of a polynomial when x = 6
In this case:
P(1) = a ∙ 1³ + b ∙ 1² + c ∙ 1 + d = 1
a + b + c + d = 1
P(2) = a ∙ 2³ + b ∙ 2² + c ∙ 2 + d = 2
8 a + 4 b + 2 c + d = 2
P(3) = a ∙ 3³ + b ∙ 3² + c ∙ 3 + d = 3
27 a + 9 b + 3 c + d = 3
P(4) = a ∙ 4³ + b ∙ 4² + c ∙ 4 + d = 5
64 a + 16 b + 4 c + d = 5
Now you must solve system of equations :
a + b + c + d = 1
8 a + 4 b + 2 c + d = 2
27 a + 9 b + 3 c + d = 3
64 a + 16 b + 4 c + d = 5
____________________
The solutions are:
a = 1 / 6 , b = -1 , c = 17 / 6 , d = -1
So your polynomial:
P(x) = ( 1 / 6 ) x³ - x² + ( 17 / 6 ) x - 1
P(6) = ( 1 / 6 ) ∙ 6³ - 6² + ( 17 / 6 ) ∙ 6 - 1 =
( 1 / 6 ) ∙ 216 - 36 + 17 - 1 =
36 - 36 + 17 - 1 = 16
P(6) = 16
Answered by
scott
P(x) = ax^3 + bx^2 + cx + d
P(1) = 1 = a + b + c + d
P(2) = 2 = 8a + 4b + 2c + d
P(3) = 3 = 27a + 9b + 3c + d
P(4) = 5 = 64a + 16b + 4c + d
solve the system of equations for the coefficients ... a , b , c , and d
P(1) = 1 = a + b + c + d
P(2) = 2 = 8a + 4b + 2c + d
P(3) = 3 = 27a + 9b + 3c + d
P(4) = 5 = 64a + 16b + 4c + d
solve the system of equations for the coefficients ... a , b , c , and d
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