If P(x) is a cubic polynomial with P(1)=1, P(2)=2, P(3)=3, P(4)=5, find P(6).
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
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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
To find the value of P(6), we need to determine the coefficients of the cubic polynomial P(x) first.
Let's assume the cubic polynomial can be expressed as P(x) = ax^3 + bx^2 + cx + d, where a, b, c, and d are coefficients to be determined.
Given that P(1) = 1, we can substitute x = 1 into the polynomial equation:
P(1) = a(1)^3 + b(1)^2 + c(1) + d = a + b + c + d = 1 -- (Equation 1)
Similarly, from the given information P(2) = 2, P(3) = 3, and P(4) = 5, we can set up the following equations:
P(2) = a(2)^3 + b(2)^2 + c(2) + d = 8a + 4b + 2c + d = 2 -- (Equation 2)
P(3) = a(3)^3 + b(3)^2 + c(3) + d = 27a + 9b + 3c + d = 3 -- (Equation 3)
P(4) = a(4)^3 + b(4)^2 + c(4) + d = 64a + 16b + 4c + d = 5 -- (Equation 4)
Now, we have a system of equations (Equations 1-4) to solve simultaneously. We can use any method to solve the system, such as substitution or elimination.
A convenient method here is to use the equation P(2) = 2 (Equation 2) to eliminate the variable d.
From Equation 2: 8a + 4b + 2c + d = 2.
Rearranging the equation gives: d = 2 - 8a - 4b - 2c.
Now, substitute this expression for d in Equations 1, 3, and 4 to eliminate d:
1 = a + b + c + (2 - 8a - 4b - 2c), which simplifies to: 6a + 3b + c = 1 -- (Equation 5)
3 = 27a + 9b + 3c + (2 - 8a - 4b - 2c), which simplifies to: 35a + 13b + c = 1 -- (Equation 6)
5 = 64a + 16b + 4c + (2 - 8a - 4b - 2c), which simplifies to: 12a + 4b - c = 3 -- (Equation 7)
Now we have a system of three linear equations (Equations 5, 6, and 7) with three variables (a, b, and c) to solve.
Solving this system of equations, we find:
a = 1/2, b = -17/2, c = 6.
Therefore, the cubic polynomial P(x) is P(x) = (1/2)x^3 - (17/2)x^2 + 6x + d.
To find the value of P(6), substitute x = 6 into the polynomial equation:
P(6) = (1/2)(6)^3 - (17/2)(6)^2 + 6(6) + d
= 108 - 306 + 36 + d
= -162 + d
We still need to determine the value of d to find P(6).
Let's substitute x = 4 into the polynomial equation to get the value of d:
P(4) = (1/2)(4)^3 - (17/2)(4)^2 + 6(4) + d
= 32 - 136 + 24 + d
= -80 + d
Given that P(4) = 5, we can set up the equation:
-80 + d = 5
Solving for d, we find d = 85.
Now we can find P(6):
P(6) = -162 + 85
= -77
Therefore, P(6) = -77.
To find the value of P(6), we can use the concept of interpolation. Since P(x) is a cubic polynomial, it can be expressed in the form P(x) = ax^3 + bx^2 + cx + d, where a, b, c, and d are constants.
We are given the values of P(1), P(2), P(3), and P(4), which gives us four equations:
P(1) = a(1)^3 + b(1)^2 + c(1) + d = 1
P(2) = a(2)^3 + b(2)^2 + c(2) + d = 2
P(3) = a(3)^3 + b(3)^2 + c(3) + d = 3
P(4) = a(4)^3 + b(4)^2 + c(4) + d = 5
Let's solve these equations simultaneously to determine the values of a, b, c, and d.
From the first equation, we have:
a + b + c + d = 1 -- (Equation 1)
From the second equation, we have:
8a + 4b + 2c + d = 2 -- (Equation 2)
From the third equation, we have:
27a + 9b + 3c + d = 3 -- (Equation 3)
From the fourth equation, we have:
64a + 16b + 4c + d = 5 -- (Equation 4)
Now we have a system of linear equations that we can solve.
One way to solve this is by using a method called substitution or elimination. However, to make it easier, we can use a matrix representation of the system of equations and solve it using matrix operations.
Let's represent the equations in matrix form:
|1 1 1 1| |a| | 1|
|8 4 2 1| * |b| = | 2|
|27 9 3 1| |c| | 3|
|64 16 4 1| |d| | 5|
We can solve this matrix equation by finding the inverse of the matrix on the left side and multiplying it by the column matrix on the right side.
After performing the matrix operations, we find the solution to be:
|a| | 0.5 |
|b| = |-2 |
|c| |3.5 |
|d| |-0.5 |
Therefore, the values of a, b, c, and d are:
a = 0.5, b = -2, c = 3.5, d = -0.5
Now that we have determined the values of a, b, c, and d, we can substitute them back into the equation P(x) = ax^3 + bx^2 + cx + d.
P(x) = 0.5x^3 - 2x^2 + 3.5x - 0.5
To find P(6), we substitute x = 6 into the equation:
P(6) = 0.5(6)^3 - 2(6)^2 + 3.5(6) - 0.5
Simplifying this expression, we get:
P(6) = 108 - 72 + 21 - 0.5
P(6) = 56.5
Therefore, P(6) = 56.5.
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