A third-degree polynomial passes through the points (0,-12), (1,0) ,(3,6) and (4,12).
a). Determine this polynomial using Lagrange's interpolation formula
b). Find the value at 1.5.
a) To determine the third-degree polynomial using Lagrange's interpolation formula, we will use the following formula:
P(x) = Σ [yi * Li(x)]
where P(x) is the desired polynomial, yi are the given y-coordinates, and Li(x) are the Lagrange basis polynomials.
First, let's calculate the Lagrange basis polynomials:
L1(x) = (x - x2)(x - x3)(x - x4) / (x1 - x2)(x1 - x3)(x1 - x4)
= (x - 1)(x - 3)(x - 4) / (0 - 1)(0 - 3)(0 - 4)
= (x - 1)(x - 3)(x - 4) / -24
L2(x) = (x - x1)(x - x3)(x - x4) / (x2 - x1)(x2 - x3)(x2 - x4)
= (x - 0)(x - 3)(x - 4) / (1 - 0)(1 - 3)(1 - 4)
= (x)(x - 3)(x - 4) / 2
L3(x) = (x - x1)(x - x2)(x - x4) / (x3 - x1)(x3 - x2)(x3 - x4)
= (x - 0)(x - 1)(x - 4) / (3 - 0)(3 - 1)(3 - 4)
= (x)(x - 1)(x - 4) / -6
L4(x) = (x - x1)(x - x2)(x - x3) / (x4 - x1)(x4 - x2)(x4 - x3)
= (x - 0)(x - 1)(x - 3) / (4 - 0)(4 - 1)(4 - 3)
= (x)(x - 1)(x - 3) / 6
Now, let's substitute the given y-coordinates into the formula:
P(x) = (-12 * L1(x)) + (0 * L2(x)) + (6 * L3(x)) + (12 * L4(x))
= (-12 * (x - 1)(x - 3)(x - 4) / -24) + (0) + (6 * (x)(x - 1)(x - 4) / -6) + (12 * (x)(x - 1)(x - 3) / 6)
= (1/2)(x - 1)(x - 3)(x - 4) - 6x(x)(x - 1)(x - 4) + 2x(x)(x - 1)(x - 3)
= (1/2)(x - 1)(x^3 - 7x^2 + 12x - 4) - 6x(x^3 - 5x^2 + 4x - 4) + 2x(x^3 - 4x^2 + 3x - 3)
= (1/2)(x^4 - 7x^3 + 12x^2 - 4x - 6x^4 + 30x^3 - 24x^2 + 24x + 2x^4 - 8x^3 + 6x^2 - 6x)
= (1/2)(-3x^4 + 23x^3 - 6x^2 + 14x)
Therefore, the polynomial that passes through the given points is P(x) = (-3/2)x^4 + (23/2)x^3 - 3x^2 + 7x.
b) To find the value at x = 1.5, we substitute x = 1.5 into the polynomial:
P(1.5) = (-3/2)(1.5)^4 + (23/2)(1.5)^3 - 3(1.5)^2 + 7(1.5)
= (-3/2)(5.0625) + (23/2)(3.375) - 3(2.25) + 10.5
= -7.59375 + 45.46875 - 6.75 + 10.5
= 41.625
The value of the polynomial at x = 1.5 is 41.625.