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.
To determine the third-degree polynomial passing through the given points using Lagrange's interpolation formula, we start by writing down the formula for a third-degree polynomial:
P(x) = L0(x)y0 + L1(x)y1 + L2(x)y2 + L3(x)y3
where L0(x), L1(x), L2(x), and L3(x) are the Lagrange basis polynomials and y0, y1, y2, and y3 are the corresponding y-values of the points.
Next, we calculate the Lagrange basis polynomials:
L0(x) = (x - x1)(x - x2)(x - x3) / (x0 - x1)(x0 - x2)(x0 - x3)
= (x - 1)(x - 3)(x - 4) / (0 - 1)(0 - 3)(0 - 4)
= (x - 1)(x - 3)(x - 4) / (-1)(-3)(-4)
L1(x) = (x - x0)(x - x2)(x - x3) / (x1 - x0)(x1 - x2)(x1 - x3)
= (x - 0)(x - 3)(x - 4) / (1 - 0)(1 - 3)(1 - 4)
= (x)(x - 3)(x - 4) / 1(-2)(-3)
L2(x) = (x - x0)(x - x1)(x - x3) / (x2 - x0)(x2 - x1)(x2 - x3)
= (x - 0)(x - 1)(x - 4) / (3 - 0)(3 - 1)(3 - 4)
= (x)(x - 1)(x - 4) / 3(2)(-1)
L3(x) = (x - x0)(x - x1)(x - x2) / (x3 - x0)(x3 - x1)(x3 - x2)
= (x - 0)(x - 1)(x - 3) / (4 - 0)(4 - 1)(4 - 3)
= (x)(x - 1)(x - 3) / 4(3)(1)
Substituting the given points into the Lagrange interpolation formula:
P(x) = L0(x)(-12) + L1(x)(0) + L2(x)(6) + L3(x)(12)
= ((x - 1)(x - 3)(x - 4) / (-1)(-3)(-4))(-12)
+ ((x)(x - 3)(x - 4) / 1(-2)(-3))(0)
+ ((x)(x - 1)(x - 4) / 3(2)(-1))(6)
+ ((x)(x - 1)(x - 3) / 4(3)(1))(12)
Simplifying the expression:
P(x) = 2(x - 1)(x - 3)(x - 4) + 3x(x - 3)(x - 4) - x(x - 1)(x - 4) + (3/2)x(x - 1)(x - 3)
Now we can substitute x = 1.5 to find the value at that point:
P(1.5) = 2(1.5 - 1)(1.5 - 3)(1.5 - 4) + 3(1.5)(1.5 - 3)(1.5 - 4) - (1.5)(1.5 - 1)(1.5 - 4) + (3/2)(1.5)(1.5 - 1)(1.5 - 3)