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
wrong again.
4.5x^3 - 21.5x^2 + 39x - 6 passes through
(0,-6), (1,16) ,(3,39) and (4,94)
To find the third-degree polynomial using Lagrange's interpolation formula, we need to find the coefficients of the polynomial by solving a system of equations.
The Lagrange's interpolation formula for a third-degree polynomial passing through (x0, y0), (x1, y1), (x2, y2), and (x3, y3) is given by:
P(x) = y0 * L0(x) + y1 * L1(x) + y2 * L2(x) + y3 * L3(x)
where:
L0(x) = ((x - x1)(x - x2)(x - x3)) / ((x0 - x1)(x0 - x2)(x0 - x3))
L1(x) = ((x - x0)(x - x2)(x - x3)) / ((x1 - x0)(x1 - x2)(x1 - x3))
L2(x) = ((x - x0)(x - x1)(x - x3)) / ((x2 - x0)(x2 - x1)(x2 - x3))
L3(x) = ((x - x0)(x - x1)(x - x2)) / ((x3 - x0)(x3 - x1)(x3 - x2))
Substituting the given points into the formula, we have:
P(x) = (-12) * L0(x) + 0 * L1(x) + 6 * L2(x) + 12 * L3(x)
where:
L0(x) = ((x - 1)(x - 3)(x - 4)) / ((0 - 1)(0 - 3)(0 - 4))
L1(x) = ((x - 0)(x - 3)(x - 4)) / ((1 - 0)(1 - 3)(1 - 4))
L2(x) = ((x - 0)(x - 1)(x - 4)) / ((3 - 0)(3 - 1)(3 - 4))
L3(x) = ((x - 0)(x - 1)(x - 3)) / ((4 - 0)(4 - 1)(4 - 3))
Simplifying the expressions of L0(x), L1(x), L2(x), and L3(x), we have:
L0(x) = (x - 1)(x - 3)(x - 4) / (-24)
L1(x) = x(x - 3)(x - 4) / 3
L2(x) = x(x - 1)(x - 4) / 6
L3(x) = x(x - 1)(x - 3) / 6
Substituting these expressions back into the polynomial equation, we have:
P(x) = (-12) * (x - 1)(x - 3)(x - 4) / (-24) + 6 * x(x - 1)(x - 4) / 6 + 12 * x(x - 1)(x - 3) / 6
Simplifying further, we have:
P(x) = (1/2) * (x - 1)(x - 3)(x - 4) + x(x - 1)(x - 4) + 2x(x - 1)(x - 3)
Expanding the polynomial, we have:
P(x) = (1/2) * (x^3 - 8x^2 + 17x - 12) + (x^3 - 5x^2 + 4x) + (2x^3 - 12x^2 + 18x)
Combining like terms, we have:
P(x) = (1/2) * x^3 - 4x^2 + (17/2) * x - 6 + x^3 - 5x^2 + 4x + 2x^3 - 12x^2 + 18x
P(x) = 4.5x^3 - 21.5x^2 + 39x - 6
Therefore, the polynomial that passes through the points (0, -12), (1, 0), (3, 6), and (4, 12) is P(x) = 4.5x^3 - 21.5x^2 + 39x - 6.
My apologies for the mistake. Let's use Lagrange's interpolation formula again to find the correct polynomial.
The Lagrange's interpolation formula for a third-degree polynomial passing through (x0, y0), (x1, y1), (x2, y2), and (x3, y3) is given by:
P(x) = y0 * L0(x) + y1 * L1(x) + y2 * L2(x) + y3 * L3(x)
where:
L0(x) = ((x - x1)(x - x2)(x - x3)) / ((x0 - x1)(x0 - x2)(x0 - x3))
L1(x) = ((x - x0)(x - x2)(x - x3)) / ((x1 - x0)(x1 - x2)(x1 - x3))
L2(x) = ((x - x0)(x - x1)(x - x3)) / ((x2 - x0)(x2 - x1)(x2 - x3))
L3(x) = ((x - x0)(x - x1)(x - x2)) / ((x3 - x0)(x3 - x1)(x3 - x2))
Substituting the given points into the formula, we have:
P(x) = (-6) * L0(x) + 16 * L1(x) + 39 * L2(x) + 94 * L3(x)
where:
L0(x) = ((x - 1)(x - 3)(x - 4)) / ((0 - 1)(0 - 3)(0 - 4))
L1(x) = ((x - 0)(x - 3)(x - 4)) / ((1 - 0)(1 - 3)(1 - 4))
L2(x) = ((x - 0)(x - 1)(x - 4)) / ((3 - 0)(3 - 1)(3 - 4))
L3(x) = ((x - 0)(x - 1)(x - 3)) / ((4 - 0)(4 - 1)(4 - 3))
Simplifying the expressions of L0(x), L1(x), L2(x), and L3(x), we have:
L0(x) = -(1/12) * (x - 1)(x - 3)(x - 4)
L1(x) = (1/2) * x(x - 3)(x - 4)
L2(x) = -(1/2) * x(x - 1)(x - 4)
L3(x) = (1/12) * x(x - 1)(x - 3)
Substituting these expressions back into the polynomial equation, we have:
P(x) = (-6) * (-(1/12) * (x - 1)(x - 3)(x - 4)) + 16 * ((1/2) * x(x - 3)(x - 4)) + 39 * (-(1/2) * x(x - 1)(x - 4)) + 94 * ((1/12) * x(x - 1)(x - 3))
Simplifying further, we have:
P(x) = (1/2) * x^3 - 9x^2 + (73/2) * x - 6
Therefore, the correct polynomial that passes through the points (0, -6), (1, 16), (3, 39), and (4, 94) is P(x) = (1/2) * x^3 - 9x^2 + (73/2) * x - 6.