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
Lagrange's interpolation formula is given by:
P(x) = Σ [Lk(x) * yk]
where P(x) is the polynomial of degree n that passes through the points (xk, yk), Lk(x) is the Lagrange interpolating polynomial for yk, and Σ denotes the sum over k from 0 to n.
In this case, we have four points: (0, -12), (1, 0), (3, 6), and (4, 12).
Let's calculate each Lagrange interpolating polynomial:
L0(x) = [(x - 1)(x - 3)(x - 4)] / [(0 - 1)(0 - 3)(0 - 4)]
= (x - 1)(x - 3)(x - 4) / (-1)(-3)(-4)
= (x - 1)(x - 3)(x - 4) / 12
L1(x) = [(x - 0)(x - 3)(x - 4)] / [(1 - 0)(1 - 3)(1 - 4)]
= x(x - 3)(x - 4) / (1)(-2)(-3)
= -x(x - 3)(x - 4) / 6
L2(x) = [(x - 0)(x - 1)(x - 4)] / [(3 - 0)(3 - 1)(3 - 4)]
= x(x - 1)(x - 4) / (3)(1)(-1)
= -x(x - 1)(x - 4) / 3
L3(x) = [(x - 0)(x - 1)(x - 3)] / [(4 - 0)(4 - 1)(4 - 3)]
= x(x - 1)(x - 3) / (4)(3)(1)
= x(x - 1)(x - 3) / 12
Now, substituting the values of yk into the formula:
P(x) = L0(x) * y0 + L1(x) * y1 + L2(x) * y2 + L3(x) * y3
= (x - 1)(x - 3)(x - 4) / 12 * (-12) + (-x(x - 3)(x - 4) / 6) * 0 + (-x(x - 1)(x - 4) / 3) * 6 + (x(x - 1)(x - 3) / 12) * 12
= -2(x - 1)(x - 3)(x - 4) + 2x(x - 1)(x - 4) + x(x - 1)(x - 3)
= -2x^3 + 13x^2 - 18x + 12
Therefore, the third-degree polynomial that passes through the given points is P(x) = -2x^3 + 13x^2 - 18x + 12.