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Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 4 / 6B / b / Solution

Appending the node

- 2

to the existing Newton interpolation polynomial gives

p_{4} (x) = p_{3} (x) + a_{4} x (x - 1) (x - 2) (x - 3) .

At the new node,

p_{3} (- 2) = - 82, (- 2) (- 3) (- 4) (- 5) = 120.

The condition

p_{4} (- 2) = 10

therefore gives

a_{4} = \frac{10 - ( - 82 )}{120} = \frac{23}{30} .

Hence

p_{4} (x) = 4 x - 3 x (x - 1) + \frac{7}{3} x (x - 1) (x - 2) + \frac{23}{30} x (x - 1) (x - 2) (x - 3) .

Solved by gpt-5.6-sol high.

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