Past exam of the options II course of the University of Oxford / 2007 / P / P3 / d / Solution

The cubic Hermite interpolant to the exponential at zero and one is

h (x) = (3 - e) x^{3} + (2 e - 5) x^{2} + x + 1.

Its endpoint values are one and

e

, and its endpoint derivatives are also one and

e

. The remainder formula gives

e^{x} - h (x) = e^{ξ} x^{2} (x - 1)^{2} /24

for an interior

ξ

. On this interval

e^{ξ} \leq e

and

x (1 - x) \leq 1/4

, so

∣ e^{x} - h (x) ∣ \leq \frac{e}{24} (\frac{1}{4})^{2} = \frac{e}{384} .

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