Question

From the following data find f'(5).
x= 1 3 4 6
F(x)= 14 2 8 9

Answer 1

Given data has unequal intervals, hence we use Lagrange's interpolation formula to find the polynomial f(x) that fit for the data and find f'(5). The Lagrange's interpolation formula is given by,

$y =f(x)= \dfrac{(x-x_{1})(x-x_{2})\cdots(x-x_{n})}{(x_{0}-x_{1})(x_{0}-x_{2})\cdots(x_{0}-x_{n})}\cdot y_{0}\\ ~~~~+\dfrac{(x-x_{0})(x-x_{2})\cdots(x-x_{n})}{(x_{1}-x_{0})(x_{1}-x_{2})\cdots(x_{1}-x_{n})}\cdot y_{1} + \cdots + \\ ~~~~\dfrac{(x-x_{0})(x-x_{1})\cdots(x-x_{n-1})}{(x_{n}-x_{0})(x_{n}-x_{2})\cdots(x_{n}-x_{n-1})}\cdot y_{n}$

Here,

$x_{0} = 1, x_{1} = 3, x_{2} = 4, x_{3} = 6; \\ y_{0} = 14, y_{1} = 2, y_{2} = 8, y_{3}=9.$

$f(x)= \dfrac{(x-3)(x-4)(x-6)}{(-2)(-3)(-5)}\cdot 14+\\ ~~~~~~~~~~~~~~\dfrac{(x-1)(x-4)(x-6)}{(2)(-1)(-3)}\cdot 2 +\\~~~~~~~~~~~~~~~ \dfrac{(x-1)(x-3)(x-6)}{(3)(1)(-2)}\cdot 4+ \\~~~~~~~~~~~~~~~ \dfrac{(x-1)(x-3)(x-4)}{(5)(3)(2)}\cdot 9$

$~~~~~~~~~~=\dfrac{-7}{15}(x^{3}-13x^{2}+54x-72)+\\ ~~~~~~~~~~~~~~~~~~~~\dfrac{1}{3}(x^{3}-11x^{2}+34x-24) - \\~~~~~~~~~~~~~~~~~\dfrac{4}{3}(x^{3}-10x^{2}+27x-18) + \\ ~~~~~~~~~~~~~\dfrac{3}{10}(x^{3}-8x^{2}+19x-12)$

This reduces to the polynomial,

$f(x) = -\dfrac{1}{6}(7x^{3}-80x^{2}+265x-276)\\$

Differentiating with respect to 'x',

$f'(x) = -\dfrac{1}{6}(21x^{2}-160x+265)\\$

Therefore, $f'(5) = -\dfrac{1}{6}(21\cdot 5^{2}-160\cdot 5+265) = \dfrac{5}{3} = 1.6667$