Use Lagrange polynomial estimate (2) for the given data.
x -2 -1 0 4
f(x) -2 4 1 8
Lagrange polynomial:
"P(x)=f_0L_0(x)+f_1L_1(x)+f_2L_2(x)+f_3L_3(x)"
"L_0(x)=\\frac{(x-x_1)(x-x_2)(x-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)}=\\frac{(x+1)x(x-4)}{(-2+1)(-2)(-2-4)}=-\\frac{(x+1)x(x-4)}{12}"
"L_1(x)=\\frac{(x-x_0)(x-x_2)(x-x_3)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)}=\\frac{(x+2)x(x-4)}{(-1+2)(-1)(-1-4)}=\\frac{(x+2)x(x-4)}{5}"
"L_2(x)=\\frac{(x-x_0)(x-x_1)(x-x_3)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)}=\\frac{(x+2)(x+1)(x-4)}{(0+2)(0+1)(0-4)}=-\\frac{(x+2)(x+1)(x-4)}{8}"
"L_3(x)=\\frac{(x-x_0)(x-x_1)(x-x_2)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)}=\\frac{(x+2)(x+1)x}{4(4+2)(4+1)}=\\frac{(x+2)(x+1)x}{120}"
"f(2)=P(2)=2\\frac{2(2+1)(2-4)}{12}+4\\frac{2(2+2)(2-4)}{5}-\\frac{(2+2)(2+1)(2-4)}{8}+8\\frac{2(2+2)(2+1)}{120}="
"=-2-12.8+3+1.6=-10.2"
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