[tex]f(x) = x^4-3x^2+x+7\\ \\f'(x) = 4x^3-6x+1\\ \\ \\ S =\frac{1}{2-x_1}+\frac{1}{2-x_2}+\frac{1}{2-x_3}+\frac{1}{2-x_4} = \\ \\ = \frac{(2-x_2)(2-x_3)(2-x_4)+(2-x_1)(2-x_3)(2-x_4)+...+(2-x_1)(2-x_2)(2-x_3)}{(2-x_1)(2-x_2)(2-x_3)(2-x_4)} =\\ \\ = \dfrac{f'(2)}{f(2)}\\ \\\\ f(x) = (x-x_1)(x-x_2)(x-x_3)(x-x_4) \\ \\ f'(x) = (x-x_2)(x-x_3)(x-x_4)+(x-x_1)(x-x_3)(x-x_4)+...+\\ +(x-x_1)(x-x_2)(x-x_3)[/tex]
Pentru x = 2 avem:
[tex]f(2) = (2-x_1)(2-x_2)(2-x_3)(2-x_4)= 16-12+2+7 = 13 \\\\ f'(2) = 4\cdot 8-12+1 = 21[/tex]
[tex]\Rightarrow \boxed{S = \dfrac{21}{13}}[/tex]
