f(x) = x3 + 0,66 x - 1,17; xo = 0,8
f '(x)= 3x^2 +0,66
Also \( x_1 = x_0 - \frac{f(x_0)}{f '(x_0)}=0,8 - \frac{f(0,8)}{f'(0,8)}=0,8 - \frac{-0,13}{2,58} =0,85503\)
\( x_2 = x_1- \frac{f(x_1)}{f '(x_1)}=0,85503 - \frac{f(0,85503)}{f'(0,85503)}=0,85503 - \frac{0,00622}{2,8295} =0,84819\)
\( x_3 = x_2- \frac{f(x_1)}{f '(x_1)}=0,84819 - \frac{f(0,84819)}{f'(0,84819)}=0,84819 - \frac{0,000012}{2,81827} =0,84818\)
Also auf 4 Stellen genau 0,8482.