use newton's method with initial approximation x1 = −2 to find x2, the second approximation to the root of the equation x3 +2x + 5 = 0. (round your answer to four decimal places.)

Question

use newton's method with initial approximation x1 = −2 to find x2, the second approximation to the root of the equation x3 +2x + 5 = 0. (round your answer to four decimal places.)

Answer 1

Here's the calculation from WolframAlpha.Com. Note the general step is to replace x by x – f(x)/f'(x):

x_0 = -2

x_1 = x_0 - (x_0^3 + 2 x_0 + 5)/(3 x_0^2 + 2)

x_1 = -2 - (-0.5)

x_1 = -1.5

x_2 = x_1 - (x_1^3 + 2 x_1 + 5)/(3 x_1^2 + 2)

x_2 = -1.5 - (-0.157143)

x_2 = -1.34286

x_3 = x_2 - (x_2^3 + 2 x_2 + 5)/(3 x_2^2 + 2)

x_3 = -1.34286 - (-0.014473)

x_3 = -1.32838

x_4 = x_3 - (x_3^3 + 2 x_3 + 5)/(3 x_3^2 + 2)

x_4 = -1.32838 - (-0.000115279)

x_4 = -1.32827

x_5 = x_4 - (x_4^3 + 2 x_4 + 5)/(3 x_4^2 + 2)

x_5 = -1.32827 - (-7.26162×10^-9)

x_5 = -1.32827