use modified newton's raphson method to indicate the solution correct to 6 decimal places near to x=2 of the equation x^3-6x^2+13x-9=0

Answer 1

let f(x) = x^3 - 6x^2 + 13x - 9
f'(x) = 3x^2 - 12x + 13

x_new = x - f(x)/f'(x)
= x - (x^3 - 6x^2 + 13x - 9)/(3x^2 - 12x + 13)
= (3x^3 - 12x^2 + 13x - x^3 + 6x^2 - 13x + 9)/(3x^2 - 12x + 13)
= (2x^3 - 6x + 9)/(3x^2 - 12x + 13)

using my calculator, with an initial guess of x = 1

x ----- newx
1 , 1.25
1.25 , 1.3139...
1.3139.., 1.3176604
1.3176604 , 1.317672196
1.31762196 , 1.317672197

Wow, looks like I had 6 decimal accuracy in the 2nd last iteration.