Use Newton’s Method with initial approximation x1 = 2 to solve the equation x5 +x= 6 . Go to x3,

Question

and tell me how good you think the approximation is. Round all answers to 3 decimal places.

Answer 1

let f(x) = x^5+x-6

newton's method says
x_n+1 = x_n - f(x_n)/f'(x_n)

f'(x) = 5x^4 + 1

so x2 = x1 - f(x1)/f'(x1)
= 2 - 28/81
= 1.65432

then x3 = 1.65432 - .209237
= 1.44508

x4 = 1.3685
x5 = 1.35942
x6 = 1.359304105
x7 = 1.359304086
x8 = 1.359304086 Ok then !!!