Asked by Soojung
Use Newton’s Method with initial approximation x1 = 2 to solve the equation x5 +x= 6 . Go to x3,
and tell me how good you think the approximation is. Round all answers to 3 decimal places.
and tell me how good you think the approximation is. Round all answers to 3 decimal places.
Answers
Answered by
Reiny
let f(x) = x^5+x-6
newton's method says
x<sub>n+1</sub> = x<sub>n</sub> - f(x<sub>n</sub>)/f'(x<sub>n</sub>)
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 !!!
newton's method says
x<sub>n+1</sub> = x<sub>n</sub> - f(x<sub>n</sub>)/f'(x<sub>n</sub>)
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 !!!
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