Hi this is a Newton's method problem.
Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given equation. (Give your answer to four decimal places.)
x^5 + 5 = 0, x1 = −1

When I finished, I got 624.7625. but it is wrong! :(

Question

Hi this is a Newton's method problem.
Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given equation. (Give your answer to four decimal places.)
x^5 + 5 = 0,    x1 = −1

When I finished, I got 624.7625. but it is wrong! :(

anonymous · Answer

you have to round it up when you move the decimal if the number next to it is grater then 5 you round up 1

dumbcow · Answer

x_n+1 = x_n - f(x)/f'(x)

f'(x) = 5x^4

x2 = -1 -4/5 = -1.8
x3 = -1.8 -(-1.8^5 +5)/5(-1.8)^4) = -1.5353

anonymous · Answer

thank you! wow i realized that I didnt have x1 as NEGATIVE 1. oops!! xD

dumbcow · Answer

:)