Posted by **Olivia** on Monday, October 31, 2011 at 11:54pm.

Using Newton's method, approximate the root of the equation x^3+x+2=0 with the initial guess x1=-1 gives: x2=? and x3=? answers are not 0 or -1

- Calculus 1 Newton's Method -
**Reiny**, Tuesday, November 1, 2011 at 9:14am
let f(x) = y = x^3 + x +2

dy/dx = 3x^2 + 1

newton's formula

x _{new} = x _{old} - f(x_{old}/f '(x_{old})

x2 = -1 - (-1-1+2)/(3+1) = 1-0 = -1

since x1 = x1 exactly, x = -1 is a root

- Calculus 1 Newton's Method -
**MathMate**, Tuesday, November 1, 2011 at 9:14am
Check for typo's in the question.

I believe there is only one real root for the given equation at -1.

So your initial guesses have to be complex, such as 1+i. Convergence will depend on the form of the iteration equation.

You can try

f(x)=(-2-x)^(1/3)

f(x)=-2-x^3

and all kinds of other ones.

The one that seems to converge best is

first make

x^3=-2-x

divide by x and take the square root to get

f(x)=sqrt(-1-2/x)

With a starting value of 1+i, ou should converge quite well, as the iterations alternate between the targeted root of (1/2)±sqrt(7)/2.

Once you have converged to one, you can take the conjugate for the other, without having to do the same things all over again.

- Calculus 1 Newton's Method-corr -
**MathMate**, Tuesday, November 1, 2011 at 9:16am
the targeted root of (1/2)±sqrt(7)/2**i**.

- Calculus 1 Newton's Method -
**Reiny**, Tuesday, November 1, 2011 at 9:16am
My last line should say:

since x2 = x1 exactly, x = -1 is a root

- Calculus 1 Newton's Method-corr -
**MathMate**, Tuesday, November 1, 2011 at 9:17am
I'll get it right this time!

the targeted root of (1/2)±**i**sqrt(7)/2.

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