# Calculus 1 Newton's Method

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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 - ,

let f(x) = y = x^3 + x +2
dy/dx = 3x^2 + 1

newton's formula

x new = x old - f(xold/f '(xold)

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

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

• Calculus 1 Newton's Method - ,

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 - ,

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

• Calculus 1 Newton's Method - ,

My last line should say:

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

• Calculus 1 Newton's Method-corr - ,

I'll get it right this time!
the targeted root of (1/2)±isqrt(7)/2.

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