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

Question

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

Answer 1

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

Answer 2

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.

Answer 3

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

Answer 4

My last line should say:

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

Answer 5

I'll get it right this time!

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

Answer 6

To approximate the root of the equation using Newton's method, we need to follow these steps:

Step 1: Start with an initial guess for the root. In this case, the given initial guess is x1 = -1.

Step 2: Calculate the derivative of the equation with respect to x. In this case, the derivative of f(x) = x^3 + x + 2 is f'(x) = 3x^2 + 1.

Step 3: Use the initial guess and the derivative value to calculate the next approximation using the formula:

x2 = x1 - (f(x1) / f'(x1))

Step 4: Repeat the process to find x3 using the formula:

x3 = x2 - (f(x2) / f'(x2))

Now, let's apply these steps to find x2 and x3.

Step 1: Given initial guess x1 = -1

Step 2: Calculate the derivative f'(x) = 3x^2 + 1

Step 3: Calculate x2 using the formula:

x2 = x1 - (f(x1) / f'(x1))

Substituting the values:

x2 = -1 - ((-1^3 + -1 + 2) / (3*(-1)^2 + 1))
= -1 - ((-2) / 4)
= -1 + 0.5
= -0.5

Step 4: Calculate x3 using the formula:

x3 = x2 - (f(x2) / f'(x2))

Substituting the values:

x3 = -0.5 - ((-0.5^3 + -0.5 + 2) / (3*(-0.5)^2 + 1))

To calculate this expression, you would substitute the values and perform the calculations. The result will be the approximation for x3.

Note: Newton's method is an iterative process, which means you can continue these steps to get better approximations as long as the calculations don't result in zero or -1.