Question

Use Newton's method to approximate a root of the equation (2 x^3 + 4 x + 4 =0) as follows.
Let (x_1 = -1\) be the initial approximation.
The second approximation (x_2) is ?
and the third approximation (x_3) is ?

Answer 1

First of all, why not reduce it to
x^3 + 2x + 2 = 0

Your iteration formula should be
x<sub>new</b> = (2x^3 - 2)/(3x^2 + 2)

so for a start of x=-1
we get
x<sub>1</b> = -1.333333
x<sub>2</b> = -.91919
x<sub>3</b> = -.78357

BTW, x<sub>5</b> = -.770917
which is correct to 7 decimal places.

Answer 2

looks like I forgot to close my subscript code
The end should read:

Your iteration formula should be
xnew = (2x^3 - 2)/(3x^2 + 2)

so for a start of x=-1
we get
x1 = -1.333333
x2 = -.91919
x3 = -.78357

BTW, x5 = -.770917
which is correct to 7 decimal places.

Answer 3

using by newton's method an approximate value the positive root