Posted by Salman on Wednesday, November 11, 2009 at 4:31am.
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 ?

Calculus  Reiny, Wednesday, November 11, 2009 at 8:14am
First of all, why not reduce it to
x^3 + 2x + 2 = 0
Your iteration formula should be
x_{new = (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.}

Calculus  Reiny, Wednesday, November 11, 2009 at 1:21pm
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.

3 x  6x + 4 = 0  rajkumar, Wednesday, October 21, 2015 at 10:26am
using by newton's method an approximate value the positive root
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