# Calculus

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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 -

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

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.

• Calculus -

looks like I forgot to close my subscript code

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 -

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

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