Question

Use Newton's method to approximate the indicated root of the equation correct to six decimal places? The root of `f(x) =x^4 − 2x^3 + 3x^2 − 6 = 0` in the interval [1, 2]

Answer 1

This reference for Newtons Method gives us the equation:

`x_(n+1)= x_n - f(x_n)/(f'(x_n))" [1]"`

where `f(x) = 0` and `f'(x)` is the derivative

Explanation:

Given: `f(x) =x^4 − 2x^3 + 3x^2 − 6 = 0" [2]"`

`f'(x) = 4x^3-6x^2+6x" [3]"`

Substitute equation [2] and [3] into equation [1]:

`x_(n+1)= x_n - (x_n^4 − 2x_n^3 + 3x_n^2 − 6)/(4x_n^3-6x_n^2+6x_n)" [4]"`

I recommend that you use an Excel spread sheet.

Enter 1 into cell A1

Enter the following into cell A2:

=A1-(A1^4-2A1^3+3A1^2-6)/(4A1^3-6A1^2+6*A1)

Please observe that this is the Excel language equivalent of equation [4].

Copy the contents of cell A2 into cells A3 through A10.

Please observe that the contents of the cells converge on the number 1.596072; this is the root within the specified interval.