Use Newton's method to approximate the indicated root of the equation correct to six decimal places?

in the interval [1,2] what is the root of x^4 − 2x^3 + 3x^2 − 3 x4−2x3+3x2−3

1 Answer
Steve M
Jan 12, 2017

x=1.2113x=1.2113 to 4dp

Explanation:

Let f(x) = x^4-2x^3+3x^2-3f(x)=x4−2x3+3x2−3 Then our aim is to solve f(x)=0f(x)=0 in the interval 1 le x le 21≤x≤2

First let us look at the graphs:
graph{x^4-2x^3+3x^2-3 [-5, 5, -15, 15]}

We can see there is one solution in the interval 1 le x le 21≤x≤2 (along with a further solution in -1 lt x lt 0−1<x<0)

We can find the solution numerically, using Newton-Rhapson method

\ \ \ \ \ \ \f(x) = x^4-2x^3+3x^2-3
:. f'(x) = 4x^3-6x^2+6x

The Newton-Rhapson method uses the following iterative sequence

{ (x_0,=1), ( x_(n+1), = x_n - f(x_n)/(f'(x_n)) ) :}

Then using excel working to 8dp we can tabulate the iterations as follows:

enter image source here

We could equally use a modern scientific graphing calculator as most new calculators have an " Ans " button that allows the last calculated result to be used as the input of an iterated expression.

And we conclude that the solution is x=1.2113 to 4dp

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