Question

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`

Answer 1

`x=1.2113` to 4dp

Explanation:

Let `f(x) = x^4-2x^3+3x^2-3` Then our aim is to solve `f(x)=0` in the interval `1 le x le 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 2` (along with a further solution in `-1 lt x lt 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:

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