Question

How do you use Newton's method to find the approximate solution to the equation `2x^3+x+4=0`?

Answer 1

Using N-R Iteration we get the solution is `x=-1.4505` to 4dp.

Explanation:

Let `f(x) = 2x^3+x^2+4` Then our aim is to solve `f(x)=0`.

First let us look at the graphs:
graph{2x^3+x^2+4 [-4, 4, -10, 10]}

We can see there is one solution in the interval `-2 le x le 0`.

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

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

The Newton-Rhapson method uses the following iterative sequence

`{ (x_1,=-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=-1.45054017` to 8dp.