Question

Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given equation. (Round your answer to four decimal places.) What is x3?

(1/3)x^3+(1/2)x^2+1=0

x1 = -3

Answer 1

Using N-R Iteration we get (to 4dp):

`x_2 = -2.4167`
`x_3 = -2.1875`

Continuing the iteration we get convergence to 8dp to the solution `-2.14937621`

Explanation:

Let `f(x) = 1/3x^3+1/2x^2+1` Then our aim is to solve `f(x)=0`.

First let us look at the graphs:
graph{1/3x^3+1/2x^2+1 [-4, 4, -10, 10]}

We can see there is one solution in the interval `-4 le x le -0`.

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

`\ \ \ \ \ \ \f(x) = 1/3x^3+1/2x^2+1`
`:. f'(x) = x^2+x`

The Newton-Rhapson method uses the following iterative sequence

`{ (x_1,=-3), ( 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=-2.14937621` to 8dp.