How do you use Newton's method to find the approximate solution to the equation 2x^3+x+4=02x3+x+4=0?

1 Answer
Steve M
Feb 5, 2017

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

Explanation:

Let f(x) = 2x^3+x^2+4f(x)=2x3+x2+4 Then our aim is to solve f(x)=0f(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 0−2≤x≤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:

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=-1.45054017 to 8dp.

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