Question

Use Newton's method with initial approximation x1 = −2 to find x2, the second approximation to the root of the equation x^3 + x + 5 = 0?

Answer 1

After a one iteration a second estimation is:

`x_2 = -1.75000000`

And iterating until we get convergence, we conclude that the solution (to 8dp) are:

`x = -1.51598023`

Explanation:

We have:

`x^3+x+5=0`

Let:

`f(x) = x^3+x+5`

Our aim is to solve `f(x)=0`. First let us look at the graph:
graph{x^3+x+5 [-4, 4, -5, 10]}

We can see that there is one solutions in the interval `-2 lt x lt 0`

To find the solution numerically, using Newton-Rhapson method we use the following iterative sequence

`{ (x_1,=x_0), ( x_(n+1), = x_n - f(x_n)/(f'(x_n)) ) :}`

Therefore we need the derivative:

`\ \ \ \ \ \ \f(x) = 3x^2+1`

Then using excel working to 8dp with `x_0=-2` we can tabulate the iterations as follows:

`x_1 = -1`
`x_2 = -1.75000000`

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 iterating until we get convergence, we conclude that the solution (to 8dp) are:

`x = -1.51598023`