Question

How do you use Newton's method to find the approximate solution to the equation `2x^5+3x=2`?

Answer 1

`x~~0.610`
(See below for Newton Method of approximation).

Explanation:

Noting that if `x=0` then `2x^5+3x < 2`
and if `x=1` then `2x^5+3x > 2`

we can start with "bracketing" values Low`=0` and High`=1`

At each iteration we evaluate the mid-point and adjust either the Low or High closing in the brackets about the solution value.

Here is what the first 10 iterations look like in a spreadsheet form:

Answer 2

`x=0.610246` to 6dp

Explanation:

Let `f(x) = 2x^5+3x-2` Then our aim is to solve `f(x)=0`

First let us look at the graphs:
graph{2x^5+3x-2 [-10, 10, -5, 4.995]}

We can see there is one solution in the interval `0 < x < 1`.

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

`f(x) = 2x^5+3x-2 => f'(x) = 10x^4+3`, and using the Newton-Rhapson method we use the following iterative sequence

<IMAGE>

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

Then using excel working to 6dp we can tabulate the iterations as follows:

And we conclude that the remaining solution is `x=0.610246` to 6dp