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

Question

8017 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-16T12:58:57

$x \approx 0.610$ $x~~0.610$
(See below for Newton Method of approximation).

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

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

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:
enter image source here

Answer 2 · 2016-12-16T13:15:56

$x = 0.610246$ $x=0.610246$ to 6dp

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

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 $0 < x < 1$ $0 < x < 1$ .

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:

enter image source here

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

How do you use Newton's method to find the approximate solution to the equation 2x^5+3x=22x5+3x=22x^5+3x=2?