How do you use Newton's method to find the approximate solution to the equation $x^{4} = x + 1, x < 0$ $x^4=x+1,x<0$ ?

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Answer 1 · 2016-12-29T21:53:25

$x = - 0.72449$ $x=-0.72449$ to 5dp

We have:

$x^{4} = x + 1 \Rightarrow x^{4} - x - 1 = 0$ $x^4=x+1 => x^4-x-1 = 0$

Let $f (x) = x^{4} - x - 1$ $f(x) = x^4-x-1$ Then our aim is to solve $f (x) = 0$ $f(x)=0$

First let us look at the graph:
graph{x^4-x-1 [-3, 3, -5, 8]}

We can see there is one solution in the interval $- 1 < x < 0$ $-1 < x < 0$ and a solution in $1 < x < 2$ $1 < x < 2$

In order to find the solution numerically, using Newton-Rhapson method, we need the derivative $f'(x)$

$f(x) = x^4-x-1 => f'(x) = 4x^3-1$ ,

and the Newton-Rhapson method uses the following iterative sequence

${ (x_0,=0), ( x_(n+1), = x_n - f(x_n)/(f'(x_n)) ) :}$

Then using excel working to 5dp we can tabulate the iterations as follows:

enter image source here

And we conclude that a solution is $x=-0.72449$ to 5dp

How do you use Newton's method to find the approximate solution to the equation x^4=x+1,x<0x4=x+1,x<0x^4=x+1,x<0?