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

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Steve Share
Dec 29, 2016

$x = - 0.72449$ to 5dp

#### Explanation:

We have:

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

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

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 $- 1 < x < 0$ and a solution in $1 < x < 2$

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

$f \left(x\right) = {x}^{4} - x - 1 \implies f ' \left(x\right) = 4 {x}^{3} - 1$,

and the Newton-Rhapson method uses the following iterative sequence

$\left\{\begin{matrix}{x}_{0} & = 0 \\ {x}_{n + 1} & = {x}_{n} - f \frac{{x}_{n}}{f ' \left({x}_{n}\right)}\end{matrix}\right.$

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

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

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