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

We have:

# x^4=x+1 => x^4-x-1 = 0 #

Let

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

In order to find the solution numerically, using Newton-Rhapson method, we need the derivative

# 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:

And we conclude that a solution is

#x=-0.72449# to 5dp

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To use Newton's method to find an approximate solution to the equation (x^4 = x + 1), where (x < 0), follow these steps:

- Choose an initial guess (x_0) for the solution.
- Calculate the derivative (f'(x)) of the function (f(x) = x^4 - x - 1).
- Use the formula for Newton's method iteration: (x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}).
- Iterate using the formula until you reach a satisfactory approximation, i.e., until (|x_{n+1} - x_n|) is sufficiently small.

Given that (x < 0), choose a suitable initial guess for (x_0), then apply the Newton's method iteration formula to find the approximate solution. Repeat the process until convergence is achieved.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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