# How do you use Newton's method to find the approximate solution to the equation #x^3+5x-10=0#?

Let

First let us look at the graphs:

graph{x^3+5x-10 [-5, 5, -20, 15]}

We can see there is one solution in the interval

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

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

And we conclude that the remaining solution is

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To use Newton's method to find the approximate solution to the equation (x^3 + 5x - 10 = 0):

- Choose an initial guess, (x_0).
- Iterate using the formula: (x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}), where (f(x)) is the function and (f'(x)) is its derivative.
- Repeat the iteration until the desired level of accuracy is reached or until the process converges.

For the given equation, (f(x) = x^3 + 5x - 10). Taking the derivative, (f'(x) = 3x^2 + 5).

Choose an initial guess, (x_0), and substitute it into the iteration formula 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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