# How do you use Newton's method to find the approximate solution to the equation #x^3-10x+4=0, 0<x<1#?

Iterate until you get the desired accuracy.

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To use Newton's method to find the approximate solution to the equation (x^3 - 10x + 4 = 0) in the interval (0 < x < 1):

- Choose an initial guess, let's call it (x_0), within the given interval.
- Use the formula for Newton's method to iteratively refine the estimate for the root: [x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}]
- Repeat step 2 until the desired level of accuracy is achieved or until the iterations converge.

Given (f(x) = x^3 - 10x + 4), we need to find (f'(x)) to apply Newton's method.

[f'(x) = 3x^2 - 10]

Now, choose an initial guess, let's say (x_0 = 0.5), within the interval (0 < x < 1).

Substitute (x_0) into (f(x)) and (f'(x)) to start the iterations:

[f(0.5) = (0.5)^3 - 10(0.5) + 4 = -6.375] [f'(0.5) = 3(0.5)^2 - 10 = -8.75]

Apply Newton's method: [x_1 = 0.5 - \frac{-6.375}{-8.75} \approx 0.7286]

Repeat the process with (x_1) until the desired accuracy is achieved or until convergence.

Iterating further: [f(0.7286) \approx 0.3902] [f'(0.7286) \approx -6.0627]

[x_2 = 0.7286 - \frac{0.3902}{-6.0627} \approx 0.6868]

Continue this process until the desired accuracy is reached.

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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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