Use Newton's method to approximate the indicated root of the equation correct to six decimal places. The root of x^4 − 2x^3 + 5x^2 − 5 = 0 in the interval [1, 2]
To use Newton's method to approximate the root of the equation x^4 - 2x^3 + 5x^2 - 5 = 0 in the interval [1, 2] to six decimal places, follow these steps:
1. Start by finding the derivative of the function. Differentiating x^4 - 2x^3 + 5x^2 - 5 with respect to x, we get: 4x^3 - 6x^2 + 10x.
2. Choose an initial guess, x_0, within the interval [1, 2] as a starting point for the method. Let's choose x_0 = 1.5.
3. Calculate the next approximation, x_1, using the formula: x_1 = x_0 - (f(x_0) / f'(x_0)), where f(x) is the original function and f'(x) is its derivative.
Plugging in the values, we have:
x_1 = 1.5 - ((1.5^4 - 2(1.5)^3 + 5(1.5)^2 - 5) / (4(1.5)^3 - 6(1.5)^2 + 10(1.5)))
Simplifying the expression, we get:
x_1 ≈ 1.414263
4. Repeat the process iteratively until the desired level of accuracy is achieved. Calculate subsequent approximations, x_2, x_3, and so on, using the same formula as in step 3.
5. Keep iterating until the difference between consecutive approximations is less than the desired tolerance (epsilon). In this case, we want the approximation to be correct to six decimal places. Therefore, continue to iterate the formula until |x_(n+1) - x_n| < 0.000001 (epsilon).
6. The final approximation, which is correct to six decimal places, will be the value of x when the iteration process stops.
Keep in mind that Newton's method may not always converge or may converge to a different root, depending on the initial guess and the behavior of the function.
To use Newton's method to approximate the root of the equation x^4 − 2x^3 + 5x^2 − 5 = 0 in the interval [1, 2], we'll need to follow these steps:
Step 1: Write the equation in the form f(x) = 0:
f(x) = x^4 − 2x^3 + 5x^2 − 5
Step 2: Find the derivative of f(x), denoted as f'(x):
f'(x) = 4x^3 - 6x^2 + 10x
Step 3: Choose an initial approximation, denoted as x₀, within the interval [1, 2]. Let's select x₀ = 1.5.
Step 4: Use the formula of Newton's method to find the next approximation, denoted as x₁:
x₁ = x₀ - f(x₀)/f'(x₀)
Step 5: Repeat step 4 until the desired accuracy is achieved. In this case, let's continue until the approximation is correct to six decimal places.
By following these steps, we can proceed with the calculations.