Use Newton's method to approximate the indicated root of the equation correct to six decimal places? The root of #f(x) =x^4 − 2x^3 + 3x^2 − 6 = 0# in the interval [1, 2]

Answer 1

This reference for Newtons Method gives us the equation:

#x_(n+1)= x_n - f(x_n)/(f'(x_n))" [1]"#

where #f(x) = 0# and #f'(x)# is the derivative

Given: #f(x) =x^4 − 2x^3 + 3x^2 − 6 = 0" [2]"#
#f'(x) = 4x^3-6x^2+6x" [3]"#

Substitute equation [2] and [3] into equation [1]:

#x_(n+1)= x_n - (x_n^4 − 2x_n^3 + 3x_n^2 − 6)/(4x_n^3-6x_n^2+6x_n)" [4]"#

I recommend that you use an Excel spread sheet.

Enter 1 into cell A1

Enter the following into cell A2:

=A1-(A1^4-2A1^3+3A1^2-6)/(4A1^3-6A1^2+6*A1)

Please observe that this is the Excel language equivalent of equation [4].

Copy the contents of cell A2 into cells A3 through A10.

Please observe that the contents of the cells converge on the number 1.596072; this is the root within the specified interval.

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Answer 2

To approximate the root of ( f(x) = x^4 - 2x^3 + 3x^2 - 6 = 0 ) in the interval [1, 2] using Newton's method, follow these steps:

  1. Start with an initial guess, let's say ( x_0 = 1.5 ) (since it lies in the interval [1, 2]).
  2. Iterate using the formula: [ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} ]
  3. Calculate ( f(x_n) ) and ( f'(x_n) ) at each iteration.
  4. Repeat the iterations until the desired accuracy is achieved.

Performing the iterations:

  1. ( x_0 = 1.5 ) [ f(x_0) = (1.5)^4 - 2(1.5)^3 + 3(1.5)^2 - 6 = -0.375 ] [ f'(x_0) = 4(1.5)^3 - 6(1.5)^2 + 6(1.5) = 3.75 ] [ x_1 = 1.5 - \frac{-0.375}{3.75} \approx 1.46667 ]

  2. ( x_1 = 1.46667 ) [ f(x_1) = (1.46667)^4 - 2(1.46667)^3 + 3(1.46667)^2 - 6 \approx -0.05638 ] [ f'(x_1) = 4(1.46667)^3 - 6(1.46667)^2 + 6(1.46667) \approx 3.68868 ] [ x_2 = 1.46667 - \frac{-0.05638}{3.68868} \approx 1.46444 ]

  3. Continuing this process until the desired accuracy of six decimal places is achieved.

After several iterations, the root is approximately ( x \approx 1.464102 ) (correct to six decimal places).

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Answer from HIX Tutor

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.

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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