Use Newton's method to approximate the indicated root of the equation correct to six decimal places?

in the interval [1,2] what is the root of #x^4 − 2x^3 + 3x^2 − 3 #

Answer 1

#x=1.2113# to 4dp

Let #f(x) = x^4-2x^3+3x^2-3# Then our aim is to solve #f(x)=0# in the interval #1 le x le 2#

First let us look at the graphs:
graph{x^4-2x^3+3x^2-3 [-5, 5, -15, 15]}

We can see there is one solution in the interval #1 le x le 2# (along with a further solution in #-1 lt x lt 0#)

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

# \ \ \ \ \ \ \f(x) = x^4-2x^3+3x^2-3 #
# :. f'(x) = 4x^3-6x^2+6x #

The Newton-Rhapson method uses the following iterative sequence

# { (x_0,=1), ( x_(n+1), = x_n - f(x_n)/(f'(x_n)) ) :} #

Then using excel working to 8dp we can tabulate the iterations as follows:

We could equally use a modern scientific graphing calculator as most new calculators have an " Ans " button that allows the last calculated result to be used as the input of an iterated expression.

And we conclude that the solution is #x=1.2113# to 4dp

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Answer 2

Sure, please provide the equation and the initial guess for the root.

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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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