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Newtons method to estimate solution to eq.

  1. Nov 24, 2008 #1
    1. The problem statement, all variables and given/known data
    Use Newton's method to estimate the requested solution of the equation. Start with given value of x0 and give x2 as the estimated solution.

    2. Relevant equations x0 = 2 ( this is the first guess numb. you start with)
    the equation you use is.... x = x(guuess #) - f(xg)/f'(xg)

    f(x) = x^4 -6x +3 and f'(x) = 4(x)^3 -6

    3. The attempt at a solution
    I have done the entire problem. several times.
    when x ='s 2, I got 1. 731

    when x ='s 1.731 I got 1.623

    when x ='s 1.623 I got 1.604 for my final answer.

    perhaps this answer is correct , and the multiple choice answer is wrong.

    If you are familiar with newtons method- then this won't take too long for you.
    You just plug and chug.
  2. jcsd
  3. Nov 24, 2008 #2
    Your answer is right. The next digits are
    I just punched into maple and got this answer.
  4. Nov 24, 2008 #3


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    That is correct to three decimal places. What are the choices?
  5. Nov 24, 2008 #4


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    Wouldn't that give you [itex]x_0=2[/itex] , [itex]x_1 \approx 1.731[/itex], [itex]x_2 \approx 1.623[/itex] and [itex]x_3 \approx 1.605[/itex]?

    soo if you are asked to find x2...would that not be 1.623?
  6. Nov 24, 2008 #5
    my full answer is 1.604938639

    when pluging the answers back into the equation and dividing
    by the derrivative, I only was tacking my answer to 3 decimal places
    keeping in mind to round up the 3rd decimal place.

    the choices are 1.600 and 1.604
  7. Nov 24, 2008 #6
    no. I think you are confusing x0 with x1. the order goes: x0
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