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anonymous

  • one year ago

Use Newton's method with initial approximation x1 = 1 to find x2, the second approximation to the root of the equation x4 − x − 8 = 0. I worked it out and got 3.6666667 but it was wrong, please help!

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  1. jim_thompson5910
    • one year ago
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    Newton's Method \[\Large x_{n+1} = x_n - \frac{f(x_n)}{f \ '(x_n)}\]

  2. jim_thompson5910
    • one year ago
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    If n = 1, then \[\Large x_{n+1} = x_n - \frac{f(x_n)}{f \ '(x_n)}\] \[\Large x_{1+1} = x_1 - \frac{f(x_1)}{f \ '(x_1)}\] \[\Large x_{2} = 1 - \frac{f(1)}{f \ '(1)}\] So you need to compute both f(1) and f ' (1). Do you know how to do that?

  3. anonymous
    • one year ago
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    yeah, i worked everything out i got x^4-x-8/4x^3-1 and I plugged in 1 for the x and then got -8/3, then i did 1-(-8/3) and got 3.66667?

  4. jim_thompson5910
    • one year ago
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    Yeah I'm getting that too. Maybe they want you to round a very specific way?

  5. jim_thompson5910
    • one year ago
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    I doubt they want it as a fraction, but who knows really.

  6. anonymous
    • one year ago
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    that's what i was thinking, that maybe they wanted it in fraction form, but it doesn't specify...

  7. jim_thompson5910
    • one year ago
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    I've never seen newton's method approximations as fractions. Usually they are in decimal form. Can you post a screenshot of the full problem?

  8. anonymous
    • one year ago
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    Use Newton's method with initial approximation x1 = 1 to find x2, the second approximation to the root of the equation x4 − x − 8 = 0. x2 =3.67 Incorrect: Your answer is incorrect.

  9. anonymous
    • one year ago
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    That's all of it

  10. jim_thompson5910
    • one year ago
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    Hmm so odd and frustrating. It would be nice if they added "round to 5 decimal places" or something.

  11. anonymous
    • one year ago
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    exactly! and i only have two tries so im scared to use my last one and get it wrong, since i dont know how im supposed to type the answer

  12. jim_thompson5910
    • one year ago
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    The only thing you can do really is look back at previous correct problems to see what the computer wants. Or try to guess it anyway.

  13. jim_thompson5910
    • one year ago
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    Or ask the teacher. I would ask for points back if you get it wrong since this problem is unfair.

  14. anonymous
    • one year ago
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    yeah I'll try to do that, thanks for checking my answer though!

  15. jim_thompson5910
    • one year ago
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    no problem

  16. anonymous
    • one year ago
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    it was in fraction form lol

  17. jim_thompson5910
    • one year ago
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    so strange, but I guess computers tend to be that way

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