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iop360

  • 3 years ago

Need verification: Apply newton's method with initial approximation x1=-4 to find the 2nd and 3rd approximations x2 and x3 to the real root of the equation

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  1. iop360
    • 3 years ago
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    \[x^3 -9x +28 = 0\] my answer is -4 for both x2 and x3.. can anyone verify this?

  2. ksaimouli
    • 3 years ago
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    can u explain ur steps

  3. iop360
    • 3 years ago
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    \[x2 = x1 - \frac{ f(x1) }{ f'(x1) }\] \[x3 = x2 - \frac{ f(x2) }{ f'(x2) }\]

  4. ksaimouli
    • 3 years ago
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    i am getting 0

  5. ksaimouli
    • 3 years ago
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    x1

  6. iop360
    • 3 years ago
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    \[x2 = (-4) - (\frac{ (-4)^3 -9(-4) +28 }{ 3(-4) - 9 })\]

  7. iop360
    • 3 years ago
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    which is equal to -4, and repeats for x3

  8. iop360
    • 3 years ago
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    wait there is a power next to the -4 on the bottom

  9. iop360
    • 3 years ago
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    i just forgot that, but i still get -4 for x2

  10. iop360
    • 3 years ago
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    so its (-4) - (0/39)....right?

  11. ksaimouli
    • 3 years ago
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    hmm but i got x1=0 so should'nt the x2=0-F(0)/(f'(0))

  12. iop360
    • 3 years ago
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    x1 is already given to be x1 = -4

  13. iop360
    • 3 years ago
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    or is initial approximation x0?

  14. ksaimouli
    • 3 years ago
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    ohh yup then ur right

  15. iop360
    • 3 years ago
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    im confused though..so is initial approximation (-4) x0 or x1?

  16. iop360
    • 3 years ago
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    i know it says x1 in the question, but im not sure

  17. iop360
    • 3 years ago
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    i think it is probably x1

  18. ksaimouli
    • 3 years ago
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    it is x1 clearly stated in question

  19. iop360
    • 3 years ago
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    do you get -4 for both x2 and x3 now when you try it?

  20. ksaimouli
    • 3 years ago
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    yup

  21. iop360
    • 3 years ago
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    ok.. thanks

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