iop360
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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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iop360
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\[x^3 -9x +28 = 0\]
my answer is -4 for both x2 and x3.. can anyone verify this?
ksaimouli
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can u explain ur steps
iop360
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\[x2 = x1 - \frac{ f(x1) }{ f'(x1) }\]
\[x3 = x2 - \frac{ f(x2) }{ f'(x2) }\]
ksaimouli
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i am getting 0
ksaimouli
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x1
iop360
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\[x2 = (-4) - (\frac{ (-4)^3 -9(-4) +28 }{ 3(-4) - 9 })\]
iop360
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which is equal to -4, and repeats for x3
iop360
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wait there is a power next to the -4 on the bottom
iop360
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i just forgot that, but i still get -4 for x2
iop360
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so its (-4) - (0/39)....right?
ksaimouli
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hmm but i got x1=0 so should'nt the x2=0-F(0)/(f'(0))
iop360
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x1 is already given to be x1 = -4
iop360
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or is initial approximation x0?
ksaimouli
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ohh yup then ur right
iop360
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im confused though..so is initial approximation (-4) x0 or x1?
iop360
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i know it says x1 in the question, but im not sure
iop360
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i think it is probably x1
ksaimouli
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it is x1 clearly stated in question
iop360
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do you get -4 for both x2 and x3 now when you try it?
ksaimouli
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yup
iop360
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ok.. thanks