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

Question

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

anonymous · Answer

$$x^3 -9x +28 = 0$$
my answer is -4 for both x2 and x3.. can anyone verify this?

ksaimouli · Answer

can u explain ur steps

anonymous · Answer

$$x2 = x1 - \frac{ f(x1) }{ f'(x1) }$$
$$x3 = x2 - \frac{ f(x2) }{ f'(x2) }$$

ksaimouli · Answer

i am getting 0

ksaimouli · Answer

x1

anonymous · Answer

$$x2 = (-4) - (\frac{ (-4)^3 -9(-4) +28 }{ 3(-4) - 9 })$$

anonymous · Answer

which is equal to -4, and repeats for x3

anonymous · Answer

wait there is a power next to the -4 on the bottom

anonymous · Answer

i just forgot that, but i still get -4 for x2

anonymous · Answer

so its (-4) - (0/39)....right?

ksaimouli · Answer

hmm but i got  x1=0 so should'nt the x2=0-F(0)/(f'(0))

anonymous · Answer

x1 is already given to be x1 = -4

anonymous · Answer

or is initial approximation x0?

ksaimouli · Answer

ohh yup then ur right

anonymous · Answer

im confused though..so is initial approximation (-4)  x0 or x1?

anonymous · Answer

i know it says x1 in the question, but im not sure

anonymous · Answer

i think it is probably  x1

ksaimouli · Answer

it is x1 clearly stated in question

anonymous · Answer

do you get -4 for both x2 and x3 now when you try it?

ksaimouli · Answer

yup

anonymous · Answer

ok.. thanks