## anonymous 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

1. anonymous

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

2. ksaimouli

can u explain ur steps

3. anonymous

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

4. ksaimouli

i am getting 0

5. ksaimouli

x1

6. anonymous

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

7. anonymous

which is equal to -4, and repeats for x3

8. anonymous

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

9. anonymous

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

10. anonymous

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

11. ksaimouli

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

12. anonymous

x1 is already given to be x1 = -4

13. anonymous

or is initial approximation x0?

14. ksaimouli

ohh yup then ur right

15. anonymous

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

16. anonymous

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

17. anonymous

i think it is probably x1

18. ksaimouli

it is x1 clearly stated in question

19. anonymous

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

20. ksaimouli

yup

21. anonymous

ok.. thanks