## 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!

1. jim_thompson5910

Newton's Method $\Large x_{n+1} = x_n - \frac{f(x_n)}{f \ '(x_n)}$

2. jim_thompson5910

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

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

Yeah I'm getting that too. Maybe they want you to round a very specific way?

5. jim_thompson5910

I doubt they want it as a fraction, but who knows really.

6. anonymous

that's what i was thinking, that maybe they wanted it in fraction form, but it doesn't specify...

7. jim_thompson5910

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

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

That's all of it

10. jim_thompson5910

Hmm so odd and frustrating. It would be nice if they added "round to 5 decimal places" or something.

11. anonymous

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

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

Or ask the teacher. I would ask for points back if you get it wrong since this problem is unfair.

14. anonymous

yeah I'll try to do that, thanks for checking my answer though!

15. jim_thompson5910

no problem

16. anonymous

it was in fraction form lol

17. jim_thompson5910

so strange, but I guess computers tend to be that way