Question

Use Newton’s method to find an approximate root (accurate to six decimal places). Sketch the graph and explain how you determined your initial guess.
cos x - x = 0

Answer 1

@wio can u help?

Answer 2

Okay, do you know how to do Newton's method?

Answer 3

yea this rigth? x n + 1 = xn - f(xn)/f'(xn)

Answer 4

yes

Answer 5

So what is your derivative?

Answer 6

okay i got (sin (x) + 1 = 0 )

Answer 7

Close, but you don't need the = 0 part.

Answer 8

where do you want to start?

Answer 9

What will be your initial guess?

Answer 10

so its sin(x) + 1?

Answer 11

\[
(\cos x-x)'=-\sin x - 1
\]

Answer 12

ok

Answer 13

So what is f(x)/f'(x)?

Answer 14

so the initial will be 1?

Answer 15

Sure, it doesn't matter too much.

Answer 16

ok

Answer 17

so what comes after 1?

Answer 18

2

Answer 19

lol oh you meant the next step?

Answer 20

No, I mean, what is \[
1+\frac{\cos x-x}{\sin x +1}
\]

Answer 21

where \(x=1\)

Answer 22

ohh i got 0.750

Answer 23

just keep going

Answer 24

umm ok

Answer 25

once the \[
f(x)/f'(x)
\]term is under 6 decimals, you can stop.

Answer 26

ohh ok

Answer 27

so 0.750363

Answer 28

need to keep going

Answer 29

ok 0.7503638678

Answer 30

0.750363867840243893034942306682176853246993065855359030966583... lol

Answer 31

You're not getting closer to the actual answer for some reason

Answer 32

ok

Answer 33

http://www.wolframalpha.com/input/?i=cos+x+-+x=0

Answer 34

this is the answer

Answer 35

so the actual answer is 0

Answer 36

\[x \approx 0.739085\]

Answer 37

ok

Answer 38

ok