Question

Show Newton's Method to approximate the real zero of the function in the interval [0,2]. Give the answer to the nearest .001

Answer 1

f(x)=x^5+x-1

Answer 2

@yun2thejae

Answer 3

Oh boy.

Answer 4

give me a second.

Answer 5

do you know the newton's method? I'm asking because i want to map out how much I have to explain

Answer 6

yeah \[x _{1} - \frac{ f(x _{1}) }{ f'(x _{1} )}\] @yun2thejae

Answer 7

k that is good. generally, you want to repeat newton's method a few times to get really close to the actual root. Choose a number for xsub1 so that you can find xsub2. Anyway, start with xsub1=1. Plug in the value and solve. and then repeat. so if xsub1=1 then xsub2= 1- (f(1)/f'(1))

Answer 8

so choose numbers between [0,2] for \[x _{1}\]

Answer 9

well you only need to choose one number. easiest number to pick is 1.

Answer 10

yeah i chose 1 and i got .833

Answer 11

when you find xsub2, use xsub2 and plug it in to find xsub3. et cetera et cetera.

Answer 12

now do .833-f(.833)/f'(.833)

Answer 13

or just keep pressing enter in ur calculator until it stops producing new numbers lol

Answer 14

ok the number it stopped producing at was .7548776662

Answer 15

do it once or twice more and use that as your answer. yeah. when I did this in class, we weren't allowed to use a graphing calculator at all during the semester

Answer 16

so my answer is .755 :)

Answer 17

but then for the other question for this part of my final \[f(x)=\sqrt[3]{x-1} \] my teacher put no answer, does that mean that this particular equation doesn't work with Newton's Method? :|

Answer 18

@yun2thejae

Answer 19

might mean that there are no real roots.

Answer 20

realy? ok

Answer 21

I'm not really. sure. that would be my first assumption.

Answer 22

newtons method does not wok with the above equation even though there is clearly a root at x=1

Answer 23

oh ok thx :)