Question

Approximate the zero of the function in the indicated interval to six decimal places.

f(x) = x^3 - x - 1

for my x0 is use 1.5

But my problem is that for every approximation i get a totally different number

Answer 1

f(x)=3.375-1.5-1=.875

Answer 2

i am going to bet you are supposed to use newton's method

Answer 3

is that right?

Answer 4

yes sorry i forgot to indicate that part

Answer 5

first guess is 1.5 then take
\[1.5-\frac{f(1.5)}{f'(.15)}\]

Answer 6

i know how to do it its just that for every approximation i get a different number

Answer 7

ok a spread sheet is nice, or we can make a formula. or you can just do it step by step

Answer 8

yeah i did it in spread sheet form well not on excel or anything but on paper

Answer 9

\[f(1.5)=.875\]
\[f'(1.5)=5.75\]
\[1.5-\frac{.875}{5.75}=1.347826087\]

Answer 10

that is
\[x_1\]

Answer 11

now
\[1.347826087-\frac{f(1.347826087)}{f'(1.347826067)}=1.3252\]

Answer 12

rounded but next two decimals are 0

Answer 13

that is \[x_2\]

Answer 14

then you can do it again. alternatively you can just get a formula for
\[x-\frac{x^3-x-1}{3x^2-1}=\frac{2x^3+1}{3x^2-1}\]

Answer 15

and keep plugging the numbers is that

Answer 16

a pain no matter which way you do it, but i will do the last one for you

Answer 17

ok u skippedd a step what after i found f(x1) and f'(x1) then i divided the two using the formula f(x1)/f'(x1)

Answer 18

i mean f(x1)/-f'(x1)

Answer 19

right and then subtract that from \[x_1\]

Answer 20

it is
\[x_2=x_1-\frac{f(x_1)}{f'(x_1)}\]

Answer 21

then i do 1.5 minus that

Answer 22

yes

Answer 23

but we dont get the same answer

Answer 24

i wrote the answer above. then finally the next one is 1.324718147

Answer 25

really?

Answer 26

start with 1.5

Answer 27

got 1.652173913 for my first approximation

Answer 28

then subtract
\[\frac{f(1.5)}{f'(1.5)}\]

Answer 29

oh i see what you are doing.

Answer 30

it is
\[xn=x_{n-1}-\frac{f(x_{n-1})}{f'(x_{n-1})}\]

Answer 31

you had a -f'(x) in the denominator so when you subtracted you really added!

Answer 32

use the formula i wrote
\[x-\frac{f(x)}{f'(x)}\]

Answer 33

compute f(x), and f'(x) then divide and then subtract

Answer 34

denominator should just be f'(x) and then subtract. not -f'(x)

Answer 35

i wrote the answer above and checked it. it gives 0 to 6 decimal places

Answer 36

ok

Answer 37

good! you were putting -f'(x) ind the denominator and then subtracting, which is like adding.

Answer 38

lol my lecturer only did one example so i wasnt sure if that was suppose to be there THANK YOU SATELLITE

Answer 39

yw!