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.
x^3 + 4x^2 – 3x + 1 = 0

Answer 1

So, you need to find a good starting value, an initial guess.

Answer 2

ok

Answer 3

1 seems like a good one, because it's already relatively close to your zero

Answer 4

okay

Answer 5

but if one is a good guess, you can maybe even improve it by trying for values like x=(1/2), but that is a matter of taste rather than rigor. Do you know how to setup the Newton Method from there?

Answer 6

\[\Large x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} \]
So you need to compute the derivative of the function as well, since it's an polynomial that will be easy, you have your initial guess:
\[\Large x_0=1 \]
for instance.

Answer 7

ok

Answer 8

I got (x(3x + 8= 3) for the derivative

Answer 9

Usually you don't factor it, but you're on the right track:

\[\Large f(x)=x^3+4x^2-3x+1 \]
\[\Large f'(x)=3x^2+8x-3 \]
You don't need to set f'(x)=0, if that's what you did above.

Now plug them back into the equation above, try to simplify it as good as possible, will safe you a lot of work because with the Newton Method you usually need to plug in quite a few variables:

\[\Large x_{n+1}=\frac{x_nf'(x_n)-f(x_n)}{f'(x_n)} \]

Answer 10

So you want to get an easy to evaluate polynomial with algebraic simplification and then start plugging in your values, starting with your initial guess, next you take the output of that input and plug it once again into the algorithm, you can continue this algorithm until you're close enough to the root of it.

Answer 11

I will be back in a few, if you still require help with this problem then.

Answer 12

ok

Answer 13

ok so 1 - ((1)^3 + 4(1)^2 - 3(1) + 1)/3((1)^2 + 8(1) - 3) = 1 - (1 + 16 - 3 + 1)/(9 + 8 - 3)

Answer 14

is taht correct?

Answer 15

i got 15/14

Answer 16

I meant -1/14

Answer 17

is it correct?

Answer 18

@Spacelimbus are u there?

Answer 19

so now how do i graph it?

Answer 20

@myko can u help?

Answer 21

http://en.wikipedia.org/wiki/File:NewtonIteration_Ani.gif

Answer 22

so all of all of those being graphed are correct?

Answer 23

yes, just each time it's a better aproximation

Answer 24

ok thnx

Answer 25

ok thnx