Using newtons method, solve the equation, x^3-4x-12=0 accuratey to six decimal places?

Im having extreme troubles with this can anyone please help(Angry)

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- Aug 4th 2008, 04:43 AMNoAsherelolNewtons Method Help
Using newtons method, solve the equation, x^3-4x-12=0 accuratey to six decimal places?

Im having extreme troubles with this can anyone please help(Angry)

- Aug 4th 2008, 06:03 AMCaptainBlack
Newton's method uses the recursion:

$\displaystyle x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$

to find a root of f(x)=0, so here we put:

$\displaystyle f(x)=x^3-4x-12$

$\displaystyle f'(x)=3x^2-4$

Then:

$\displaystyle x_{n+1}=x_n-\frac{x_n^3-4x_n-12}{3x^2-4}$

Now we need a guess to get us started, we observe that $\displaystyle f(x)$ is negative at $\displaystyle x=2$ and positive at $\displaystyle x=3$ (and nearer to zero at $\displaystyle x=3$) so we take $\displaystyle x_0=3$ as our starting value.

You will find this converges to the required accuracy after about 3 itterations.

RonL