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
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