# Thread: Newton's Method of convergence

1. ## Newton's Method of convergence

I'm stuck with this whole question on convergence -

f(x) = xe^(-x) + 3.
For what values of xo ("x naught") does Newton's method converge to the root, and for what values of xo will Newton's method fail to find the root?

Thanks

2. Hi

$f(x) = xe^{-x} + 3$
$f'(x) = (1-x)e^{-x}$

Newton's method starting from $x_0$ will give

$x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}$

If $f'(x_0) = 0$ you will not be able to calculate $x_1$

3. Thanks,

so am i right in saying that Newton's method converges to the root for xo not equal to 1? Or greater than or equal to 1?