Given a function $f: x\mapsto e^x - x- 2$, we try to find solutions to the equation $f(x) = 0$ with Newton's method. It is known that this equation has 2 solutions $\xi_1$ and $\xi_2$, with $\xi_1 > 0$ and $\xi_2 < 0$. Now, given any $x_0\in [\xi_1, +\infty)$ or $x_0\in(-\infty, \xi_2]$, I can prove that the Newton iteration sequence $(x_k)$ defined by
$$ x_{k+1} = x_k - \frac{f(x_k)}{f'(x_k)}\quad \forall k\in\mathbb{N} $$
will converge to $\xi_1$ (when $x_0 \geqslant \xi_1$) or $\xi_2$ (when $x_0 \leqslant \xi_2$), respectively. Now, what if I choose the starting point $x_0$ on $(\xi_2, 0)\cup(0, \xi_1)$? How could I prove that that the Newton iteration sequences starting from these points will converge/diverge? Any help will be much appreciated! :)
The function is strictly convex, the tangent thus always below the function. The function value at a tangent root is consequently positive. This then implies that if $f(x_0)>0$ that then the sequence of Newton iterates has $f(x_k)>0$ and it converges monotonously towards the closest root.
If now $f(x_0)<0$ and $f'(x_0)\ne 0$, that is, in the case in question with $x_0\ne 0$, the tangent root will again lead to $f(x_1)>0$ and from then on the iteration follows the above case. Now confirm that $x_1$ has the same sign as $x_0$ to determine to which root the sequence converges.