Let $f(x)=\frac12 e^{-x}-x$. Find an approximate root of $f(x)$ using
A) Fixed point iteration method (3 iterations, $x_0 = 1$).
B) Newton’s method (3 iterations, $x_0 = 1$).
C) Find, with proof, some interval $I$ for which the fixed point method is guaranteed to converge if the initial guess $x_0\in I$.
I have done the first two parts of this question but I am just unsure of how to answer the third! Any help is appreciated.
$f(x) = \dfrac{e^{-x}}{2} - x$
a)
$$\begin{align} &0 = \dfrac{e^{-x}}{2} - x, \ \ \ x_0 = 1 \\ &x = \dfrac{e^{-x}}{2}, \ \ \ x_1 = \dfrac{e^{-1}}{2} = 0.1839397 \\ &x_{n + 1} = \dfrac{e^{-x_n}}{2}, \ \ \ x_2 = \dfrac{e^{-0.1839397}}{2} = 0.415993 \\ &x_3 = \dfrac{e^{-0.415993}}{2} = 0.3298425 \end{align}$$
b)
$$\begin{align} &x_{n + 1} = x_n - \dfrac{f(x_n)}{f'(x_n)} \ \ \ f'(x) = \dfrac{-e^{-x}}{2} - 1 \\ &x_0 = 1 \\ &x_1 = 1 - \dfrac{\dfrac{e^{-1}}{2} - 1}{\dfrac{-e^{-1}}{2} - 1} = 0.3107248 \\ &x_2 = 0.3107248 - \dfrac{\dfrac{e^{-0.31072}}{2} - 0.31072}{\dfrac{-e^{-0.31072}}{2} - 1} = 0.35151126 \\ &x_3 = 0.35151126 - \dfrac{\dfrac{e^{-0.35151126}}{2} - 0.35151126}{\dfrac{-e^{-0.35151126}}{2} - 1} = 0.35173370 \end{align}$$
The fixed point iteration is $x_{k+1}=\phi(x_k)$ where $\phi(x) = {1 \over 2} e^{-x}$ and you are looking for a fixed point of $\phi$.
Note that $\phi(x) \ge 0$, $\phi'(x) = -{1 \over 2} e^{-x}$ and for $x \ge 0$ we have $|\phi'(x)| \le {1 \over 2} < 1$. Hence $\phi:[0,\infty) \to [0,\infty)$ and it is a contraction there.
Indeed, it is not hard to see that if $c > - \log 2$ then $\phi$ is a contraction on $[c,\infty)$, but this is underkill.
In any event, since $\phi(x) \ge 0$ it follows that after the first iteration we have $x_1 \ge 0$ and so we see from the above that if the iteration is started anywhere on the real line then it will converge to the unique root.
A little work shows that Newton's method applied to $f$ will also converge from any starting point (since $f$ is strictly convex and has a solution).