The equation \(\mathrm { e } ^ { - x } - 2 + \sqrt { x } = 0\) has a single root, \(\alpha\).
Show that \(\alpha\) lies between 3 and 4 .
Use the recurrence relation \(x _ { n + 1 } = \left( 2 - e ^ { - x _ { n } } \right) ^ { 2 }\), with \(x _ { 1 } = 3.5\), to find \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
The diagram below shows parts of the graphs of \(y = \left( 2 - \mathrm { e } ^ { - x } \right) ^ { 2 }\) and \(y = x\), and a position of \(x _ { 1 }\).
On the diagram, draw a staircase or cobweb diagram to show how convergence takes place, indicating the positions of \(x _ { 2 }\) and \(x _ { 3 }\) on the \(x\)-axis.
\includegraphics[max width=\textwidth, alt={}, center]{063bbfa5-df49-44a1-8143-5e076397f63f-03_1100_1402_881_367}