5 The equation \(3 - 2 \ln x - x = 0\) has a root near \(x = 1.8\).
A student proposes to use the iterative formula \(\mathrm { x } _ { \mathrm { n } + 1 } = \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right) = 3 - 2 \ln \mathrm { x } _ { \mathrm { n } }\) to find this root.
The diagram shows the graphs of \(\mathrm { y } = \mathrm { x }\) and \(\mathrm { y } = \mathrm { g } ( \mathrm { x } )\) for values of \(x\) from - 2 to 6 .
\includegraphics[max width=\textwidth, alt={}, center]{4023e87c-34b1-4abd-9acc-ede5e4d68c7f-05_913_917_502_233}

1. With reference to the graph, explain why it might not be possible to use the student's iterative formula to find the root near \(x = 1.8\).
2. Use the relaxed iteration \(\mathrm { x } _ { \mathrm { n } + 1 } = \lambda \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right) + ( 1 - \lambda ) \mathrm { x } _ { \mathrm { n } }\), with \(\lambda = 0.475\) and \(x _ { 0 } = 2\), to determine the root correct to \(\mathbf { 6 }\) decimal places. A student uses the same relaxed iteration with the same starting value. Some analysis of the iterates is carried out using a spreadsheet, which is shown in the table below.

   \(r\)	difference	ratio
   0
   1	- 0.1834898
   2	- 0.0049137	0.02678
   3	\(- 6.44 \mathrm { E } - 06\)	0.00131
   4	\(- 3.862 \mathrm { E } - 09\)	0.0006
   5	\(- 2.313 \mathrm { E } - 12\)	0.0006

3. Explain what the analysis tells you about the order of convergence of this sequence of approximations.