2.
$$\mathrm { g } ( x ) = \mathrm { e } ^ { x - 1 } + x - 6$$
- Show that the equation \(\mathrm { g } ( x ) = 0\) can be written as
$$x = \ln ( 6 - x ) + 1 , \quad x < 6$$
The root of \(\mathrm { g } ( x ) = 0\) is \(\alpha\).
The iterative formula
$$x _ { n + 1 } = \ln \left( 6 - x _ { n } \right) + 1 , \quad x _ { 0 } = 2$$
is used to find an approximate value for \(\alpha\). - Calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\) to 4 decimal places.
- By choosing a suitable interval, show that \(\alpha = 2.307\) correct to 3 decimal places.