2 A curve is defined by the equation \(y = \left( x ^ { 2 } - 4 \right) \ln ( x + 2 )\) for \(x \geqslant 3\).
The curve intersects the line \(y = 15\) at a single point, where \(x = \alpha\).
- Show that \(\alpha\) lies between 3.5 and 3.6.
- Show that the equation \(\left( x ^ { 2 } - 4 \right) \ln ( x + 2 ) = 15\) can be arranged into the form
$$x = \pm \sqrt { 4 + \frac { 15 } { \ln ( x + 2 ) } }$$
(2 marks)
- Use the iteration
$$x _ { n + 1 } = \sqrt { 4 + \frac { 15 } { \ln \left( x _ { n } + 2 \right) } }$$
with \(x _ { 1 } = 3.5\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
(2 marks)