8. A curve has the equation \(y = x ^ { 2 } - \sqrt { 4 + \ln x }\).
- Show that the tangent to the curve at the point where \(x = 1\) has the equation
$$7 x - 4 y = 11$$
The curve has a stationary point with \(x\)-coordinate \(\alpha\).
- Show that \(0.3 < \alpha < 0.4\)
- Show that \(\alpha\) is a solution of the equation
$$x = \frac { 1 } { 2 } ( 4 + \ln x ) ^ { - \frac { 1 } { 4 } }$$
- Use the iteration formula
$$x _ { n + 1 } = \frac { 1 } { 2 } \left( 4 + \ln x _ { n } \right) ^ { - \frac { 1 } { 4 } }$$
with \(x _ { 0 } = 0.35\), to find \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to 5 decimal places.
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