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 iterative formula
$$x _ { n + 1 } = \frac { 1 } { 2 } \left( 4 + \ln x _ { n } \right) ^ { - \frac { 1 } { 4 } }$$
with \(x _ { 0 } = 0.35\), to find \(\alpha\) correct to 5 decimal places.
You should show the result of each iteration.