7.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{e28350e9-5090-4079-97da-e669ef9a5a7a-16_621_799_246_630}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{figure}
Figure 1 shows a sketch of the curve \(C\) with equation
$$y = \frac { 4 x ^ { 2 } + x } { 2 \sqrt { x } } - 4 \ln x \quad x > 0$$
- Show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 12 x ^ { 2 } + x - 16 \sqrt { x } } { 4 x \sqrt { x } }$$
The point \(P\), shown in Figure 1, is the minimum turning point on \(C\).
- Show that the \(x\) coordinate of \(P\) is a solution of
$$x = \left( \frac { 4 } { 3 } - \frac { \sqrt { x } } { 12 } \right) ^ { \frac { 2 } { 3 } }$$
- Use the iteration formula
$$x _ { n + 1 } = \left( \frac { 4 } { 3 } - \frac { \sqrt { x _ { n } } } { 12 } \right) ^ { \frac { 2 } { 3 } } \quad \text { with } x _ { 1 } = 2$$
to find (i) the value of \(x _ { 2 }\) to 5 decimal places,
(ii) the \(x\) coordinate of \(P\) to 5 decimal places.