6.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{cb92f7b6-2ba5-4703-9595-9ba8570fc52b-09_1152_1006_285_374}
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\caption{Figure 4}
\end{figure}
Figure 4 shows a sketch of the curve \(C\) with equation \(y = \mathrm { f } ( x )\), where
$$f ( x ) = \frac { 2 x ^ { 2 } - x } { \sqrt { x } } - 2 \ln \left( \frac { x } { 2 } \right) , \quad x > 0$$
The curve has a minimum turning point at \(Q\), as shown in Figure 4.
a. Show that \(\mathrm { f } ^ { \prime } ( x ) = \frac { 6 x ^ { 2 } - x - 4 \sqrt { x } } { 2 x \sqrt { x } }\)
b. Show that the \(x\)-coordinate of \(Q\) is the solution of
$$x = \sqrt { \frac { x } { 6 } + \frac { 2 \sqrt { x } } { 3 } }$$
To find an approximation for the \(x\)-coordinate of \(Q\), the iteration formula
$$x _ { n + 1 } = \sqrt { \frac { x _ { n } } { 6 } + \frac { 2 \sqrt { x _ { n } } } { 3 } }$$
is used.
c. Taking \(x _ { 0 } = 0.8\), find the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\).
Give your answers to 3 decimal places.