4 Fig. 4 shows the graph of \(y = x ^ { 5 } - 6 \sqrt { x } + 4\).
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{4838f71e-a1d0-4695-89d2-c7ebb47edd77-6_867_700_317_246}
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\caption{Fig. 4}
\end{figure}
There are two roots of the equation \(x ^ { 5 } - 6 \sqrt { x } + 4 = 0\). The roots are \(\alpha\) and \(\beta\), such that \(\alpha < \beta\).
- Show that \(0 < \alpha < 1\) and \(1 < \beta < 2\).
- Obtain the Newton-Raphson iterative formula
$$x _ { n + 1 } = x _ { n } - \frac { x _ { n } ^ { \frac { 11 } { 2 } } - 6 x _ { n } + 4 \sqrt { x _ { n } } } { 5 x _ { n } ^ { \frac { 9 } { 2 } } - 3 }$$
- Use the iterative formula found in part (b) with a starting value of \(x _ { 0 } = 1\) to obtain \(\beta\) correct to 6 decimal places.
- Use the iterative formula found in part (b) with a starting value of \(x _ { 0 } = 0\) to find \(x _ { 1 }\).
- Give a geometrical explanation of why the Newton-Raphson iteration fails to find \(\alpha\) in part (d).
- Obtain the iterative formula
$$x _ { n + 1 } = \left( \frac { x _ { n } ^ { 5 } + 4 } { 6 } \right) ^ { 2 }$$
- Use the iterative formula found in part (f) with a starting value of \(x _ { 0 } = 0\) to obtain \(\alpha\) correct to 6 decimal places.