4 Fig. 4 shows the graph of $y = x ^ { 5 } - 6 \sqrt { x } + 4$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{4838f71e-a1d0-4695-89d2-c7ebb47edd77-6_867_700_317_246}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}

There are two roots of the equation $x ^ { 5 } - 6 \sqrt { x } + 4 = 0$. The roots are $\alpha$ and $\beta$, such that $\alpha < \beta$.
\begin{enumerate}[label=(\alph*)]
\item Show that $0 < \alpha < 1$ and $1 < \beta < 2$.
\item Obtain the Newton-Raphson iterative formula

$$x _ { n + 1 } = x _ { n } - \frac { x _ { n } ^ { \frac { 11 } { 2 } } - 6 x _ { n } + 4 \sqrt { x _ { n } } } { 5 x _ { n } ^ { \frac { 9 } { 2 } } - 3 }$$
\item Use the iterative formula found in part (b) with a starting value of $x _ { 0 } = 1$ to obtain $\beta$ correct to 6 decimal places.
\item Use the iterative formula found in part (b) with a starting value of $x _ { 0 } = 0$ to find $x _ { 1 }$.
\item Give a geometrical explanation of why the Newton-Raphson iteration fails to find $\alpha$ in part (d).
\item Obtain the iterative formula

$$x _ { n + 1 } = \left( \frac { x _ { n } ^ { 5 } + 4 } { 6 } \right) ^ { 2 }$$
\item Use the iterative formula found in part (f) with a starting value of $x _ { 0 } = 0$ to obtain $\alpha$ correct to 6 decimal places.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Numerical Methods 2019 Q4 [13]}}