7 The value of a function, $\mathrm { y } = \mathrm { f } ( \mathrm { x } )$, and its gradient function, $\frac { \mathrm { dy } } { \mathrm { dx } }$, when $x = 2$, is given in Table 7.1.

\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Table 7.1}
\begin{tabular}{ | c | c | c | }
\hline
$x$ & $\mathrm { f } ( x )$ & $\frac { \mathrm { dy } } { \mathrm { dx } }$ \\
\hline
2 & 6 & - 2.8 \\
\hline
\end{tabular}
\end{center}
\end{table}
\begin{enumerate}[label=(\alph*)]
\item Determine the approximate value of the error when $f ( 2 )$ is used to estimate $f ( 2.03 )$.

The Newton-Raphson method is used to find a sequence of approximations to a root, $\alpha$, of the equation $\mathrm { f } ( x ) = 0$. The spreadsheet output showing the iterates, together with some further analysis, is shown in Table 7.2.

\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Table 7.2}
\begin{tabular}{|l|l|l|l|l|}
\hline
 & A & B & C & D \\
\hline
1 & r & Xr & difference & ratio \\
\hline
2 & 0 & 12 &  &  \\
\hline
3 & 1 & -13.1165572 & -25.1165572 &  \\
\hline
4 & 2 & 1.76283279 & 14.87939004 & -0.5924136 \\
\hline
5 & 3 & 2.18052157 & 0.41768878 & 0.02807163 \\
\hline
6 & 4 & 2.182419024 & 0.001897454 & 0.00454275 \\
\hline
7 & 5 & 2.182419066 & $4.13985 \mathrm { E } - 08$ & $2.1818 \mathrm { E } - 05$ \\
\hline
\end{tabular}
\end{center}
\end{table}
\item \begin{enumerate}[label=(\roman*)]
\item Explain what the values in column D tell you about the order of convergence of this sequence of approximations.
\item Without doing any further calculation, state the value of $\alpha$ as accurately as you can, justifying the precision quoted.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Numerical Methods 2023 Q7 [6]}}