11 Fig. 11.1 shows the curve with equation \(\mathrm { y } = \mathrm { g } ( \mathrm { x } )\) where \(\mathrm { g } ( x ) = x \sin x + \cos x\) and the curve of the gradient function \(\mathrm { y } = \mathrm { g } ^ { \prime } ( \mathrm { x } )\) for \(- 2 \pi \leqslant x \leqslant 2 \pi\).
\begin{figure}[h]
\captionsetup{labelformat=empty}
\caption{Fig. 11.1}
\includegraphics[alt={},max width=\textwidth]{60e1e785-c34b-48ef-a63f-13a25fee186e-08_1136_1196_459_246}
\end{figure}
- Show that the \(x\)-coordinates of the points on the curve \(y = g ( x )\) where the gradient is 1 satisfy the equation \(\frac { 1 } { x } - \cos x = 0\).
Fig. 11.2 shows part of the curve with equation \(y = \frac { 1 } { x } - \cos x\).
\begin{figure}[h]
\captionsetup{labelformat=empty}
\caption{Fig. 11.2}
\includegraphics[alt={},max width=\textwidth]{60e1e785-c34b-48ef-a63f-13a25fee186e-09_678_1363_424_239}
\end{figure} - Use the Newton-Raphson method with a suitable starting value to find the smallest positive \(x\)-coordinate of a point on the curve \(y = x \sin x + \cos x\) where the gradient is 1 .
You should write down at least the following.
- The iteration you use
- The starting value
- The solution correct to \(\mathbf { 4 }\) decimal places
- Explain why \(x _ { 1 } = 3\) is not a suitable starting value for the Newton-Raphson method in part (b).