6.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{96948fd3-5438-4e95-b41b-2f649ca8dfac-16_565_844_217_552}
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\caption{Figure 3}
\end{figure}
Figure 3 shows a sketch of curve \(C _ { 1 }\) with equation \(y = 5 \mathrm { e } ^ { x - 1 } + 3\)
and curve \(C _ { 2 }\) with equation \(y = 10 - x ^ { 2 }\)
The point \(P\) lies on \(C _ { 1 }\) and has \(y\) coordinate 18
- Find the \(x\) coordinate of \(P\), writing your answer in the form \(\ln k\), where \(k\) is a constant to be found.
The curve \(C _ { 1 }\) meets the curve \(C _ { 2 }\) at \(x = \alpha\) and at \(x = \beta\), as shown in Figure 3.
- Using a suitable interval and a suitable function that should be stated, show that to 3 decimal places \(\alpha = 1.134\)
The iterative equation
$$x _ { n + 1 } = - \sqrt { 7 - 5 \mathrm { e } ^ { x _ { n } - 1 } }$$
is used to find an approximation to \(\beta\).
Using this iterative formula with \(x _ { 1 } = - 3\)
- find the value of \(x _ { 2 }\) and the value of \(\beta\), giving each answer to 6 decimal places.