5.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{f3272b4c-d8dc-4f32-add9-153de90f4d0a-10_620_622_210_662}
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\caption{Figure 1}
\end{figure}
Figure 1 shows a sketch of part of the curve with equation
$$y = 6 \ln ( 2 x + 3 ) - \frac { 1 } { 2 } x ^ { 2 } + 4 \quad x > - \frac { 3 } { 2 }$$
The curve cuts the negative \(x\)-axis at the point \(P\), as shown in Figure 1.
- Show that the \(x\) coordinate of \(P\) lies in the interval \([ - 1.25 , - 1.2 ]\)
The curve cuts the positive \(x\)-axis at the point \(Q\), also shown in Figure 1.
Using the iterative formula
$$x _ { n + 1 } = \sqrt { 12 \ln \left( 2 x _ { n } + 3 \right) + 8 } \text { with } x _ { 1 } = 6$$ - find, to 4 decimal places, the value of \(x _ { 2 }\)
- find, by continued iteration, the \(x\) coordinate of \(Q\). Give your answer to 4 decimal places.
The curve has a maximum turning point at \(M\), as shown in Figure 1.
- Using calculus and showing each stage of your working, find the \(x\) coordinate of \(M\).