4.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{d3ba2776-eedb-48f0-834f-41aa454afba3-06_675_1118_205_406}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{figure}
Figure 1 shows a sketch of part of the curve with equation \(y = g ( x )\), where
$$\mathrm { g } ( x ) = \left| 4 \mathrm { e } ^ { 2 x } - 25 \right| , \quad x \in \mathbb { R }$$
The curve cuts the \(y\)-axis at the point \(A\) and meets the \(x\)-axis at the point \(B\). The curve has an asymptote \(y = k\), where \(k\) is a constant, as shown in Figure 1
- Find, giving each answer in its simplest form,
- the \(y\) coordinate of the point \(A\),
- the exact \(x\) coordinate of the point \(B\),
- the value of the constant \(k\).
The equation \(\mathrm { g } ( x ) = 2 x + 43\) has a positive root at \(x = \alpha\)
- Show that \(\alpha\) is a solution of \(x = \frac { 1 } { 2 } \ln \left( \frac { 1 } { 2 } x + 17 \right)\)
The iteration formula
$$x _ { n + 1 } = \frac { 1 } { 2 } \ln \left( \frac { 1 } { 2 } x _ { n } + 17 \right)$$
can be used to find an approximation for \(\alpha\)
- Taking \(x _ { 0 } = 1.4\) find the values of \(x _ { 1 }\) and \(x _ { 2 }\)
Give each answer to 4 decimal places.
- By choosing a suitable interval, show that \(\alpha = 1.437\) to 3 decimal places.
\includegraphics[max width=\textwidth, alt={}, center]{d3ba2776-eedb-48f0-834f-41aa454afba3-07_2258_47_315_37}