2 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), where
$$\mathrm { f } ( x ) = \left( \mathrm { e } ^ { x } - 2 \right) ^ { 2 } - 1 , x \in \mathbb { R } .$$
The curve crosses the \(x\)-axis at O and P , and has a turning point at Q .
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{27f6c723-b199-48f1-ab18-22cc0b4b017b-2_866_979_576_573}
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\caption{Fig. 9}
\end{figure}
- Find the exact \(x\)-coordinate of P .
- Show that the \(x\)-coordinate of Q is \(\ln 2\) and find its \(y\)-coordinate.
- Find the exact area of the region enclosed by the curve and the \(x\)-axis.
The domain of \(\mathrm { f } ( x )\) is now restricted to \(x \geqslant \ln 2\).
- Find the inverse function \(\mathrm { f } ^ { - 1 } ( x )\). Write down its domain and range, and sketch its graph on the copy of Fig. 9.