9 The function \(\mathrm { f } ( x ) = \ln \left( 1 + x ^ { 2 } \right)\) has domain \(- 3 \leqslant x \leqslant 3\).
Fig. 9 shows the graph of \(y = \mathrm { f } ( x )\).
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{3efea8db-9fa1-47a8-89b8-e4888f87a313-4_540_943_477_550}
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\caption{Fig. 9}
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- Show algebraically that the function is even. State how this property relates to the shape of the curve.
- Find the gradient of the curve at the point \(\mathrm { P } ( 2 , \ln 5 )\).
- Explain why the function does not have an inverse for the domain \(- 3 \leqslant x \leqslant 3\).
The domain of \(\mathrm { f } ( x )\) is now restricted to \(0 \leqslant x \leqslant 3\). The inverse of \(\mathrm { f } ( x )\) is the function \(\mathrm { g } ( x )\).
- Sketch the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\) on the same axes.
State the domain of the function \(\mathrm { g } ( x )\).
Show that \(\mathrm { g } ( x ) = \sqrt { \mathrm { e } ^ { x } - 1 }\).
- Differentiate \(\mathrm { g } ( x )\). Hence verify that \(\mathrm { g } ^ { \prime } ( \ln 5 ) = 1 \frac { 1 } { 4 }\). Explain the connection between this result and your answer to part (ii).