1 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 2 x ^ { 2 } - 1 } { x ^ { 2 } + 1 }\) for the domain \(0 \leqslant x \leqslant 2\).
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\includegraphics[alt={},max width=\textwidth]{b7588524-8a5e-42af-8b52-29cdddc09eeb-1_976_1208_450_514}
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\caption{Fig. 9}
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- Show that \(\mathrm { f } ^ { \prime } ( x ) = \frac { 6 x } { \left( x ^ { 2 } + 1 \right) ^ { 2 } }\), and hence that \(\mathrm { f } ( x )\) is an increasing function for \(x > 0\).
- Find the range of \(\mathrm { f } ( x )\).
- Given that \(\mathrm { f } ^ { \prime \prime } ( x ) = \frac { 6 - 18 x ^ { 2 } } { \left( x ^ { 2 } + 1 \right) ^ { 3 } }\), find the maximum value of \(\mathrm { f } ^ { \prime } ( x )\).
The function \(\mathrm { g } ( x )\) is the inverse function of \(\mathrm { f } ( x )\).
- Write down the domain and range of \(\mathrm { g } ( x )\). Add a sketch of the curve \(y = \mathrm { g } ( x )\) to a copy of Fig. 9 .
- Show that \(\mathrm { g } ( x ) = \sqrt { \frac { x + 1 } { 2 - x } }\).