Modulus function transformations

9. $$\mathrm { f } ( x ) = 2 \ln ( x ) - 4 , \quad x > 0 , \quad x \in \mathbb { R }$$

1. Sketch, on separate diagrams, the curve with equation
   1. \(y = \mathrm { f } ( x )\)
   2. \(y = | \mathrm { f } ( x ) |\) On each diagram, show the coordinates of each point at which the curve meets or cuts the axes. On each diagram state the equation of the asymptote.
2. Find the exact solutions of the equation \(| \mathrm { f } ( x ) | = 4\) $$\mathrm { g } ( x ) = \mathrm { e } ^ { x + 5 } - 2 , \quad x \in \mathbb { R }$$
3. Find \(\mathrm { gf } ( x )\), giving your answer in its simplest form.
4. Hence, or otherwise, state the range of gf.

3 The diagram shows the curve \(y = \mathrm { f } ( x )\). Points \(A , B , C\) and \(D\) on the curve have coordinates ( \(- 1,0 ) , ( 2,0 )\), \(( 5,0 )\) and \(( 0,2 )\) respectively.
\includegraphics[max width=\textwidth, alt={}, center]{a31997f4-7890-42c1-9725-1b7058e8741f-2_593_1221_1041_406} On the copy of this diagram in the Printed Answer Book, sketch the curve \(y ^ { 2 } = \mathrm { f } ( x )\), giving the coordinates of the points where the curve crosses the axes.

7

1. Show algebraically that the function \(\mathrm { f } ( x ) = \frac { 2 x } { 1 - x ^ { 2 } }\) is odd. Fig. 7 shows the curve \(y = \mathrm { f } ( x )\) for \(0 \leqslant x \leqslant 4\), together with the asymptote \(x = 1\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{28ce1bcc-e9d5-4ae6-98c0-67b5b8c50bc6-4_730_817_431_607} \captionsetup{labelformat=empty} \caption{Fig. 7}
   \end{figure}
2. Use the copy of Fig. 7 to complete the curve for \(- 4 \leqslant x \leqslant 4\).