2 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), which has a \(y\)-intercept at \(\mathrm { P } ( 0,3 )\), a minimum point at \(\mathrm { Q } ( 1,2 )\), and an asymptote \(x = - 1\).
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{f7049002-f97a-4c83-a7d6-eba28e3b589a-1_904_937_785_604}
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\caption{Fig. 9}
\end{figure}
- Find the coordinates of the images of the points P and Q when the curve \(y = \mathrm { f } ( x )\) is transformed to
(A) \(y = 2 \mathrm { f } ( x )\),
(B) \(y = \mathrm { f } ( x + 1 ) + 2\).
You are now given that \(\mathrm { f } ( x ) = \frac { x ^ { 2 } + 3 } { x + 1 } , x \neq - 1\). - Find \(\mathrm { f } ^ { \prime } ( x )\), and hence find the coordinates of the other turning point on the curve \(y = \mathrm { f } ( x )\).
- Show that \(\mathrm { f } ( x - 1 ) = x - 2 + \frac { 4 } { x }\).
- Find \(\int _ { a } ^ { b } \left( x - 2 + \frac { 4 } { x } \right) \mathrm { d } x\) in terms of \(a\) and \(b\).
Hence, by choosing suitable values for \(a\) and \(b\), find the exact area enclosed by the curve \(y = \mathrm { f } ( x )\), the \(x\)-axis, the \(y\)-axis and the line \(x = 1\).