15.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{f7994129-07ee-4f6d-9531-08a15a38b794-30_551_1026_219_523}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{figure}
Figure 5 shows a sketch of the curve with equation \(y = \mathrm { f } ( x )\), where
$$\mathrm { f } ( x ) = \frac { 4 \sin 2 x } { \mathrm { e } ^ { \sqrt { 2 } x - 1 } } , \quad 0 \leqslant x \leqslant \pi$$
The curve has a maximum turning point at \(P\) and a minimum turning point at \(Q\) as shown in Figure 5.
- Show that the \(x\) coordinates of point \(P\) and point \(Q\) are solutions of the equation
$$\tan 2 x = \sqrt { 2 }$$
- Using your answer to part (a), find the \(x\)-coordinate of the minimum turning point on the curve with equation
- \(y = \mathrm { f } ( 2 x )\).
- \(y = 3 - 2 \mathrm { f } ( x )\).