14.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{15b5fb91-3bc6-4167-afb9-91879ebbfc96-26_545_1029_164_571}
\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
$$y = 3 - 2 f ( x )$$
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