Figure 5 shows a sketch of the curve with equation $y = f(x)$, where $f(x) = \frac{4 \sin 2x}{e^{\sqrt{x} - 1}}, \quad 0 \leq x \leq \pi$ The curve has a maximum turning point at $P$ and a minimum turning point at $Q$ as shown in Figure 5 - Edexcel - A-Level Maths Pure - Question 1 - 2017 - Paper 1

For the equation $y = f(2x)$, we substitute $2x$ into the original function, which changes the equation of the turning points:

Letting $2x = k$, we have:

$\tan k = \sqrt{2}$

The general solutions of this equation can be derived as:

$k = \frac{\pi}{4} + n\pi$

Substituting for $k$, we find:

$x = \frac{1}{2}\left(\frac{\pi}{4} + n\pi\right)$

The minimum occurs at:

$x = \frac{1}{2}\cdot \frac{\pi}{4} = \frac{\pi}{8}$

By evaluating, we obtain:

$x \approx 0.392$

For the equation $y = 3 - 2f(x)$, the critical points are unaffected by the coefficient of $2$ since it merely reflects the function.

Using the original function $f(x)$, and knowing that:

$\tan 2x = \sqrt{2}$

We will derive:

$x = \frac{\pi}{4} + n\pi$

Finding the smallest positive solution yields the minimum turning point at:

$x = \frac{\pi}{8}$

However, for the first quadrant, we consider:

$x \approx 0.478$