Lời giải:

\( f'(x) = -2\sqrt{2} \sin 2x + 4 \cos x \)

\(= -4\sqrt{2} \sin x \cos x + 4 \cos x \)

\(=4 \cos x (1 - \sqrt{2} \sin x) = 0 \)

\( \Leftrightarrow \left[ \begin{array}{} \cos x = 0 \\ \sin x = \frac{\sqrt{2}}{2}\end{array} \right. \text{trên đoạn}  \left[ 0, \frac{\pi}{2} \right] \)

\( \Leftrightarrow \left[ \begin{array}{} x = \frac{\pi}{2} \\ x = \frac{\pi}{4} \end{array} \right. \)

\( \max\limits_{[0, \frac{\pi}{2}]} f(x) = \max \left\{ f(0), f\left(\frac{\pi}{4}\right), f\left(\frac{\pi}{2}\right) \right\} = \max \left\{ \sqrt{2}, 4 - \sqrt{2}, 2\sqrt{2} \right\} = 2\sqrt{2} \)

\( \min\limits_{[0, \frac{\pi}{2}]} f(x) = \min \left\{ f(0), f\left(\frac{\pi}{4}\right), f\left(\frac{\pi}{2}\right) \right\} = \min \left\{ \sqrt{2}, 4 - \sqrt{2}, 2\sqrt{2} \right\} = \sqrt{2} \)

Cách 2: Chạy bảng:

Start: \( 0 \), End: \( \frac{\pi}{2} \), Step: \( \frac{\pi}{12} \)

Cách 3: \( y = -2\sqrt{2} \sin^2 x + 4 \sin x + \sqrt{2} \)

Đặt \( t = \sin x \): \( 0 \leq x \leq \frac{\pi}{2} \Rightarrow 0 \leq t \leq 1 \)

Hàm số chuyển thành: \( g(t) = -2\sqrt{2} t^2 + 4t + \sqrt{2} \)

Tính đạo hàm của \( g(t) \): \( g'(t) = -4\sqrt{2} t + 4 = 0 \Leftrightarrow t = \frac{\sqrt{2}}{2} \)

Giá trị lớn nhất của \( y \):  

\( \text{Max } y = \max\limits_{[0, 1]} g(t) = \max \left\{ g(0), g(1), g\left(\frac{\sqrt{2}}{2}\right) \right\} = \max \left\{ \sqrt{2}, 4 - \sqrt{2}, 2\sqrt{2} \right\} = 2\sqrt{2} \)