Lời giải:

• \( D = \mathbb{R} \)
• Đặt \( t = \sin x + \cos x = \sqrt{2} \sin \left( x + \frac{\pi}{4} \right) \)  \(\Rightarrow -\sqrt{2} \leq t \leq \sqrt{2}\)

\(t^2= 1+2 \sin x \cos x \Rightarrow \sin x \cos x = \frac{t^2-1}{2} \)

• \( y = t + \frac{t^2 - 1}{2} = \frac{1}{2}(t^2 + 2t - 1) = g(t) \)
• \( g'(t) = t + 1 = 0 \Leftrightarrow t = -1 \)
• \( g(-\sqrt{2}) = \frac{1 - 2\sqrt{2}}{2}, \quad g(\sqrt{2}) = \frac{1 + 2\sqrt{2}}{2}, \quad g(-1) = -1 \)
• Min \( y = \min\limits_{[-\sqrt{2}, \sqrt{2}]} g(t) = -1 \) khi \( \sin\left(x + \frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} \)
• Max \( y = \max\limits_{[-\sqrt{2}, \sqrt{2}]} g(t) = \frac{1 + 2\sqrt{2}}{2} \) khi \( \sin\left(x + \frac{\pi}{4}\right) = 1\)