Đáp án:
\[\begin{array}{l}
f{\left( x \right)_{\min }} = \sqrt 2 \Leftrightarrow \sin x = \pm 1\\
f{\left( x \right)_{\max }} = 2 \Leftrightarrow \sin x = 0
\end{array}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
f\left( x \right) = \sqrt {1 + \sin x} + \sqrt {1 - \sin x} \\
\Rightarrow {f^2}\left( x \right) = 1 + \sin x + 2\sqrt {1 - {{\sin }^2}x} + 1 - \sin x\\
\Rightarrow {f^2}\left( x \right) = 2 + 2\sqrt {1 - {{\sin }^2}x} \\
- 1 \le \sin x \le 1 \Leftrightarrow 0 \le {\sin ^2}x \le 1\\
\Rightarrow 0 \le 1 - \sin x \le 1\\
\Leftrightarrow 2 \le 2 + 2\sqrt {1 - {{\sin }^2}x} \le 4\\
\Leftrightarrow 2 \le {f^2}\left( x \right) \le 4\\
\Leftrightarrow \sqrt 2 \le f\left( x \right) \le 2\\
\Rightarrow f{\left( x \right)_{\min }} = \sqrt 2 \Leftrightarrow \sin x = \pm 1\\
f{\left( x \right)_{\max }} = 2 \Leftrightarrow \sin x = 0
\end{array}\)
