Đáp án:
Max y=3 khi $x=\dfrac{\pi}2+k2\pi$
Min y=-1 khi $x=\dfrac{{ - 3\pi }}{4} + k2\pi$
Giải thích các bước giải:
\(\begin{array}{l} - 1 \le \sin \left( {x + \dfrac{\pi }{4}} \right) \le 1\\ \Leftrightarrow - 2 \le 2\sin \left( {x + \dfrac{\pi }{4}} \right) \le 2\\ \Leftrightarrow - 2 + 1 \le 2\sin \left( {x + \dfrac{\pi }{4}} \right) + 1 \le 2 + 1\\ \Leftrightarrow - 1 \le 2\sin \left( {x + \dfrac{\pi }{4}} \right) + 1 \le 3\\ \Rightarrow Min\,y = - 1 \Leftrightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) = - 1\\ \Leftrightarrow x + \dfrac{\pi }{4} = \dfrac{{ - \pi }}{2} + k2\pi \Leftrightarrow \dfrac{{ - 3\pi }}{4} + k2\pi \\ Max\,\,y = 3 \Leftrightarrow \,\,\sin \left( {x + \dfrac{\pi }{4}} \right) = 1\\ \Leftrightarrow x + \dfrac{\pi }{4} = \dfrac{\pi }{2} + k2\pi \Leftrightarrow \dfrac{\pi }{4} + k2\pi \end{array}\)
