Đáp án:
$\begin{cases}\min y = 5 \Leftrightarrow x = - \dfrac{\pi}{4} + k2\pi\\\max y = 9 \Leftrightarrow x = \dfrac{3\pi}{4} +k2\pi\end{cases} \quad (k \in \Bbb Z)$
Giải thích các bước giải:
$y = 7 - 2\cos\left(x + \dfrac{\pi}{4}\right)$
Ta có:
$-1 \leq \cos\left(x + \dfrac{\pi}{4}\right) \leq 1$
$\to -2 \leq -2\cos\left(x + \dfrac{\pi}{4}\right) \leq 2$
$\to 5 \leq 7 -2\cos\left(x + \dfrac{\pi}{4}\right) \leq 9$
Hay $5 \leq y \leq 9$
Vậy $\min y = 5 \Leftrightarrow \cos\left(x + \dfrac{\pi}{4}\right) = 1 \Leftrightarrow x = - \dfrac{\pi}{4} + k2\pi$
$\max y = 9 \Leftrightarrow \cos\left(x + \dfrac{\pi}{4}\right) = -1 \Leftrightarrow x = \dfrac{3\pi}{4} +k2\pi \quad (k \in \Bbb Z)$
