Đáp án:

$a)\quad \begin{cases}\min y = - 3 \Leftrightarrow x =- \dfrac{3\pi}{4} + k2\pi\\\max y = 3\Leftrightarrow  x = \dfrac{\pi}{4} + k2\pi\end{cases}\quad (k\in\Bbb Z)$

$b)\quad \begin{cases}\min y = 4\sqrt2 -1\Leftrightarrow x = -\dfrac{\pi}{2} + k2\pi\\\max y = 7 \Leftrightarrow x = \dfrac{\pi}{2} + k2\pi\end{cases}\quad (k\in\Bbb Z)$

Giải thích các bước giải:

$a)\quad y = 3\sin\left(x + \dfrac{\pi}{4}\right)$

Ta có:

$\quad - 1 \leqslant \sin\left(x + \dfrac{\pi}{4}\right) \leqslant 1$

$\Leftrightarrow - 3 \leqslant 3\sin\left(x + \dfrac{\pi}{4}\right) \leqslant 3$

Vậy $\min y = - 3 \Leftrightarrow \sin\left(x + \dfrac{\pi}{4}\right)= -1 \Leftrightarrow x =- \dfrac{3\pi}{4} + k2\pi$

$\max y = 3 \Leftrightarrow \sin\left(x + \dfrac{\pi}{4}\right)= 1 \Leftrightarrow x = \dfrac{\pi}{4} + k2\pi\quad (k\in\Bbb Z)$

$b)\quad y = 4\sqrt{\sin x +3} - 1$

Ta có:

$\quad -1\leqslant \sin x \leqslant 1$

$\Leftrightarrow 2\leqslant \sin x + 3 \leqslant 4$

$\Leftrightarrow 4\sqrt2 \leqslant 4\sqrt{\sin x +3} \leqslant 8$

$\Leftrightarrow 4\sqrt2 -1\leqslant 4\sqrt{\sin x +3} -1\leqslant 7$

Vậy $\min y = 4\sqrt2 -1\Leftrightarrow \sin x = -1\Leftrightarrow x = -\dfrac{\pi}{2} + k2\pi$

$\max y = 7 \Leftrightarrow \sin x = 1 \Leftrightarrow x = \dfrac{\pi}{2} + k2\pi\quad (k\in\Bbb Z)$

Đáp án:

Giải thích các bước giải:

 a) `y=3sin (x+\frac{\pi}{4})`

Ta có: `-1 \le sin (x+\frac{\pi}{4}) \le 1`

`⇔ -3 \le sin (x+\frac{\pi}{4}) \le 3`

`⇒ -3 \le y \le 3`

Vậy `y_{min}=-3` khi `x=\frac{-3\pi}{4}+k2\pi\ (k \in \mathbb{Z})`

`y_{max}=3` khi `x=\frac{\pi}{4}+k2\pi\ (k \in \mathbb{Z})`

b) `y=4\sqrt{sin\ x+3}-1`

Ta có: `-1 \le sin\ x \le 1`

`⇔ 2 \le sin\ x+3 \le 4`

`⇔ \sqrt{2} \le \sqrt{sin\ x+3} \le 2`

`⇔ 4\sqrt{2} \le 4\sqrt{sin\ x+3} \le 8`

`⇔ 4\sqrt{2}-1 \le 4\sqrt{sin\ x+3}-1 \le 7`

`⇒ 4\sqrt{2}-1 \le y \le 7`

Vậy `y_{min}=4\sqrt{2}-1` khi `x=\frac{-\pi}{2}+k2\pi\ (k \in \mathbb{Z})`

`y_{max}=7` khi `x=\frac{\pi}{2}+k2\pi\ (k \in \mathbb{Z})`