$\begin{cases}\min y = -1 \Leftrightarrow x = -\dfrac{\pi}{6} + k\dfrac{2\pi}{3}\\\max y = \Leftrightarrow x = \dfrac{\pi}{6} + k\dfrac{2\pi}{3}\end{cases}\quad(k\in\Bbb Z)$
Giải thích các bước giải:
$y = 3\sin3x + 2$
Ta có:
$\quad -1 \leq \sin3x \leq 1$
$\to -3\leq 3\sin3x \leq 3$
$\to -1 \leq 3\sin3x +2 \leq 5$
$\to -1\leq y \leq 5$
Vậy $\begin{cases}\min y = -1 \Leftrightarrow \sin3x = -1\Leftrightarrow x = -\dfrac{\pi}{6} + k\dfrac{2\pi}{3}\\\max y = 5 \Leftrightarrow \sin3x = 1 \Leftrightarrow x = \dfrac{\pi}{6} + k\dfrac{2\pi}{3}\end{cases}\,\,(k\in\Bbb Z)$
`y = 3sin 3x + 2`
`text{Ta có}`
`-1 <= sin 3x <= 1`
`-> -3 <= 3sin 3x <= 3`
`-> -1 <= 3sin 3x + 2 <= 5`
`-> -1 <= y <= 5`
`text{Vậy}`
`y_{Max} = 5`
`y_{Min} = -1`
Đáp án:
$\begin{cases}\min y = -1 \Leftrightarrow x = -\dfrac{\pi}{6} + k\dfrac{2\pi}{3}\\\max y = \Leftrightarrow x = \dfrac{\pi}{6} + k\dfrac{2\pi}{3}\end{cases}\quad(k\in\Bbb Z)$
Giải thích các bước giải:
$y = 3\sin3x + 2$
Ta có:
$\quad -1 \leq \sin3x \leq 1$
$\to -3\leq 3\sin3x \leq 3$
$\to -1 \leq 3\sin3x +2 \leq 5$
$\to -1\leq y \leq 5$
Vậy $\begin{cases}\min y = -1 \Leftrightarrow \sin3x = -1\Leftrightarrow x = -\dfrac{\pi}{6} + k\dfrac{2\pi}{3}\\\max y = 5 \Leftrightarrow \sin3x = 1 \Leftrightarrow x = \dfrac{\pi}{6} + k\dfrac{2\pi}{3}\end{cases}\,\,(k\in\Bbb Z)$