tìm gtln ,gtnn của y=2sin2x +3 với x thuộc [-pi/8;pi/3]

$$\eqalign{ & y = 2\sin 2x + 3 \cr & x \in \left[ { - {\pi \over 8};{\pi \over 3}} \right] \Rightarrow 2x \in \left[ { - {\pi \over 4};{{2\pi } \over 3}} \right] \cr & \Rightarrow \sin 2x \in \left[ { - {{\sqrt 2 } \over 2};1} \right] \cr & \Rightarrow - {{\sqrt 2 } \over 2} \le \sin 2x \le 1 \cr & \Leftrightarrow - \sqrt 2 \le 2\sin 2x \le 2 \cr & \Leftrightarrow - \sqrt 2 + 3 \le 2\sin 2x + 3 \le 5 \cr & \Rightarrow \min y = - \sqrt 2 + 3 \cr & \,\,\,\,\,\,\,\max y = 5 \cr} $$

Đáp án:

$\left\{ \begin{array}{l} \min\,y\,\, = 3 - \sqrt 2 \,\,\,\,khi\,\,x=\{-\dfrac{\pi}8+k\pi;\dfrac{5\pi}8+k\pi\}\\ \max\,\,y = \,5\,\,khi\,\,\,\,x=\dfrac{\pi}4+k\pi. \end{array} \right.$ $(k\in\mathbb Z)$

Lời giải:

\(\begin{array}{l} y = 2\sin 2x + 3\,,\,\,\\ x \in \left[ { - \dfrac{\pi }{8};\,\,\dfrac{\pi }{3}} \right] \Rightarrow 2x \in \left[ { - \dfrac{\pi }{4};\,\,\dfrac{{2\pi }}{3}} \right] \Rightarrow - \dfrac{{\sqrt 2 }}{2} \le \sin 2x \le 1\\ \Rightarrow - \sqrt 2 \le 2\sin 2x \le 1\\ \Leftrightarrow 3 - \sqrt 2 \le 2\sin 2x + 3 \le 5\\ \Rightarrow \left\{ \begin{array}{l} \min\,y\,\, = 3 - \sqrt 2 \,\,\,\,khi\,\,\sin 2x = - \dfrac{{\sqrt 2 }}{2}\Leftrightarrow x=\{-\dfrac{\pi}8+k\pi;\dfrac{5\pi}8+k\pi\}\\ \max\,\,y = \,5\,\,khi\,\,\,\,\sin \,2x = 1\Leftrightarrow x=\dfrac{\pi}4+k\pi. \end{array} \right. \end{array}\) $(k\in\mathbb Z)$.