Đáp án: $GTLN_y=\sqrt{2}, GTNN_y=-1$
Giải thích các bước giải:
Ta có:
$y=\sin x-\cos x$
$\to y=\sqrt{2}\cdot (\sin x\cdot \dfrac{1}{\sqrt{2}}-\cos x\cdot \dfrac{1}{\sqrt{2}})$
$\to y=\sqrt{2}\cdot (\sin x\cdot \cos(\dfrac{\pi}{4})-\cos x\cdot \cos(\dfrac{\pi}{4}))$
$\to y=\sqrt{2}\cdot \sin(x-\dfrac{\pi}{4})$
Vì $x\in[0,3]\to$Ta xét $2$ trường hợp:
$I): x\in[0, \dfrac34\pi]$
$\to x-\dfrac{\pi}{4}\in[-\dfrac{\pi}{4}, \dfrac{\pi}{2}]$
$\to \sin(x-\dfrac{\pi}{4})\in[-1,\sqrt{2}](*)$
$II): x\in(\dfrac34\pi, 3]$
$\to x-\dfrac{\pi}{4}\in[ \dfrac{\pi}{2}, 3-\dfrac{\pi}{4}]$
$\to \sin(x-\dfrac{\pi}{4})\in(\sin(3-\dfrac{\pi}{4}), 1)(**)$
Từ $(*), (**)$
$\to GTLN_y=1\to x=0, GTNN_y=\sqrt{2}\to x=\dfrac34\pi$
