Đáp án:
$y_{\min}=-3$ khi $x=\pi+k2\pi\,\,(k\in\mathbb Z)$
$y_{\max}=2\sqrt 2-3$ khi $x=k2\pi\,\,(k\in\mathbb Z)$
Giải thích các bước giải:
$y=2\sqrt{\cos x+1}-3$
Ta có:
$-1\le \cos x\le 1$
$⇒0\le \cos x+1\le 2$
$⇒0\le \sqrt{\cos x+1}\le \sqrt 2$
$⇒0\le 2\sqrt{\cos x+1}\le 2\sqrt 2$
$⇒-3\le 2\sqrt{\cos x+1}-3\le 2\sqrt 2-3$
$⇒-3\le y\le 2\sqrt 2-3$
$y\ge -3⇒y_{\min}=-3$
Dấu "=" xảy ra khi:
$2\sqrt{\cos x+1}-3=-3$
$⇔\sqrt{\cos x+1}=0$
$⇔\cos x+1=0$
$⇔\cos x=-1$
$⇔x=\pi+k2\pi\,\,(k\in\mathbb Z)$
Vậy $y_{\min}=-3$ khi $x=\pi+k2\pi\,\,(k\in\mathbb Z)$
$y\le 2\sqrt 2-3⇒y_{\max}=2\sqrt 2-3$
Dấu "=" xảy ra khi:
$2\sqrt{\cos x+1}-3=2\sqrt 2-3$
$⇔\sqrt{\cos x+1}=\sqrt 2$
$⇔\cos x+1=2$
$⇔\cos x=1$
$⇔x=k2\pi\,\,(k\in\mathbb Z)$
Vậy $y_{\max}=2\sqrt 2-3$ khi $x=k2\pi\,\,(k\in\mathbb Z)$.
