Y= cos^2 x - sinx -3 Tìm gtln gtnn của hàm số

Đáp án:

$\min y = -4;\,\max y = -\dfrac74$

Giải thích các bước giải:

$y =\cos^2x -\sin x -3$

$\to y = 1 -\sin^2x -\sin x -3$

$\to y = -\left(\sin^2x +2\cdot\dfrac12\sin x +\dfrac14\right) -\dfrac74$

$\to y = -\left(\sin x +\dfrac12\right)^2 -\dfrac74$

Ta có:

$\quad -1\leq \sin x \leq 1$

$\to -\dfrac12 \leq \sin x +\dfrac12\leq \dfrac32$

$\to 0 \leq \left(\sin x +\dfrac12\right)^2\leq \dfrac94$

$\to -\dfrac94 \leq -\left(\sin x +\dfrac12\right)^2 \leq 0$

$\to -4 \leq -\left(\sin x +\dfrac12\right)^2 -\dfrac74 \leq -\dfrac74$

$\to -4 \leq y \leq -\dfrac74$

Vậy $\min y = -4;\,\max y = -\dfrac74$