Toán Lớp 11: Tìm GTNN,GTLN của hàm số:
y=sin^2 x-4sinx-5
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Toán Lớp 11: Tìm GTNN,GTLN của hàm số:
y=sin^2 x-4sinx-5
Comments ( 2 )
GTLN:y = 0\,khi:x = \dfrac{{ – \pi }}{2} + k2\pi \\
GTNN:y = – 8\,khi:x = \dfrac{\pi }{2} + k2\pi
\end{array} \right.$
y = {\sin ^2}x – 4\sin x – 5\\
= {\sin ^2}x – 4\sin x + 4 – 9\\
= {\left( {\sin x – 2} \right)^2} – 9\\
Do: – 1 \le \sin x \le 1\\
\Leftrightarrow – 3 \le \sin x – 2 \le – 1\\
\Leftrightarrow 1 \le {\left( {\sin x – 2} \right)^2} \le 9\\
\Leftrightarrow – 8 \le {\left( {\sin x – 2} \right)^2} – 9 \le 0\\
\Leftrightarrow – 8 \le y \le 0\\
\Leftrightarrow \left\{ \begin{array}{l}
GTLN:y = 0\,khi:x = \dfrac{{ – \pi }}{2} + k2\pi \\
GTNN:y = – 8\,khi:x = \dfrac{\pi }{2} + k2\pi
\end{array} \right.
\end{array}$