Toán Tìm GTLN GTNN của hàm số y=căn2 sin(x\2-pi\4)-1 14/09/2021 By Josephine Tìm GTLN GTNN của hàm số y=căn2 sin(x\2-pi\4)-1
$\begin{array}{l} – 1 \le \sin \left( {\frac{x}{2} – \frac{\pi }{4}} \right) \le 1 \Rightarrow – \sqrt 2 \le \sqrt 2 \sin \left( {\frac{x}{2} – \frac{\pi }{4}} \right) \le \sqrt 2 \\ \Rightarrow – \sqrt 2 – 1 \le \sqrt 2 \sin \left( {\frac{x}{2} – \frac{\pi }{4}} \right) – 1 \le \sqrt 2 – 1 \end{array}$ $\begin{array}{l} \Rightarrow \min y = – \sqrt 2 – 1\,khi\,\\ \sin \left( {\frac{x}{2} – \frac{\pi }{4}} \right) = – 1 \Leftrightarrow \frac{x}{2} – \frac{\pi }{4} = – \frac{\pi }{2} + k2\pi \Leftrightarrow x = – \frac{\pi }{2} + k4\pi \\ \Rightarrow \max y = \sqrt 2 – 1\,khi\,\\ \sin \left( {\frac{x}{2} – \frac{\pi }{4}} \right) = 1 \Leftrightarrow \frac{x}{2} – \frac{\pi }{4} = \frac{\pi }{2} + k2\pi \Leftrightarrow x = \frac{{3\pi }}{2} + k4\pi \end{array}$ Trả lời
$\begin{array}{l}
– 1 \le \sin \left( {\frac{x}{2} – \frac{\pi }{4}} \right) \le 1 \Rightarrow – \sqrt 2 \le \sqrt 2 \sin \left( {\frac{x}{2} – \frac{\pi }{4}} \right) \le \sqrt 2 \\
\Rightarrow – \sqrt 2 – 1 \le \sqrt 2 \sin \left( {\frac{x}{2} – \frac{\pi }{4}} \right) – 1 \le \sqrt 2 – 1
\end{array}$
$\begin{array}{l}
\Rightarrow \min y = – \sqrt 2 – 1\,khi\,\\
\sin \left( {\frac{x}{2} – \frac{\pi }{4}} \right) = – 1 \Leftrightarrow \frac{x}{2} – \frac{\pi }{4} = – \frac{\pi }{2} + k2\pi \Leftrightarrow x = – \frac{\pi }{2} + k4\pi \\
\Rightarrow \max y = \sqrt 2 – 1\,khi\,\\
\sin \left( {\frac{x}{2} – \frac{\pi }{4}} \right) = 1 \Leftrightarrow \frac{x}{2} – \frac{\pi }{4} = \frac{\pi }{2} + k2\pi \Leftrightarrow x = \frac{{3\pi }}{2} + k4\pi
\end{array}$