gtnn và gtln y=cos2x trên [-pi/3 pi/6]

Đáp án:

\(\begin{array}{l}
\max y = 1 \Leftrightarrow x = 0\\
\min y =  - \dfrac{1}{2} \Leftrightarrow x =  - \dfrac{\pi }{3}
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
x \in \left[ { - \dfrac{\pi }{3};\dfrac{\pi }{6}} \right] \Rightarrow 2x \in \left[ { - \dfrac{{2\pi }}{3};\dfrac{\pi }{3}} \right]\\
 \Rightarrow y = \cos 2x \in \left[ { - \dfrac{1}{2};1} \right]\\
 \Rightarrow \max y = 1 \Leftrightarrow \cos 2x = 1 \Leftrightarrow 2x = k2\pi  \Leftrightarrow x = k\pi \,\,\left( {k \in Z} \right)\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \in \left[ { - \dfrac{\pi }{3};\dfrac{\pi }{6}} \right] \Leftrightarrow x = 0\\
\,\,\,\,\,\,\min y =  - \dfrac{1}{2} \Leftrightarrow \cos 2x =  - \dfrac{1}{2} \Leftrightarrow 2x =  \pm \dfrac{{2\pi }}{3} + k2\pi  \Leftrightarrow x =  \pm \dfrac{\pi }{3} + k\pi \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \in \left[ { - \dfrac{\pi }{3};\dfrac{\pi }{6}} \right] \Leftrightarrow x =  - \dfrac{\pi }{3}
\end{array}\)