tìm cực trị y=sin2x - x

Đáp án:

\(\eqalign{ & Ham\,\,so\,\,dat\,\,cuc\,\,dai\,\,tai\,\,x = {\pi \over 6} + k\pi \cr & Ham\,\,so\,\,dat\,\,cuc\,\,tieu\,\,tai\,\,x = - {\pi \over 6} + k\pi \cr} \)

Giải thích các bước giải:

$$\eqalign{ & y = \sin 2x - x \cr & y' = 2\cos 2x - 1 = 0 \cr & \Leftrightarrow \cos 2x = {1 \over 2} \cr & \Leftrightarrow \left[ \matrix{ 2x = {\pi \over 3} + k2\pi \hfill \cr 2x = - {\pi \over 3} + k2\pi \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x = {\pi \over 6} + k\pi \hfill \cr x = - {\pi \over 6} + k\pi \hfill \cr} \right. \cr & y'' = - 4\sin 2x \cr & y''\left( {{\pi \over 6} + k\pi } \right) = - 4\sin \left( {{\pi \over 3} + k2\pi } \right) = - 4\sin {\pi \over 3} = - 2\sqrt 3 < 0 \cr & \Rightarrow Ham\,\,so\,\,dat\,\,cuc\,\,dai\,\,tai\,\,x = {\pi \over 6} + k\pi \cr & y''\left( { - {\pi \over 6} + k\pi } \right) = - 4\sin \left( { - {\pi \over 3} + k2\pi } \right) = - 4\sin \left( { - {\pi \over 3}} \right) = 2\sqrt 3 > 0 \cr & \Rightarrow Ham\,\,so\,\,dat\,\,cuc\,\,tieu\,\,tai\,\,x = - {\pi \over 6} + k\pi \cr} $$