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:

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