tìm cực trị y=sin2x – x

tìm cực trị
y=sin2x – x

Share

1 Answer

  1. 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:

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

    • 0
Leave an answer

Leave an answer

Browse