求y=x+|sin2x|单调区间

2个回答

  • y=x+|sin2x|等价于

    y=x+sin2x kπ≤x≤kπ+π/2

    y=x-sin2x kπ+π/2≤x≤(k+1)π

    对该曲线方程求导,得

    y'=1+2cos2x kπ≤x≤kπ+π/2

    y'=1-2cos2x kπ+π/2≤x≤(k+1)π

    当y'>0时,该函数单调增,即

    1+2cos2x>0 kπ≤x≤kπ+π/2

    1-2cos2x>0 kπ+π/2≤x≤(k+1)π

    即±cos2x>-1/2

    在kπ≤x≤kπ+π/2时,当且仅当kπ≤x≤kπ+π/3时,

    cos2x>cos2π/3,即cos2x>-1/2

    kπ+π/2≤x≤(k+1)π时,当且仅当kπ+π/2≤x≤kπ+5π/6时,

    cos2x-1/2,

    因此y=x+|sin2x|单调增区间是[kπ,kπ+π/3]U[kπ+π/2,kπ+5π/6],

    单调减区间是[kπ+π/3,kπ+π/2]U[kπ+5π/6,kπ+π]