Question

y=x+|sin2x|,求函数单调区间，求详细过程，谢谢各位!

Answer 1

1. 当2kπ<=2x<=2kπ+π,即kπ<=x<=kπ+π/2时

   y=x+sin2x ,y'=1+2cos2x ，

   (1) 当 2kπ-2π/3<=2x<=2kπ+2π/3,即 kπ-π/3<=x<=kπ+π/3，y'>0，

   (2)当 2kπ+2π/3<=2x<=2kπ+4π/3,即 kπ+π/3<=x<=kπ+2π/3，y'<0，

   所以，x∈[kπ+π/3,kπ+π/2],y为增函数，x∈[kπ,kπ+π/3],y为减函数

2. 当2kπ-π<=2x<=2kπ,即kπ-π/2<=x<=kπ时

   y=x-sin2x ,y'=1-2cos2x ，

   (1) 当 2kπ-2π/3<=2x<=2kπ+2π/3,即 kπ-π/3<=x<=kπ+π/3，y'<0，

   (2) 当 2kπ+2π/3<=2x<=2kπ+4π/3,即 kπ+π/3<=x<=kπ+2π/3，y'>0，

   所以，x∈[kπ-π/2,kπ-π/3],y为增函数,x∈[kπ-π/3,kπ]，y为减函数

综上，x∈[kπ-π/2,kπ-π/3]∪[kπ+π/3,kπ+π/2],y为增函数，x∈[kπ-π/3,kπ]∪[kπ,kπ+π/3],y为减函数。

Answer 2

追答