求下列函数的单调增区间,最值,x的集合 (1) y=sin(2x-π/3) (2) y=3sin(-2x+π/3)
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/28 04:13:36
![求下列函数的单调增区间,最值,x的集合 (1) y=sin(2x-π/3) (2) y=3sin(-2x+π/3)](/uploads/image/z/11506720-40-0.jpg?t=%E6%B1%82%E4%B8%8B%E5%88%97%E5%87%BD%E6%95%B0%E7%9A%84%E5%8D%95%E8%B0%83%E5%A2%9E%E5%8C%BA%E9%97%B4%2C%E6%9C%80%E5%80%BC%2Cx%E7%9A%84%E9%9B%86%E5%90%88+%281%29+y%3Dsin%282x-%CF%80%2F3%29+%282%29+y%3D3sin%28-2x%2B%CF%80%2F3%29)
求下列函数的单调增区间,最值,x的集合 (1) y=sin(2x-π/3) (2) y=3sin(-2x+π/3)
求下列函数的单调增区间,最值,x的集合 (1) y=sin(2x-π/3) (2) y=3sin(-2x+π/3)
求下列函数的单调增区间,最值,x的集合 (1) y=sin(2x-π/3) (2) y=3sin(-2x+π/3)
因为sinx的递增区间为2kπ-π/2≤x≤2kπ+π/2,k∈Z.
所以sin(2x-π/3)的递增区间为2kπ-π/2≤2x-π/3≤2kπ+π/2,k∈Z.
解出x得递增区间(亲,请写成区间形式).
因为sinx的递减区间为2kπ+π/2≤x≤2kπ+3π/2,k∈Z.
所以sin(2x-π/3)的递减区间为2kπ+π/2≤2x-π/3≤2kπ+3π/2,k∈Z.
解出x得递减区间(亲,请写成区间形式).
当2x-π/3=2kπ+π/2,k∈Z时,y max=1;
当2x-π/3=2kπ-π/2,k∈Z时,y max=-1.
2.
因为sin(-x)的单减区间为2kπ-π/2≤x≤2kπ+π/2,k∈Z.
所以sin(-2x+π/3)的单减区间为2kπ-π/2≤-2x+π/3≤2kπ+π/2,k∈Z.
解出x得递增区间(亲,请写成区间形式).
因为sin(-x)的单增区间为2kπ+π/2≤x≤2kπ+3π/2,k∈Z.
所以sin(-2x+π/3)的单增区间为2kπ+π/2≤-2x+π/3≤2kπ+3π/2,k∈Z.
解出x得单增区间(亲,请写成区间形式).
当-2x+π/3=2kπ-π/2,k∈Z时,y max=3;
当-2x+π/3=2kπ+π/2,k∈Z时,y max=-3.
希望你从中“悟出”这类问题的解法来!