题目

求函数y=2sin(-x)的单调区间. 答案：思路分析:可依据y=sinx的单调区间来求本题函数的单调区间.解：y=2sin(-x)=-2sin(x-),∵y=sinu(u∈R)的递增,递减区间分别为［2kπ-,2kπ+］(k∈Z),［2kπ+,2kπ+］(k∈Z),∴函数y=-2sin(x-)的递增,递减区间分别由下面的不等式确定:2kπ+≤x-≤2kπ+(k∈Z),2kπ-≤x-≤2kπ+(k∈Z),得2kπ+≤x≤2kπ+(k∈Z∈Z),2kπ-≤x≤2kπ+(k∈Z).∴函数y=2si ----Do you know the result of the ________ race? ----Yes. The winner is a boy ________ Lin Feng from Class 4.A．100-metres, calledB．100-metre; callingC．100-metre; calledD．100-metres, calling