Question

函数f(x)=sinx减根号3cosx(x属于[负派，0])的单调递增区间是？急

Answer 1

解：
f(x)
＝sinx－√3cosx
＝2(sinx•1/2－cosx•√3/2)
＝2sin(x－π/3)
由：－π/2＋2kπ≤x－π/3≤π/2＋2kπ
得：－π/6＋2kπ≤x≤5π/6＋2kπ
∵x∈[－π，0]
取交集，得：[－π/6，0]

Answer 2

f(x)=sinx－√3cosx=2sin(x－π/3)。因x∈[－π，0]，则x－π/3∈[－4π/3，－π/3]，增区间是[－π/6，0]。

Answer 3

f(x)=sinx-根号3cosx=2(1/2sinx-(根号3)/2cosx)=2sin(x-π/3)
-π/2+2kπ<=x-π/3<=π/2+2kπ
-π/6+2kπ<=x<=5π/6+2kπ
又因为-π<=x<=0
所以-π/6<=x<=0

Answer 4

对它求导数，找出零点，也就是分区间点，就是导数等于0的点，大于零的部分就是单调增区间