Question

一道数学题，急需，在线等

函数f(x)=2sin(x-π/3)(x∈[0,π])的单调增区间
打错了，是是函数f(x)=2sin(x-π/3)(x∈[-π，0])的单调增区间

Answer 1

π/3<=x<=5π/6 因为 0<=x<=π -π/3<=x-π/3<=2π/3 -π/3<=x-π/3<=π/2时 函数单调递增 即π/3<=x<=5π/6

Answer 2

-π/2+2kπ<(x-π/3)<π/2+2kπ
-π/6+2kπ<x<5π/6+2kπ
又x∈[-π，0]
所以-π/6<x<0
【<好都表示小于等于，电脑用的不熟，不会小于等于号】

Answer 3

2kπ-π/2<=x-π/3<=2kπ+π/2
这是递增区间
2kπ-1/6π<=x<=2kπ+5/6π
【0，5/6π】都是哦

Answer 4

∵0<=x<=π
∴-π/3≤x-π/3≤2π/3
又∵单调递增
∴-π/2≤x-π/3≤π/2
得0≤x≤5π/6

Answer 5

[π/3，5π/6）