Question

已知函数fx=根号2sin(2x+3分之π)的单调性和最小正周期，还有在[0,,2分之π】的值域

详细步骤！！！！因为很急！所以谢谢！

Answer 1

f(x)=√2sin(2x+π/3)
=√2sin(2x+2π+π/3)
=√2sin[2(x+π)+π/3]
∴最小正周期=π
∵2kπ-π/2<2x+π/3<2kπ+π/2
即 kπ-5π/12<x<kπ+π/12
∴f(x)在kπ-5π/12<x<kπ+π/12单增
∵2kπ+π/2<2x+π/3<2kπ+3π/2
即 kπ+π/12<x<kπ+7π/12
∴f(x)在kπ+π/12<x<kπ+7π/12单减
∵0<x<π/2
即 2π/3<2x+π/3<5π/3
f(x)在2x+π/3∈（2π/3，3π/2)递减，在2x+π/3∈（3π/2，5π/3）递增，
在2x+π/3=2π/3 函数值=√6/2，最大
在2x+π/3=3π/2时，函数值最小=-√2
∴函数在[0,,2分之π】的值域：[-√2, √6/2]

Answer 2

2x。=2π，x。=π
最小正周期: π
主值单调递增区间：
-π/2<=2x+3分之π<=π/2
-5π/6<=2x<=π/6
-5π/12<=x<=π/12
主值单调递减区间：
。。。

在[0,,2分之π】区间
3分之π<=π+3分之π
则 -根号3<=f(x)<=2