已知函数f(x)=2sin(2x/3 + π/6) -1,其中x∈[0,π],求f(x)的值域和单调增区间
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∵0≤x≤π,∴0≤2x/3≤2π/3,π/6≤2x/3 + π/6≤5π/6
∴1/2≤sin(2x/3 + π/6)≤1
∴0≤2sin(2x/3 + π/6) -1≤1
∴f(x)的值域:[0,1]
.π/6≤2x/3 + π/6≤π/2单增
∴f(x)的单增区间：[0,π/2]
π/2≤2x/3 + π/6≤5π/6单减
∴f(x)的单减区间：[π/2,π]
======以下答案可供参考======
供参考答案1:
-π/2-πf(x)单调增区间 [0,π/2]
f(x)值域 f(π/2)=1 f(0)=0 f(π)=-1 f(x)值域 [-1,1]
供参考答案2:
o--1 已知函数f(x)=2sin(2x/3 + π/6) -1,其中x∈[0,π],求f(x)的值域和单调增区间(图1)