函数f(x)=2sin(x-π4),x∈[-π,0]的单调递增区间为______.-数学

题目简介

函数f(x)=2sin(x-π4),x∈[-π,0]的单调递增区间为______.-数学

题目详情

函数f(x)=2sin(x-
π
4
)
,x∈[-π,0]的单调递增区间为______.
题型:填空题难度:偏易来源:盐城二模

答案

∵x∈[-π,0]
∴x-class="stub"π
4
∈[-class="stub"5π
4
,-class="stub"π
4
],
令z=x-class="stub"π
4
,则z∈[-class="stub"5π
4
,-class="stub"π
4
],
∵正弦函数y=sinz在[-class="stub"π
2
,-class="stub"π
4
]上单调递增,
∴由-class="stub"π
2
≤x-class="stub"π
4
≤-class="stub"π
4
得:
-class="stub"π
4
≤x≤0.
∴函数f(x)=2sin(x-class="stub"π
4
)在x∈[-π,0]的单调递增区间为[-class="stub"π
4
,0].
故答案为[-class="stub"π
4
,0].

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