求y=3tan(π6-x4)的周期及单调区间.-数学

题目简介

求y=3tan(π6-x4)的周期及单调区间.-数学

题目详情

求y=3tan(
π
6
-
x
4
)的周期及单调区间.
题型:解答题难度:中档来源:不详

答案

y=3tan(class="stub"π
6
-class="stub"x
4
)=-3tan(class="stub"x
4
-class="stub"π
6
),
∴T=class="stub"π
|ω|
=4π,
∴y=3tan(class="stub"π
6
-class="stub"x
4
)的周期为4π.
由kπ-class="stub"π
2
class="stub"x
4
-class="stub"π
6
<kπ+class="stub"π
2
,得4kπ-class="stub"4π
3
<x<4kπ+class="stub"8π
3
(k∈Z),
y=3tan(class="stub"x
4
-class="stub"π
6
)在(4kπ-class="stub"4π
3
,4kπ+class="stub"8π
3
)(k∈Z)内单调递增.
∴y=3tan(class="stub"π
6
-class="stub"x
4
)在(4kπ-class="stub"4π
3
,4kπ+class="stub"8π
3
)(k∈Z)内单调递减.

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