已知函数f(x)=2sin(2x-π/3)求函数的值域,周期,单调区间rt
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/28 08:23:56
已知函数f(x)=2sin(2x-π/3)求函数的值域,周期,单调区间rt
已知函数f(x)=2sin(2x-π/3)求函数的值域,周期,单调区间
rt
已知函数f(x)=2sin(2x-π/3)求函数的值域,周期,单调区间rt
(1)sin(2x-π/3)∈[-1,1] ∴2sin(2x-π/3)∈[-2,2]
即函数值域是[-2,2]
(2)周期T=2π/2 =π
(3)由-π/2 +2kπ《2x- π/3《π/2 +2kπ
得-π/12 +kπ《x《5π/12 +kπ
∴函数的单调增区间是[-π/12 +kπ,5π/12 +kπ] k∈Z
由π/2 +2kπ《2x- π/3《3π/2 +2kπ
得由5π/12+ kπ《x《11π/12 +kπ
∴函数的单调减区间是[5π/12 +kπ,11π/12 +kπ] k∈Z
附加:对称轴 和 对称中心也帮你解答了 这4点是一个体系!
(4)由2x- π/3=π/2 +kπ 得函数的对称轴方程 x=5π/12 +kπ/2
由2x- π/3=kπ 得x=π/6+kπ/2 ∴对称中心坐标是(π/6+kπ/2 ,0) k∈Z
f(x)=2sin(2x-π/3)
∵-1≤sin(2x-π/3)≤1
∴-2≤2sin(2x-π/3)≤2
值域为[-2,2]
周期T=2π/w=2π/2=π
单调区间:
2kπ-π/2≤2x-π/3≤2kπ+π/2
2kπ-π/6≤2x≤2kπ+5π/6
kπ-π/12≤x≤kπ+5π/12 k∈Z
即递增区间为[...
全部展开
f(x)=2sin(2x-π/3)
∵-1≤sin(2x-π/3)≤1
∴-2≤2sin(2x-π/3)≤2
值域为[-2,2]
周期T=2π/w=2π/2=π
单调区间:
2kπ-π/2≤2x-π/3≤2kπ+π/2
2kπ-π/6≤2x≤2kπ+5π/6
kπ-π/12≤x≤kπ+5π/12 k∈Z
即递增区间为[kπ-π/12,kπ+5π/12] (k∈Z)
2kπ+π/2≤2x-π/3≤2kπ+3π/2
2kπ+5π/6≤2x≤2kπ+11π/6
kπ+5π/12≤x≤kπ+11π/12 k∈Z
即递减区间为[kπ+5π/12,kπ+11π/12 ] (k∈Z)
收起
值域±2。
周期π。
[-π/12±π,7π/12π±π]递增,
[7π/12±π,13π/12±π]递减。