作业帮y=sin(π/6-3x),x∈[0,π/4]的单调区间.
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/28 14:06:36
y=sin(π/6-3x),x∈[0,π/4]的单调区间.
y=sin(π/6-3x),x∈[0,π/4]的单调区间.
y=sin(π/6-3x),x∈[0,π/4]的单调区间.
令t=π/6-3x,则t∈[-7π/12,π/6]
y=sin(t),t∈[-7π/12,π/6]
所以
y=sin(t),t∈[-7π/12,-π/2]单调递减t∈[-π/2,π/6]单调递增
y=sin(π/6-3x),单调递减区间[2π/9,π/4]单调递增区间[0,2π/9]
y=sin(π/6-3x)
=cos[π/2-(π/6-3x)]
=cos(3x-π/3)
余弦值 在 -π+2kπ 到 2kπ 单调增
∴3x-π/3 ∈[-π+2kπ,2kπ]单调增 3x ∈[-2π/3+2kπ,π/3+2kπ] x∈[-2π/9+2kπ/3,π/9+2kπ/3]
x∈[0,π/4] 所以 k取0 即 x∈[-2π/9,π/9...
全部展开
y=sin(π/6-3x)
=cos[π/2-(π/6-3x)]
=cos(3x-π/3)
余弦值 在 -π+2kπ 到 2kπ 单调增
∴3x-π/3 ∈[-π+2kπ,2kπ]单调增 3x ∈[-2π/3+2kπ,π/3+2kπ] x∈[-2π/9+2kπ/3,π/9+2kπ/3]
x∈[0,π/4] 所以 k取0 即 x∈[-2π/9,π/9]
即x∈[0,π/9]
在 2kπ 到 π+2kπ 单调减
∴3x-π/3 ∈[2kπ,π+2kπ]单调减3x∈[π/3+2kπ,4π/3+2kπ] x∈[π/9+2kπ/3,4π/9+2kπ/3]
x∈[0,π/4] 所以 k取0 即 x∈[π/9,4π/9]
即x∈[π/9,π/4]
综合 [0,π/9]单调增 [π/9,π/4]单调减
收起