函数f(x)=sin(wx+fai)在它的某一个周期内单调减区间[5π/12,11π5/12] |fai|
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函数f(x)=sin(wx+fai)在它的某一个周期内单调减区间[5π/12,11π5/12] |fai|
函数f(x)=sin(wx+fai)在它的某一个周期内单调减区间[5π/12,11π5/12] |fai|
函数f(x)=sin(wx+fai)在它的某一个周期内单调减区间[5π/12,11π5/12] |fai|
函数f(x)=sin(wx+fai)在它的某一个周期内单调减区间[5π/12,11π5/12] |fai|T=π==>w=2π/π=2
∴f(x)=sin(2x+φ)
f(5π/12)=sin(2*5π/12+φ)=1==>5π/6+φ=π/2==>φ=-π/3
∴f(x)=sin(2x-π/3)
(2)解析：将y=f(x)的图像先向右平移π/6个单位
g(x)=sin(2(x-π/6)-π/3)=sin(2x-2π/3)
再将图像上的所有点的横坐标变为原来的1/2倍,纵坐标不变
T=π==>T=π/2==>w=4
∴g(x)=sin(4x-2π/3)
2kπ-π/2
函数f(x)=sin(ωx+θ)在它的某一个周期内单调减区间[5π/12,11π5/12] |θ|<π/2 ω>0｡
(1)求f(x)解析式
11π5/12-5π/12=π/2
周期=π
ω=2
π/2≤2x+θ≤3π/2
π/4-θ/2≤x≤3π/4-θ/2
π/4-θ/2=5π/12
θ=-π/3
f(x)=...
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函数f(x)=sin(ωx+θ)在它的某一个周期内单调减区间[5π/12,11π5/12] |θ|<π/2 ω>0｡
(1)求f(x)解析式
11π5/12-5π/12=π/2
周期=π
ω=2
π/2≤2x+θ≤3π/2
π/4-θ/2≤x≤3π/4-θ/2
π/4-θ/2=5π/12
θ=-π/3
f(x)=sin(2x-π/3)
(2)将y=f(x)的图像先向右平移π/6个单位,再将图像上的所有点的横坐标变为原来的1/2倍,纵坐标不变 ,所得到的图像对应的函数记为g(x),求函数g(x)在[π/8,3π/8]上的最大值和最小值
g(x)=sin[2(2x-π/3-π/6)]
g(x)=sin(4x-π)
周期π/2
3π/8-π/8=π/4
4x-π=π/2+2kπ g(x)最大
x=3π/8+kπ/2 k取0
g(3π/8)=1
4x-π=-π/2+2kπ g(x)最大
x=π/8+kπ/2 k取0
g(π/8)=-1
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