Question

已知函数fx=Asin(wx+φ) (x∈R,A>0,w>0,0<φ<π/2)的部分图像如图所示。

1、求fx解析式。2、设gx=[f(x-π/12)]^2,求函数gx在x∈[-π/6,3/π]上的最大值，并确定此时x的最大值。
求详细解题过程。

Answer 1

A=2
T=4*[π/6-(-π/6)]=4π/3w=2π/(4π/3)=1.5f(x)=2sin(1.5x+φ)2sin(1.5*π/6+φ)=2π/6+φ=π/2φ=π/3
f(x)=2sin(1.5x+π/3)

g(x)=[f(x-π/12)]^2
=[2sin(1.5(x-π/12)+π/3)]^2
=[2sin(1.5x+π/12)]^2
=2-2cos(3x+π/6)
-π/6≤x≤π/3
-π/3≤3x+π/6≤7π/6
3x+π/6=π
x=5π/18
ymax=4

追问

这是解答题，求正规解题过程。

追答

解：1）由函数图像可知：A=2
T=4*[π/6-(-π/6)]=4π/3
∴w=2π/(4π/3)=1.5
∴f(x)=2sin(1.5x+φ)
把(π/6,2)代入函数解析式得：
2sin(1.5*π/6+φ)=2
∵0<φ<π/2
∴π/6+φ=π/2
∴φ=π/3
因此，函数解析式为：f(x)=2sin(1.5x+π/3)

2）g(x)=[f(x-π/12)]^2
=[2sin(1.5(x-π/12)+π/3)]^2
=[2sin(1.5x+π/12)]^2
=2-2cos(3x+π/6)
∵-π/6≤x≤π/3
∴-π/3≤3x+π/6≤7π/6
当cos(3x+π/6)=-1时，g(x)值最大
3x+π/6=π
x=5π/18
∴ymax=4
因此，当x=5π/18时，g(x)有最大值，为4

Answer 2

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