设函数f(x)=2sin^2(wx+π/4)+2(cos^2wx)(w>1)的图像上两个相邻的最低点之间的距离为2π/3
问题描述：
设函数f(x)=2sin^2(wx+π/4)+2(cos^2wx)(w>1)的图像上两个相邻的最低点之间的距离为2π/3
1.求函数f（x）的最大值 及此时的x值
2.若函数y=g（x）的图像为y=f（x）图像向右平移π/8个单位长度,再沿y轴对称后得到,求y=g（x）的单调递减区间
答
f(x)=2sin^2(wx+π/4)+2(cos^2wx)=1-cos(2wx+π/2)+cos(2wx)+1=sin(2wx)+cos(2wx)+2
=√2 sin(2wx+π/4)+2
两个相邻的最低点之间的距离=T=2π/2w=2π/3,所以w=3/2
f(x)=√2 sin(3x+π/4)+2
（1）f(x)的最大值2+√2
此时sin(3x+π/4)=1,3x+π/4=π/2+2kπ,x=π/12+2/3 kπ, k=1,2,3...仔细点y=f（x）=√2 sin(3x+π/4)+2右平移π/8个单位长度√2 sin(3(x-π/8)+π/4)+2=√2 sin(3x-π/8)+2沿y轴对称后得到g（x）=√2 sin(π/8-3x)+2 = - √2 sin(3x-π/8)+2递减区间-π/2+2kπ