函数f(x)=根号2 sin(2x+π/4)令g(x)=f(x+π/8)-a,若g(x)在x∈[-π/6,π/3]时有两个零点,求a的取值范围
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![函数f(x)=根号2 sin(2x+π/4)令g(x)=f(x+π/8)-a,若g(x)在x∈[-π/6,π/3]时有两个零点,求a的取值范围](/uploads/image/z/3711956-68-6.jpg?t=%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3D%E6%A0%B9%E5%8F%B72+sin%EF%BC%882x%2B%CF%80%2F4%EF%BC%89%E4%BB%A4g%EF%BC%88x%EF%BC%89%3Df%EF%BC%88x%2B%CF%80%2F8%EF%BC%89-a%2C%E8%8B%A5g%EF%BC%88x%EF%BC%89%E5%9C%A8x%E2%88%88%5B-%CF%80%2F6%2C%CF%80%2F3%5D%E6%97%B6%E6%9C%89%E4%B8%A4%E4%B8%AA%E9%9B%B6%E7%82%B9%2C%E6%B1%82a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
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函数f(x)=根号2 sin(2x+π/4)令g(x)=f(x+π/8)-a,若g(x)在x∈[-π/6,π/3]时有两个零点,求a的取值范围
函数f(x)=根号2 sin(2x+π/4)令g(x)=f(x+π/8)-a,若g(x)在x∈[-π/6,π/3]时有两个零点,
求a的取值范围
函数f(x)=根号2 sin(2x+π/4)令g(x)=f(x+π/8)-a,若g(x)在x∈[-π/6,π/3]时有两个零点,求a的取值范围
函数f(x)=根号2 sin(2x+π/4)
g(x)=f(x+π/8)-a
=根号2 sin(2(x+π/8)+π/4)-a
=根号2sin(2x+π/2)-a
=根号2cos(2x)-a
x∈[-π/6,π/3] 2x∈[-π/3,2π/3] 根号2cos(2x) ∈[-√2/2,√2]
由图像可知
g(x)在x∈[-π/6,π/3]时有两个零点
a∈[√2/2,√2)