已知函数f(x)=cos(2x-π/3)+sin^2x-cos^2x,设函数g(x)=[f(x)]^2+f(x),求g(x)值域
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![已知函数f(x)=cos(2x-π/3)+sin^2x-cos^2x,设函数g(x)=[f(x)]^2+f(x),求g(x)值域](/uploads/image/z/1260984-48-4.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dcos%282x-%CF%80%2F3%29%2Bsin%5E2x-cos%5E2x%2C%E8%AE%BE%E5%87%BD%E6%95%B0g%28x%29%3D%5Bf%28x%29%5D%5E2%2Bf%28x%29%2C%E6%B1%82g%28x%29%E5%80%BC%E5%9F%9F)
已知函数f(x)=cos(2x-π/3)+sin^2x-cos^2x,设函数g(x)=[f(x)]^2+f(x),求g(x)值域
已知函数f(x)=cos(2x-π/3)+sin^2x-cos^2x,设函数g(x)=[f(x)]^2+f(x),求g(x)值域
已知函数f(x)=cos(2x-π/3)+sin^2x-cos^2x,设函数g(x)=[f(x)]^2+f(x),求g(x)值域
∵f(x)=cos(2x-π/3)+(sinx)^2-(cosx)^2
=cos(2x-π/3)-cos2x=2sin(2x-π/6)sin(π/6)=sin(2x-π/6).
∴g(x)=[sin(2x-π/6)]^2+sin(2x-π/6)=[sin(2x-π/6)+1/2]^2-1/4.
∴当sin(2x-π/6)=-1/2时,g(x)有最小值为-1/4;
当sin(2x-π/6)=1时,g(x)有最大值为(1+1/2)^2-1/4=9/4-1/4=2.
∴g(x)的值域是[-1/4,2].