Question

已知函数f(x)=cos(2x-π\3)+sin²x-cos²x

设函数g(x)=[f(x)]²+f(x)，求g(x)的值域。

Answer 1

f(x)=cos(2x-π\3)+sin²x-cos²x
=1/2cos2x+√3/2sin2x-cos2x
=√3/2sin2x-1/2cos2x
=-cos(2x+π\3)
-1<=f(x)=t<=1
g(x)=[f(x)]²+f(x)，
=t^2+t
=(t+1/2)^2-1/4
t=-1/2,g(x)min=-1/4
t=1,g(x)max=2
-1/4<=g(x)<=2

Answer 2

f(x)=cos(2x-π\3)+sin²x-cos²x
=(1/2)cos2x+(√3/2)sin2x-cos2x
=(√3/2)sin2x-(1/2)cos2x
=sin(2x-pi/6)
g(x)=[f(x)]²+f(x)=[sin(2x-pi/6)]^2+sin(2x-pi/6)
=[sin(2x-pi/6)+1/2]^2-1/4
则g(x)的最大值为1-1/4=3/4，最小值为-1/4
其值域为[-1/4,3/4]