jonesvsjack
幼苗
共回答了13个问题采纳率:84.6% 举报
f(x)=-2sinxcosx+2√3cos^2x-√3
=-sin2x+√3(2cos^2x-1)
=-sin2x+√3cos2x
=-2sin(2x-π/3)
最小正周期:2π/2=π
2x-π/3=-π/2+2kπ(k∈Z)
x=-π/12+kπ
2x-π/3=π/2+2kπ(k∈Z)
x=5π/12+kπ
2x-π/3=3π/2+2kπ(k∈Z)
x=11π/12
x∈[5π/12+kπ,11π/12+kπ](k∈Z),f(x)为递增
x∈[-π/12+kπ,5π/12+kπ](k∈Z),f(x)为递减
f(x)max=f(0)=√3
f(x)min=f(5π/12)=-2
f(x)>1
-2sin(2x-π/3)>1
sin(2x-π/3)<-1/2
2x-π/3=-π/6
2x=π/6
x=π/12
x∈(π/12+kπ,3π/4+kπ)(k∈Z)
1年前
7