lawnet
幼苗
共回答了19个问题采纳率:89.5% 举报
f(x)=2√3sinxcosx-2sin^2x
=√3sin2x-(1-cos2x)
=√3sin2x+cos2x-1
=2(√3/2*sin2x+1/2*cos2x)-1
=2sin(2x+π/6)-1
(1)函数的最小正周期T=2π/2=π
(2)
∵x∈【-π/6,π/4】
∴2x∈[-π/3,π/2]
∴2x+π/6∈[-π/6,2π/3]
∴2x+π/6=-π/6时,
f(x)取得最小值-2
2x+π/6=π/2时,
f(x)取得最大值1
1年前
1