f(x)=4sinxcos(x+π/3)+根号3

f(x)=4sinxcos（x+π/3)+√3
=4sinx(1/2cosx-√3/2sinx)+√3
=2sinxcosx-2√3sin²x+√3
=sin2x+√3（1-2sin²x）
=sin2x+√3con2x
=2sin(2x+π/3)
f(x)最小正周期T=2π/2=π
-π/2+2Kπ≤2x+π/3≤π/2+2Kπ
2、
因为x∈[-π/4,π/6] 则2x+π/3∈[-π/6,2π/3]
由图像知在2x+π/3=-π/6时取得最小值=-1
2x+π/3=π/2取得最大值=2

f(x)=4sinxcos(x+π/3)+根号3=2[sin(x+x+π/3)+sin(x-x-π/3)]+根号3
=2[sin(2x+π/3)-sinπ/3]+根号3=2sin(2x+π/3)
所以f（x）的最小正周期=2π/2=π
因为-π/4<=x<=π/6,所以-π/6<=2x+π/3<=2π/3
所以当2x+π/3=π/2，即x=π/12时有最大值2
当2x+π/3=-π/6，即x=-π/4时有最小值-1