(1)
函数
f（x）=2√3sinxcosx+1-2sin²x
=√3sin2x+cos2x
=2(√3/2sin2x+1/2cos2x)
=2sin(2x+π/6)
f(x)最小正周期T=2π/2=π
由2kπ-π/2≤2x+π/6≤2kπ+π/2,k∈Z
得kπ-π/3≤x≤kπ+π/6,k∈Z
f(x)递增区间为[kπ-π/3,kπ+π/6],k∈Z
递减区间为[kπ+π/6,kπ+2π/3],k∈Z
(2)
函数y=f（x)的图像向左平移π／6个单位,
得到g(x)=f(x+π/6)=2sin[2(x+π/6)+π/6]=2sin(2x+π/2)=2cos2x
x∈[0,π/8],2x∈[0,π/4]
2x=π/4时,cos2x取得最小值√2/2
∴g(x)的最小值为√2