f(x)=√3*2sinxcosx/2+2cosxcosx/2+m=)=√3sin2x/2+(cos2x+1)/2+m=√3sin2x/2+cos2x/2+m+1/2=sin(2x+π/6)+m+1/2,最小正周期t=2π/2=π,
由正弦图像可以看出,当2x+π/6∈[-π/2+2nπ,π/2+2nπ]是属于单调递增区间.所以当x∈[-π/3+nπ,π/6+nπ]为函数f(x)的单调递增区间.
由单调递增区间可以看出,x∈[-π/6,π/6]是单调递增区间,[π/6,π/3]为单调递减区间.x∈[-π/6,π/6]时最小值为f(-π/6)=m,x∈[π/6,π/3]时最小值为f(π/3）=m+1,显然,在x∈[-π/6,π/3]时,最小值为m=2,最大值f(x)=f(π/6)=3.5