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,
由正弦圖像可以看出,當(dāng)2x+π/6∈[-π/2+2nπ,π/2+2nπ]是屬于單調(diào)遞增區(qū)間.所以當(dāng)x∈[-π/3+nπ,π/6+nπ]為函數(shù)f(x)的單調(diào)遞增區(qū)間.
由單調(diào)遞增區(qū)間可以看出,x∈[-π/6,π/6]是單調(diào)遞增區(qū)間,[π/6,π/3]為單調(diào)遞減區(qū)間.x∈[-π/6,π/6]時(shí)最小值為f(-π/6)=m,x∈[π/6,π/3]時(shí)最小值為f(π/3）=m+1,顯然,在x∈[-π/6,π/3]時(shí),最小值為m=2,最大值f(x)=f(π/6)=3.5