已知函数f(X)=(√2)sin(2X+π/4),(1)求函数f(X)的最小正周期及单调增区间;
(2)当X∈[-π/4,π/4]时,求函数f(X)的最大值及最小值.
(1)最小正周期T=2π/2=π;单增区间:由-π/2+2kπ≦2x+π/4≦π/2+2kπ,
得-3π/4≦2x≦π/4+2kπ,故单增区间为-3π/8+kπ≦x≦π/8+kπ,k∈Z.
(2).当x∈[π/4,π/4]时,maxf(x)=f(π/8)=(√2)sin(π/4+π/4)=√2;
minf(x)=f(-π/4)=(√2)sin(-π/2+π/4)=(√2)sin(-π/4)=-1.