Question

f(x)=x(0<=x<2)展开傅里叶级数（周期任选）

Answer 1

已知函数f(x)=sin(2wX兀/6)十1/2(w>0)周期兀1求w值2求函数f(x)区间[02兀/3]取值范围(1)解析：函数f(x)=sin(2wX兀/6)十1/2(w>0)周期兀所2w=2π/π=2==>w=1(2)解析：f(x)=sin(2X-π/6)+1/2单调增区间：2kπ-π/2kπ-π/6<=X<=kπ+π/3区间[02兀/3]f(0)=sin(-π/6)+1/2=0f(2π/3)=sin(4π/3-π/6)+1/2=0f(π/3)=sin(2π/3-π/6)+1/2=3/2所函数f(x)区间[02兀/3]取值范围[0,3/2]