Question

设f(x)={1+x,x<0 1,x≥0,则f[f(x)]=

设f(x)={1+x,x<0 1,x≥0,则f[f(x)]=

Answer 1

解g(1/4)=cosπ(1/4)=-√2/2 g(5/6)=g(5/6-1)=g(-1/6)=cos(-1/6)π=-√3/2 f(1/3)=f(1/3-1)+1 =f(-2/3)+1 =sin(-2/3)π+1 =-sin2π/3+1 =-√3/2+1 f(3/4)=f(3/4-1)+1 =f(-1/4)+1 =sin(-1/4)π+1 =-sinπ/4+1 =-√2/2+1 故原式 =-√2/2-√3/2+1-√3/2-√2/2+1 =。