分母1+IxI≥1,极值决定于sinx
f(x)最小=f(π/3)=1-(√3/2)/(1+π/3)=1-[3√3/(6+2π)]
f(x)最大=f(-π/3)=1+(√3/2)/(1+π/3)=1+[3√3/(6+2π)]
所以值域为[1-3√3/(6+2π),1-3√3/(6+2π)]