是嵌套函数式,
设f(x)=ax+b,则f[f(x)]=a[f(x)]+b=a(ax+b)+b=a^2x+ab+b
又f[f(x)]=4x-1
==>a^2=4,ab+b=-1解a,b
==>f(x)=-2x+1或f(x)=2x-1/3