问题 填空题 设函数f1(x)=x 12,f2(x)=x-1,f3(x)=x2,则f3[f2(f1(2012))]=______. 答案 f3[f2(f1(x))]=f3[f2(x12))]=f3[x-12]=x-1=1x所以f3[f2(f1(2012))]=12012