函数对于n∈N*,通过定义fn+1(x)=f1[fn(x)],求出f2(x)=f1[f1(x)].f3(x),f4(x),f5(x),f6(x),f7(x).所以从f1(x)到f6(x),每6个一循环.由此能求出结果.
【解析】
∵函数对于n∈N*,定义fn+1(x)=f1[fn(x)],
∴f2(x)=f1[f1(x)]=f1()==.
f3(x)=f1[f2(x)]=f1()==,
f4(x)=f1[f3(x)]=f1()==,
f5(x)=f1[f4(x)]=f1()==,
f6(x)=f1[f5(x)]=f1()==x,
f7(x)=f1[f6(x)]=f1(x)==f1(x).
所以从f1(x)到f6(x),每6个一循环.
∵60=10×6,
∴f60(x)=f6(x)=x,
故答案为:x.