函数f(x)=ax-1满足f[f(x)]﹦x,则常数a等于
问题描述：
函数f(x)=ax-1满足f[f(x)]﹦x,则常数a等于
答
∵f(x)=ax-1
∴f[f(x)]=f(ax-1)
=a(ax-1)-1
=a²x-a-1
∵f[f(x)]=x
∴a²x-a-1=x ==>a²=1,-a-1=0
==>a=-1
故a=-1.非常感谢！别客气！