函数f(x)=ax-1满足f[f(x)]﹦x,则常数a等于

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函数f(x)=ax-1满足f[f(x)]﹦x,则常数a等于
函数f(x)=ax-1满足f[f(x)]﹦x,则常数a等于

函数f(x)=ax-1满足f[f(x)]﹦x,则常数a等于
∵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.