已知f(x)是一次函数，且f(f(x))=4x-1,求f(x)
热心网友
设f(x)=ax+bf(ax+b)=4x-1a(ax+b)+b=4x-1a^2x+ab+b=4x-1a^2=4ab+b=-1得a=-2 b=1 或 a=2 b=-1/3所以f(x)=-2x+1或f(x)=2x-1/3
热心网友
设f(x)=ax+bf(f(x))=a(f(x))+b=a(ax+b)+b=a^2*x+ab+b=a^2*x+(a+1)b又 f(f(x))=4x-1则 a^2=4 故a=2或a=-2(a+1)b=-1当a=2时,b=-1/3当a=-2时,b=1故 f(x)=2x-1/3或f(x)=-2x+1