设F(x)=ax+b,
F[F(x)]=F[ax+b]=a(ax+b)+b=a^2x+(ab+b),
已知F(F(X))=2X-1,所以a^2=2,ab+b=-1.
a=±√2.
a=√2时,b=-1/(a+1)=-1/(√2+1)=1-√2,
F(x)=√2x+1-√2;
a=-√2时,b=-1/(a+1)=-1/(-√2+1)=√2+1,
F(x)=-√2x+√2+1.
答补充问题：a^2x+(ab+b)和2X-1是同一个一次函数,所以一次项系数相同,a^2=2,常数项相同,ab+b=-1.