设 f(x)=ax+b
f [ f(x)]=f(ax+b)=a(ax+b)+b=a^2x+ab+b
则 a^2=2,a=√2,a=-√2
ab+b=1
(1) a=√2
(1+√2)b=1,b=√2-1
(2) a=-√2
(1-√2)b=1,b=-√2-1
设 f(x)=ax+b
f [ f(x)]=f(ax+b)=a(ax+b)+b=a^2x+ab+b
则 a^2=2,a=√2,a=-√2
ab+b=1
(1) a=√2
(1+√2)b=1,b=√2-1
(2) a=-√2
(1-√2)b=1,b=-√2-1