Question

一次函数f(x)满足f[f(x)]=2x+1.求f(x)

Answer 1

解：设f(x)=kx+b
则f[f(x)]=k(kx+b)+b=k^2x+(kb+b)=1+2x
对应项系数相等
所以k^2=2,kb+b=1
k^2=2,k=±√2
若k=√2,b=1/(k+1)=1/(√2+1)=(√2-1)/(√2-1)(√2+1)=(√2-1)/(2-1)=√2-1
若k=-√2,b=1/(k+1)=1/(-√2+1)=(-√2-1)/(-√2-1)(-√2+1)=(-√2-1)/(2-1)=-√2-1
所以f(x)=√2x+√2-1或f(x)=-√2x-√2-1

Answer 2

因为f(x)为一次函数,故设f(x) = ax+b,
则f[f(x)] = af(x) +b =a(ax+b)+b a^2 x+ab +b
因为f[f(x)]=2x+1
所以:a^2 = 2
ab+b =1
解得:a1=根号2, a2= - 根号2
b1 = 1/(根号2+1)
b2 = 1/(-根号2+1)

所以:f(x) = 根号2x+1/(根号2+1),或f(x) = -根号2x+1/(-根号2+1),

Answer 3

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

Answer 4

根号2X 1/(根号2 1)