Question

f(f(x))=2x-1,求f(x)

Answer 1

设f(x)=kx+b
f(f(x))=2x-1
f(f(x))=f(kx+b)
=k(kx+b)+b
=k^2x+kb+b
对比有:
k^2=2
kb+b=-1
解得k1=√2,b1=1-√2
k2=-√2,b2=√2+1
f(x)=√2x+1-√2或者f(x)=-√2x+√2+1

Answer 2

设f（X）=ax+b 则 f（f（x））=f（ax+b)=a（ax+b）+b=a平方x +ab +b=2x-1
所以 a平方=2 ab+b=-1 解得a=正负根号2 b=根号2-1或根号2+1
f（x）=（根号2）X+根号2-1或 f（x）=（－根号2）X+1

Answer 3

设f(x)=ax+b =>f(f(x))=af(x)+b=a[ax+b]+b=a^2x+ab+b=2x-1
=> a^2=2 ;ab+b=-1=b(a+1)
=>a=√2时 b=-1/(1+√2)=1-√2 => f(x)=√2 x +(1-√2)
a=-√2时 b=-1/(1-√2)=1+√2 => f(x)=-√2 x +(1+√2)

Answer 4

设f（x）=ax+b
则 f(f(x))=a（ax+b）+b=2x-1
所以 a^2=2 ab+b=-1 a=根号2 b=-1/ 根号2+1 f（x）=根号2x-1/根号2+1

Answer 5

f(x)=sqar(2)+1-sqar(2)