定义在R上的函数f(x)>0,对任意x,y∈R都有f(x+y)=f(x) f(y)成立,且当x>0时,f(x)

1个回答

  • (1)令x=0,y=1,则f(0+1)=f(0)f(1),

    ∵当x>0时,f(x)>1,∴f(1)>1,∴f(0)=1;

    (2)证明:设x 1<x 2,则x 2-x 1>0

    ∵当x>0时,f(x)>1,∴f(x 2-x 1)>1

    ∴f(x 2)=f(x 2-x 1+x 1)=f(x 2-x 1)f(x 1)>f(x 1

    ∴f(x)在R上是增函数;

    (3)∵f(x)在R上是增函数,f(k•3 x) f(3 x-9 x-2)=f(k 3 x+3 x-9 x-2)<f(0),

    ∴3 2x-(1+k)•3 x+2>0对任意x∈R成立.

    ∴1+k<3 x+

    2

    3 x

    ∵3 x>0,∴3 x+

    2

    3 x ≥ 2

    2

    ∴k< 2

    2 -1 .