Question

拜托 帮忙解答一下哈

Answer 1

• 解：(1)
  令a=b=x/2
  f(x)=f(x/2)*f(x/2)=[f(x/2)]^2
  非零函数f(x)
  所以f(x)>0
  (2)
  令a=x1-x2 b=x2 且x1<x2
  f(x1-x2+x2)=f(x1-x2)*f(x2)
  f(x1)=f(x1-x2)*f(x2)
  [x1-x2<0 f(x1-x2)>1
  且f(x1)>0 f(x2)>0]
  f(x1)/f(x2)>1
  f(x1)>f(x2)
  即得当x1<x2  f(x1)>f(x2)
  所以f(x)为减函数
  (3)
  f(4)=f(2)*f(2) f(2)>0
  所以f(2)=1/4
  f(x-3)*f(5)<=1/4
  f(x-3+5)<=f(2)
  [f(x)为减函数]
  x-3+5>=2
  x>=0