题目

已知定义在(-∞,0)U(0,+∞)上的偶函数f(x)满足对任意正数x,y满足f(x×y)=f(x)×f(y),且x>1时,01.求f（1）
2.求证f(x)在(0,+∞)上是减函数
3.若f(4)=1/2,解不等式f(x)-4≥0
4.求证恒有f(x)>0

提问时间：2021-12-10

答案

(1)
put y=1
f(x)=f(x)f(1)
=> f(1) = 1
(2)
for y>x and x,y∈(0,+∞)
then y = kx where k > 1
f(y) = f(kx)
= f(k)f(x)
< f(x)
f是减函数
(3)
for |x| >1 then f(x) 1 is not solution of f(x)-4≥0
Consider |x| < 1
f(4) = 1/2
put x = 4 ,y= 1/4
f(1) = f(4)f(1/4)
1 = (1/2) f(1/4)
f(1/4) = 2
put x=y= 1/4
f(1/16) = f(1/4)f(1/4)
f(1/16) = 4
f(x)-4≥0
0 < x ≤ 1/16 or -1/16 ≤ x < 0
(4)
x ∈(0,+∞)
put y=x
f(x^2) = f(x) f(x) > 0
f(x) > 0 for x ∈(0,+∞)
for x∈(-∞,0)
x -x >0
f(x) = f(-x) > 0
恒有f(x)>0