Question

已知f(x)是定义在(0,∞)上的增函数,且f(x/y)=f(x)-f(y)

若f(6)=1,解不等式f(x+3)+f(1/x)≤2.急在线等！

Answer 1

令x=y
则f(x/x)=f(x)-f(x)
f(1)=0
f(1/x)=f(1)-f(x)=-f(x)
f(x+3)+f(1/x)=f(x+3)-f(x)=f((x+3)/x)<=2
令x=36,y=6
f(36/6)=f(36)-f(6)
f(6)=f(36)-f(6)
f(36)=2f(6)=2
因为f(x)增，f((x+3)/x)<=2=f(36)
表示(x+3)/x<=36
x+3<=36x (x>0)
35x>=3
x>=3/35