Question

设f(x)是定义在负无穷到正无穷(不等于0)上的函数，且f(-x)+f(x)=0,当x>0,f(x)=x/1-2^x

解不等式:f(x)>-x/3

Answer 1

令x<0,则-x>0,∴f(x)=-f(-x)=-(-x)/[1-2^(-x)]=(x*2^x)/(2^x-1)
x/(1-2^x) (x>0)
∴f(x)=
(x*2^x)/(2^x-1) (x<0)

当x>0时，2^x>1 => 1-2^x<0，则f(x)>-x/3
=>x/(1-2^x)>-x/3
=>2^x-1>3
=>x>2
当x<0时，2^x<1 => 2^x-1<0，则f(x)>-x/3
=>(x*2^x)/(2^x-1)>-x/3
=>4*2^x>1
=>x>-2 ∵x<0 ∴-2<x<0
综上所述，原不等式的解集为(-2,0)∪(2,+∞)。