(1).x∈(0,+∞)时,f(x)=2^x/(4^x+1)
设0＜x1＜x2,则
f(x1)-f(x2)
=2^(x1)/(4^(x1)+1)-2^(x2)/(4^(x2)+1)
=(2^(x1+x2)-1)(2^(x2)-2^(x1))/(4^(x1)+1)(4^(x2)+1)
因为x1+x2＞0,所以2^(x1+x2)-1＞0,
因为x1＜x2,所以2^(x2)-2^(x1)＞0,
所以,f(x1)-f(x2)＞0,
即f(x1)＞f(x2)
所以,f(x)在(0,+∞)上递减
(2).由题知f(x)为奇函数,
1.x∈(0,+∞)时,
f(x)=2^x/(4^x+1)
2.x∈(-∞,0)时,
f(x)=-f(-x)
=-(2^(-x)/(4^(-x)+1))
=-(2^x/(4^x+1))
3.x∈{0}时,f(0)=-f(-0)所以f(0)=0
综上所述,
f(x)=2^x/(4^x+1),x∈(0,+∞)
f(x)=0,x∈{0}
f(x)=-(2^x/(4^x+1)),x∈(-∞,0)