f（x）=（a*2^x-1)/(1+2^x)
f(-x)=(a*2^(-x)-1)/(1+2^(-x))=(a-2^x)/(2^x+1)
f(x)是R上的奇函數(shù),f(x)=-f(-x)
（a*2^x-1)/(1+2^x)=-(a-2^x)/(2^x+1)
a*2^x-1=-a+2^x
(a-1)*(2^x+1)=0
a=1
f(x)=(2^x-1)/(2^x+1)
2)
f(x)=(2^x-1)/(2^x+1)=1-2/(2^x+1)<1
2^x+1>1
-2/(2^x+1)>-2
f(x)=1-2/(2^x+1)>1-2=-1
-1函數(shù)f(x)的值域(-1,1)
3)
設(shè)x1f(x1)-f(x2)=[1-2/(2^x1+1)]-[1-2/(2^x2+1)]
=2/(2^x2+1)-2/(2^x1+1)
=2(2^x1-2^x2)/(2^x2+1)(2^x1+1)
<0
f(x)在R上的單調(diào)增
0即f(0)即0<2x-1<4
得：1/2