1)f(-x)=-f(x)
--->[a^(-x)-1]/[2^(-x)+1]=[a/2^x-1]/[1/2^x+1)=-(a*2^x-1)/(2^x+1)
--->(a-^x)/(1+2^x)=(-a*2^x+1)/(1+2^x)
--->a-2^x=-a*2^x+1
--->a(2^x+1)-(2^x+1)=0
--->(2^x+1)(a-1)=0
2^x+1>1>0--->a=1.
2)y=(2^x-1)/(2^x+1)
--->y*2^x+y=2^x-1
--->(y-1)*2^x=-1-y
--->2^x=(1+y)/(1-y)
--->x=log[(1+y)/(1-y)] 此处及以下省略底数2.
交换x,y得到反函数f~(x)=log[(1+x)/(1-x)],-1(1+x)/k.
因为-10 & k>0(!)
--->k>1-x
--->x>1-k.