f[x]的反函数为log2^1+x/1-x,求f[x]的解析式。

解答：设y＝f(x),由条件得
x=log2[(1+y)/(1-y)](2是底数,x∈R)
将对数式写成指数式，得
(1+y)/(1-y)＝2^x,(这是2的x次方)
去分母解关于y的一次方程，得
y＝(2^x-1)/(2^x+1)(x∈R).
即f(x)＝(2^x-1)/(2^x+1)(x∈R).

f(x)-1=log2^((1+x)/(1-x));
(1+y)/(1-y)>0,1-y^2>0,-1<y<1;
long2^((1+y)/(1-y))<=0,
x=long2^((1+y)/(1-y));
2^x=(1+y)/(1-y)=(1-y+2y)/1-y;
2^x-1=2y/1-y;
y*2^x-2^x+1+2y=0;
(2^x+2)*y=2^x-1;
y=(2^x-1)/(x^2+2);(x<=0)