令t=2^x,则x≥0时,t≥1
∴f(t)=(t-1/t)/(t+1/t )
=(t^2-1)/(t^2+1)
=1-2/(t^2+1)
t≥1,则t^2+1≥2;0＜1/(t^2+1)≤1/2;0＞-2/(t^2+1)≥-1;1＞1-2/(t^2+1)≥0.
即当x≥0时,f(x)值域为[0,1)
∵f(-x)=(2^-x-2^x)/(2^-x+2^x )=-(2^x-2^-x)/(2^x+2^-x )=-f(x),f(x)是奇函数,则值域关于原点对称.则当x≤0时,f(x)值域为(-1,0]
综上,函数f(x)=(2^x-2^-x)/(2^x+2^-x )值域为(-1,1)