y=f(x)=1/[1+(2^-x)²]1+(2^-x)²=1/y(2^-x)²=(1-y)/y2^-x=√[(1-y)/y]x=-log2 √[(1-y)/y]=(1/2)[log2 y-log2 (1-y)]所以f^-1(x)=(1/2)[log2 x-log2 (1-x)]定义域满足x>01-x>0解得0