The function is
#f(x)=100/(1+2^-x)=100/(1+1/2^x)#
#AA x in RR#, #f(x)>0#
Therefore,
The domain is #x in RR#
To find the range, proceed as follows :
Let #y=100/(1+2^-x)#
#y(1+2^-x)=100#
#y+y2^-x=100#
#y2^-x=100-y#
#2^-x=(100-y)/y#
#2^x=y/(100-y)#
Taking logarithms
#xln2=ln(y/(100-y))#
#x=1/ln2ln(y/(100-y))#
In order for this equation to have solutions,
#y/(100-y)>0#
Let #f(y)=y/(100-y)#
Build a sign chart
#color(white)(aaaa)##y##color(white)(aaaaaaa)##-oo##color(white)(aaaa)##0##color(white)(aaaaaa)##100##color(white)(aaaa)##+oo#
#color(white)(aaaa)##y##color(white)(aaaaaaaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaaa)##+##color(white)(aaaa)###
#color(white)(aaaa)##100-y##color(white)(aaaaaa)##+##color(white)(aaaa)##+##color(white)(aaa)##||##color(white)(a)##-#
#color(white)(aaaa)##f(y)##color(white)(aaaaaaaaa)##-##color(white)(aaaa)##+##color(white)(aaa)##||##color(white)(a)##-#
Therefore,
#f(y)>0# when #y in (0, 100)#
The range is #y in (0, 100)#
graph{100/(1+2^-x) [-205.5, 222.2, -83.2, 130.5]}
