Question #52e03

1 Answer
Feb 7, 2018

The domain is #x in RR#. The range is #y in (0, 100)#

Explanation:

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]}