Question #52e03

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Answer 1 · 2018-02-07T16:12:21

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

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]}