# Question 52e03

Feb 7, 2018

The domain is $x \in \mathbb{R}$. The range is $y \in \left(0 , 100\right)$

#### Explanation:

The function is

$f \left(x\right) = \frac{100}{1 + {2}^{-} x} = \frac{100}{1 + \frac{1}{2} ^ x}$

$\forall x \in \mathbb{R}$, $f \left(x\right) > 0$

Therefore,

The domain is $x \in \mathbb{R}$

To find the range, proceed as follows :

Let $y = \frac{100}{1 + {2}^{-} x}$

$y \left(1 + {2}^{-} x\right) = 100$

$y + y {2}^{-} x = 100$

$y {2}^{-} x = 100 - y$

${2}^{-} x = \frac{100 - y}{y}$

${2}^{x} = \frac{y}{100 - y}$

Taking logarithms

$x \ln 2 = \ln \left(\frac{y}{100 - y}\right)$

$x = \frac{1}{\ln} 2 \ln \left(\frac{y}{100 - y}\right)$

In order for this equation to have solutions,

$\frac{y}{100 - y} > 0$

Let $f \left(y\right) = \frac{y}{100 - y}$

Build a sign chart

$\textcolor{w h i t e}{a a a a}$$y$$\textcolor{w h i t e}{a a a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a a a}$$100$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$y$$\textcolor{w h i t e}{a a a a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$#

$\textcolor{w h i t e}{a a a a}$$100 - y$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a}$$-$

$\textcolor{w h i t e}{a a a a}$$f \left(y\right)$$\textcolor{w h i t e}{a a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a}$$-$

Therefore,

$f \left(y\right) > 0$ when $y \in \left(0 , 100\right)$

The range is $y \in \left(0 , 100\right)$

graph{100/(1+2^-x) [-205.5, 222.2, -83.2, 130.5]}