Example

Find the domain of each function:

$F( x) =\log _{2}( x+3)$

$g( x) =\ln \left( \dfrac{1+x}{1-x}\!\right)$

$h( x) =\log _{1/2}\vert x\vert$

Solving inequalities is discussed in Appendix A.1, $\!\!\!\!\!\!\!\!\!\!\!$ pp. A-5 to A-8.

Solution (a) The argument of a logarithm must be positive. So to find the domain of $F( x) =\log _{2}( x+3)$ , we solve the inequality $x+3>0.$ The domain of $F$ is $\{ x | x >-3\}$ .

Interval	Test Number	Sign of $\dfrac{{1+x}}{{1-x}}$
$( -\infty ,-1)$	$-2$	Negative
$( -1,1)$	$0$	Positive
$( 1,\infty )$	$2$	Negative

(b) Since $\ln \left( \dfrac{1+x}{1-x}\!\right)$ requires $\dfrac{1+x}{ 1-x}>0$ , we find the domain of $g$ by solving the inequality $\dfrac{1+x}{1-x} >0.$ Since $\dfrac{1+x}{1-x}$ is not defined for $x=1$ , and the solution to the equation $\dfrac{1+x}{1-x}=0$ is $x=-1$ , we use $-1$ and $1$ to separate the real number line into three intervals $( -\infty ,-1)$ , $( -1,1)$ , and $( 1,\infty ) .$ Then we choose a test number in each interval, and evaluate the rational expression $\dfrac{1+x}{1-x}$ at these numbers to determine if the expression is positive or negative. For example, we chose the numbers $-2$ , $0$ , and $2$ and found that $\dfrac{1+x}{ 1-x}>0$ on the interval $( -1,1)$ . See the table on the left. So the domain of $g( x) =\ln \left( \dfrac{1+x}{1-x}\!\right)$ is $\{ x|{-}1<x<1\}$ .

(c) $\log _{1/2}\vert x\vert$ requires $\vert x\vert >0$ . So the domain of $h( x) =\log _{1/2}\vert x\vert$ is $\{ x|x \neq 0\} .$