Find the domain of each function:

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 $\{ x | x >-3\}$.

(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 $\{ x|x \neq 0\} .$