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