Find the domain of each of the following functions:

\(f( x) =x^{2}+5x\)

\(g( x) =\dfrac{ 3x}{x^{2}-4}\)

\(h( t) = \sqrt{4-3t}\)

\(F( u) =\dfrac{5u}{\sqrt{u^{2}-1}}\)

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

Solution (a) Since \(f( x) =x^{2}+5x\) is defined for any real number \(x\), the domain of \(f\) is the set of all real numbers.

(b) Since division by zero is not defined, \(x^{2}-4\) cannot be \( 0,\) that is, \(x\neq -2\) and \(x\neq 2.\) The function \(g( x) = \dfrac{3x}{x^{2}-4}\) is defined for any real number except \(x=-2\) and \(x=2.\) So, the domain of \(g\) is the set of real numbers \(\{ x|x\neq -2, x\neq 2\} \).

(c) Since the square root of a negative number is not a real number, the value of \(4-3t\) must be nonnegative. The solution of the inequality \(4-3t\geq 0\) is \(t\leq \dfrac{4}{3}\), so the domain of \(h\) is the set of real numbers \(\left\{ t |t\leq \dfrac{4}{3}\right\}\) or the interval \(\left( -\infty ,\dfrac{4}{3}\right] \).

(d) Since the square root is in the denominator, the value of \(u^{2}-1\) must be not only nonnegative, it also cannot equal zero. That is, \(u^{2}-1>0.\) The solution of the inequality \(u^{2}-1>0\) is the set of real numbers \(\{ u|u<-1\} \cup \{ u|u>1\} \) or the set \(( -\infty ,-1) \cup ( 1,\infty )\).