Find the domain of each of the following functions:

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 $( -\infty ,-1) \cup ( 1,\infty )$.