## College Algebra (10th Edition)

Domain: $\color{blue}{(-\infty, -3) \cup (-3, 1) \cup (1, +\infty)}$
Factor the denominator of the function to obtain: $f(x) = \dfrac{x}{(x+3)(x-1)}$ The value of $x$ can be any real number except the ones that will make the function undefined. The given function is undefined when the value of the denominator is zero. This is because division of zero leads to an undefined expression. Find the values of $x$ that will make the denominator zero by equating each factor of the denominator to zero and then solving each equation. $\begin{array}{ccc} &x+3=0 &\text{ or } &x-1=0 \\&x=-3 &\text{ or } &x=1 \end{array}$ Thus, the value of $x$ in the given function can be any real number except $-3$ and $1$. In interval notation: Domain: $\color{blue}{(-\infty, -3) \cup (-3, 1) \cup (1, +\infty)}$