College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 3 - Review Exercises - Page 268: 6

Answer

$(-\infty, -3) \cup (-3, 3) \cup +\infty)$

Work Step by Step

Factor the denominator to obtain: $f(x) = \dfrac{x}{(x-3)(x+3)}$ To find the domain of the given function, exclude the values of $x$ that will make the function undefined. Note that the function is undefined when the denominator is equal to $0$. Thus, the value of $x$ can be any real number except the ones that will make the denominator equal to zero. To find the values of $x$ that will make the denominator zero, equate each factor of the denominator to zero; then, solve each equation. $\begin{array}{ccc} &x-3=0 &\text{ or } &x+3-0 \\&x=3 &\text{ or } &x=-3\end{array}$ Thus, the value of $x$ can be any real number except $3$ and $-3$. In interval notation, Domain: $\color{blue}{(-\infty, -3) \cup (-3, 3) \cup +\infty)}$
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