Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.5 Rational Expressions - R.5 Exercises: 15

Answer

$\color{blue}{(-\infty, -3) \cup (-3, -2) \cup (-2, +\infty)}$

Work Step by Step

Factor the denominator to obtain: $=\dfrac{12}{(x+3)(x+2)}$ The denominator of a rational expression is not allowed to be equal to zero as it will make the expression undefined. Find the values of $x$ for which the denominator is equal to zero by equating the denominator to zero: $(x+3)(x+2)= 0$ Use the Zero-Product Property by equating each factor to zero then solving each equation to obtain: \begin{array}{ccc} &x+3 = 0 &\text{ or } &x+2=0 \\&x=-3 &\text{or} &x=-2 \end{array} This means that the value of $x$ can be any real number except $-3$ and $-2$. Therefore, the domain of the given rational expression is the set of real numbers except $-3$ and $-2$. In interval notation, the domain is: $\color{blue}{(-\infty, -3) \cup (-3, -2) \cup (-2, +\infty)}$
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