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 - Page 53: 16


$\color{blue}{(-\infty, -1) \cup (-1, 6) \cup (6, +\infty)}$

Work Step by Step

Factor the denominator to obtain: $=\dfrac{3}{(x-6)(x+1)}$ 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: $(x-6)(x+1)= 0$ Use the Zero-Product Property by equating each factor to zero then solving each equation to obtain: \begin{array}{ccc} &x-6 = 0 &\text{ or } &x+1=0 \\&x=6 &\text{or} &x=-1 \end{array} This means that the value of $x$ can be any real number except $6$ and $-1$. Therefore, the domain of the given rational expression is the set of real numbers except $6$ and $-1$. In interval notation, the domain is: $\color{blue}{(-\infty, -1) \cup (-1, 6) \cup (6, +\infty)}$
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