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: 13


$\color{blue}{(-\infty, -\frac{1}{2}) \cup (-\frac{1}{2}, 1) \cup (1, +\infty)}$

Work Step by Step

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: $(4x+2)(x-1)= 0$ Use the Zero-Product Property by equating each factor to zero then solving each equation to obtain: \begin{array}{ccc} &4x+2 = 0 &\text{ or } &x-1=0 \\&4x=-2 &\text{or} &x=1 \\&x=-\frac{1}{2} &\text{or} &x=1 \end{array} This means that the value of $x$ can be any real number except $-\frac{1}{2}$ and $1$. Therefore, the domain of the given rational expression is the set of real numbers except $-\frac{1}{2}$ and $1$. In interval notation, the domain is: $\color{blue}{(-\infty, -\frac{1}{2}) \cup (-\frac{1}{2}, 1) \cup (1, +\infty)}$
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