Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Section 7.1 - Systems of Linear Equations in Two Variables - Exercise Set - Page 823: 112


The domain of the given function is $\left( -\infty ,-6 \right)\bigcup \left( -6,6 \right)\bigcup \left( 6,\infty \right)$

Work Step by Step

The domain in which the given function $f\left( x \right)=\frac{\left( x-6 \right)}{\left( {{x}^{2}}-36 \right)}$ is defined is given as: ${{x}^{2}}-36\ne 0$ because $\frac{a}{b}$ , where $b\ne 0$ Using property the ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$, we simplify ${{x}^{2}}-36\ne 0$: ${{x}^{2}}-36=\left( x+6 \right)\left( x-6 \right)\ne 0$ And from the zero-factor property: $\left( x+6 \right)\ne 0,\left( x-6 \right)\ne 0$ Therefore, $\left( x+6 \right)\ne 0$. Add $-6$ to both sides: $\begin{align} & \left( x+6-6 \right)\ne 0-6 \\ & x=-6 \end{align}$ And $\left( x-6 \right)\ne 0$. Add $+6$ to both sides: $\begin{align} & \left( x-6+6 \right)\ne 0+6 \\ & x=+6 \end{align}$ So, the domain of the function $f\left( x \right)=\frac{\left( x-6 \right)}{\left( {{x}^{2}}-36 \right)}$ is $x\in \text{R}-\left( 6,-6 \right)$. This means that it includes all the real numbers excluding (6,-6). This can represented as: Therefore, $\left( -\infty ,-6 \right)\bigcup \left( -6,6 \right)\bigcup \left( 6,\infty \right)$. Hence, the domain of the function $f\left( x \right)=\frac{\left( x-6 \right)}{\left( {{x}^{2}}-36 \right)}$ is $\left( -\infty ,-6 \right)\bigcup \left( -6,6 \right)\bigcup \left( 6,\infty \right)$.
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