Precalculus (6th Edition) Blitzer

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

Chapter 1 - Section 1.7 - Combinations of Functions; Composite Functions - Exercise Set - Page 258: 29


The domain of the function $f\left( x \right)=\frac{2x+7}{{{x}^{3}}-5{{x}^{2}}-4x+20}$ is $\left( -\infty ,-2 \right)\cup \left( -2,2 \right)\cup \left( 2,5 \right)\cup \left( 5,\infty \right)$.

Work Step by Step

In the given function $f\left( x \right)=\frac{2x+7}{{{x}^{3}}-5{{x}^{2}}-4x+20}$ if the denominator is zero then it will make the function undefined. So, put the denominator ${{x}^{3}}-5{{x}^{2}}-4x+20$ equal to zero. Therefore, ${{x}^{3}}-5{{x}^{2}}-4x+20=0$ Factorize the equation ${{x}^{3}}-5{{x}^{2}}-4x+20=0$ $\begin{align} {{x}^{2}}\left( x-5 \right)-4\left( x-5 \right)=0 & \\ \left( {{x}^{2}}-4 \right)\left( x-5 \right)=0 & \\ \end{align}$ Case 1: $\begin{align} & {{x}^{2}}-4=0\ \ \ \ \ \ \ \ \\ & \left( x-2 \right)\left( x+\text{2} \right)\text{=}\ \text{0} \end{align}$ This implies $x=2$ and $x=-2$. Case 2: $\begin{align} x-5=0 & \\ x=5 & \\ \end{align}$ The factors of the equation are $2\ ,-2$ and $5$. Exclude the factors from the domain of the function $f\left( x \right)=\frac{2x+7}{{{x}^{3}}-5{{x}^{2}}-4x+20}$. Hence, the domain of the function $f\left( x \right)=\frac{2x+7}{{{x}^{3}}-5{{x}^{2}}-4x+20}$ is $\left( -\infty ,-2 \right)\cup \left( -2,2 \right)\cup \left( 2,5 \right)\cup \left( 5,\infty \right)$.
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