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: 44

Answer

The values of the functions are as follows: $f+g=\frac{5x-3}{{{x}^{2}}-25}$ , $f-g=\frac{1}{x-5}$ , $fg=\frac{6{{x}^{2}}-10x-4}{\left( {{x}^{2}}-25 \right)\left( {{x}^{2}}-25 \right)}$ and $\frac{f}{g}=\frac{3x+1}{2x-4}$ The domain of $\frac{f}{g}$ is $\left( -\infty ,-5 \right)\cup \left( -5,2 \right)\cup \left( 2,5 \right)\cup \left( 5,\infty \right)$ The domain of $f-g,f+g$, and $fg$ is $\left( -\infty ,-5 \right)\cup \left( -5,5 \right)\cup \left( 5,\infty \right)$.

Work Step by Step

Calculate the value of $f+g$ as shown below to get, $\begin{align} & \left( f+g \right)\left( x \right)=f\left( x \right)+g\left( x \right) \\ & =\frac{3x+1}{{{x}^{2}}-25}+\frac{2x-4}{{{x}^{2}}-25} \\ & =\frac{3x+1+2x-4}{{{x}^{2}}-25} \\ & =\frac{5x-3}{{{x}^{2}}-25} \end{align}$ Calculate the value of $f-g$ as shown below to get, $\begin{align} & \left( f-g \right)\left( x \right)=f\left( x \right)-g\left( x \right) \\ & =\frac{3x+1}{{{x}^{2}}-25}-\frac{2x-4}{{{x}^{2}}-25} \\ & =\frac{3x+1-2x+4}{{{x}^{2}}-25} \\ & =\frac{x+5}{{{x}^{2}}-25} \\ & =\frac{x+5}{\left( x-5 \right)\left( x+5 \right)} \\ & =\frac{1}{x-5} \end{align}$ Calculate the value of $fg$ as shown below to get, $\begin{align} & \left( fg \right)\left( x \right)=f\left( x \right)g\left( x \right) \\ & =\left( \frac{3x+1}{{{x}^{2}}-25} \right)\left( \frac{2x-4}{{{x}^{2}}-25} \right) \\ & =\frac{\left( 3x+1 \right)\left( 2x-4 \right)}{\left( {{x}^{2}}-25 \right)\left( {{x}^{2}}-25 \right)} \\ & =\frac{6{{x}^{2}}-12x+2x-4}{\left( {{x}^{2}}-25 \right)\left( {{x}^{2}}-25 \right)} \\ & =\frac{6{{x}^{2}}-10x-4}{\left( {{x}^{2}}-25 \right)\left( {{x}^{2}}-25 \right)} \end{align}$ Calculate the value of $\frac{f}{g}$ as shown below to get, $\begin{align} & \left( \frac{f}{g} \right)\left( x \right)=\frac{f\left( x \right)}{g\left( x \right)} \\ & =\frac{\left( \frac{3x+1}{{{x}^{2}}-25} \right)}{\left( \frac{2x-4}{{{x}^{2}}-25} \right)} \\ & =\frac{3x+1}{2x-4} \end{align}$ If the functions $f+g$ , $f-g$ and $fg$ contain division by zero, they will be undefined. So, put the denominator ${{x}^{2}}-25$ equal to zero. $\begin{align} & {{x}^{2}}-25=0 \\ & x=\pm \sqrt{25} \\ & x=\pm 5 \end{align}$ Now, the domain of the functions $f+g$ $f-g$ and $fg$ is all the real numbers except 5 and -5. Therefore, the domain of functions $f-g$ and $fg$ is $\left( -\infty ,-5 \right)\cup \left( -5,5 \right)\cup \left( 5,\infty \right)$. If the function $\frac{f}{g}$ is divided by zero, it will be undefined. So, put the denominator $2x-4$ equal to zero. $\begin{align} & 2x-4=0 \\ & 2x=4 \\ & x=2 \end{align}$ Now, the domain of the function $\frac{f}{g}$ is all the real numbers except $2$. Therefore, the domain of the function $\frac{f}{g}$ is $\left( -\infty ,-5 \right)\cup \left( -5,2 \right)\cup \left( 2,5 \right)\cup \left( 5,\infty \right)$.
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