## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 1 - Section 1.7 - Combinations of Functions; Composite Functions - Exercise Set - Page 258: 34

#### Answer

The value of the functions is as follows: $f+g=5{{x}^{2}}+x-6,f-g=-5{{x}^{2}}+x-6,fg=\ 5{{x}^{3}}-30{{x}^{2}},\frac{f}{g}=\frac{x-6}{5{{x}^{2}}}$. The domain of the functions $f+g,f-g$ , and $\ fg$ is $\left( -\infty ,\infty \right)$ and that of $\frac{f}{g}$ is $\left( -\infty ,0 \right)\cup \left( 0,\infty \right)$.

#### Work Step by Step

Calculate the value of $f+g$ as shown below to get: \begin{align} & f+g=f\left( x \right)+g\left( x \right) \\ & =\ x-6+5{{x}^{2}} \\ & =5{{x}^{2}}+x-6 \end{align} Calculate the value of $f-g$ as shown below to get: \begin{align} & f-g=f\left( x \right)-g\left( x \right) \\ & =\ x-6-5{{x}^{2}} \\ & =-5{{x}^{2}}+x-6 \end{align} Calculate the value of $fg$ as shown below to get: \begin{align} & fg=f\left( x \right)g\left( x \right) \\ & =\ \left( x-6 \right)\left( 5{{x}^{2}} \right) \\ & =5{{x}^{3}}-30{{x}^{2}} \end{align} Calculate the value of $f+g$ as shown below to get: \begin{align} & \ \frac{f}{g}=\frac{f\left( x \right)}{g\left( x \right)} \\ & =\frac{x-6}{5{{x}^{2}}} \end{align} If the function $\frac{f}{g}$ is divided by zero, it will be undefined. So, put the denominator $5x$ equal to zero. \begin{align} & 5{{x}^{2}}=0 \\ & x=0 \end{align} Now, the domain of the function $\frac{f}{g}$ is all the real numbers except 0. Therefore, the domain of $\frac{f}{g}$ is $\left( -\infty ,0 \right)\cup \left( 0,\infty \right)$. The functions $f+g,f-g,fg$ do not involve any division or do not contain even roots, so the domain of the functions $f+g,f-g,fg$ is the set of real numbers -- that is, $\left( -\infty ,\infty \right)$.

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