## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 1 - Section 1.7 - Combinations of Functions; Composite Functions - Exercise Set - Page 258: 40

#### Answer

The values of the functions are as follows: $f+g=\sqrt{x}+x-5,f-g=\sqrt{x}-x+5,fg=\sqrt{x}\left( x-5 \right)\$ and $\frac{f}{g}=\frac{\sqrt{x}}{x-5}$. The domain of the functions $f+g,f-g$ , and $fg$ is $\left[ 0,\infty \right)$ and that of $\frac{f}{g}$ is $\left[ 0,5 \right)\cup \left( 5,\infty \right)$.

#### 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) \\ & =\sqrt{x}+x-5\ \ \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) \\ & =\sqrt{x}-x+5 \end{align} Calculate the value of $fg$ as shown below to get: \begin{align} & fg=f\left( x \right)g\left( x \right) \\ & =\sqrt{x}\left( x-5 \right) \end{align} Calculate the value of $\frac{f}{g}$ as shown below to get: \begin{align} & \frac{f}{g}=\frac{f\left( x \right)}{g\left( x \right)} \\ & =\frac{\sqrt{x}}{x-5} \end{align} In the function $f\left( x \right)=\sqrt{x}$, only positive numbers have square roots that are real numbers, so the expression under the square root sign, $x$ , must be positive. Therefore, \begin{align} & \sqrt{x}\ge 0 \\ & \ \ x\ge 0 \end{align} If the function $\frac{f}{g}$ has division by zero, then it will be undefined. So, put the denominator $x-5$ equal to zero. \begin{align} & x-5=0 \\ & x=5 \end{align} This implies that $x=0$ and $x=5$. Now, the domain of the function $\frac{f}{g}$ is all the real numbers except 0 and 5. Therefore, the domain of $\frac{f}{g}$ is $\left[ 0,5 \right)\cup \left( 5,\infty \right)$. The functions $f+g,f-g,fg$ do not involve any division and do not contain even roots, so the domain of the functions $f+g,f-g,fg$ is the set of real numbers -- that is, $\left[ 0,\infty \right)$.

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