\(\text{www.the}\beta\text{etafunction.com}\)
Home
Login
Questions: Algebra BusinessCalculus
Please login to create an exam or a quiz.
For the functions \(\displaystyle f(x)=\sqrt{x + 7}\) and \(\displaystyle g(x)=x - 12\), find \(\displaystyle \left(\frac{f}{g}\right)(x)\) and the domain of \(\displaystyle \left(\frac{f}{g}\right)(x)\)
\(\displaystyle \frac{f}{g}\left(x\right)=\frac{\sqrt{x + 7}}{x - 12}\). The domain of \(\displaystyle f\) is the solution to \(\displaystyle x + 7\geq 0\). Solving gives \(\displaystyle [-7,\infty)\). The domain of \(\displaystyle g\) is all real numbers and the zero is \(\displaystyle x - 12=0 \iff x= 12\). The domain is the intersection of the domains of \(\displaystyle f\) and \(\displaystyle g\) excluding the zeros of \(\displaystyle g\). This gives \(\displaystyle [-7,12)\cup (12,\infty)\)
\begin{question}For the functions $f(x)=\sqrt{x + 7}$ and $g(x)=x - 12$, find $\left(\frac{f}{g}\right)(x)$ and the domain of $\left(\frac{f}{g}\right)(x)$
\soln{10cm}{$\frac{f}{g}\left(x\right)=\frac{\sqrt{x + 7}}{x - 12}$. The domain of $f$ is the solution to $x + 7\geq 0$. Solving gives $[-7,\infty)$. The domain of $g$ is all real numbers and the zero is $x - 12=0 \iff x= 12$. The domain is the intersection of the domains of $f$ and $g$ excluding the zeros of $g$. This gives $[-7,12)\cup (12,\infty)$}
\end{question}
\documentclass{article}
\usepackage{tikz}
\usepackage{amsmath}
\usepackage[margin=2cm]{geometry}
\usepackage{tcolorbox}
\newcounter{ExamNumber}
\newcounter{questioncount}
\stepcounter{questioncount}
\newenvironment{question}{{\noindent\bfseries Question \arabic{questioncount}.}}{\stepcounter{questioncount}}
\renewcommand{\labelenumi}{{\bfseries (\alph{enumi})}}
\newif\ifShowSolution
\newcommand{\soln}[2]{%
\ifShowSolution%
\noindent\begin{tcolorbox}[colframe=blue,title=Solution]#2\end{tcolorbox}\else%
\vspace{#1}%
\fi%
}%
\newcommand{\hideifShowSolution}[1]{%
\ifShowSolution%
%
\else%
#1%
\fi%
}%
\everymath{\displaystyle}
\ShowSolutiontrue
\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>For the functions <img class="equation_image" title=" \displaystyle f(x)=\sqrt{x + 7} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%3D%5Csqrt%7Bx%20%2B%207%7D%20" alt="LaTeX: \displaystyle f(x)=\sqrt{x + 7} " data-equation-content=" \displaystyle f(x)=\sqrt{x + 7} " /> and <img class="equation_image" title=" \displaystyle g(x)=x - 12 " src="/equation_images/%20%5Cdisplaystyle%20g%28x%29%3Dx%20-%2012%20" alt="LaTeX: \displaystyle g(x)=x - 12 " data-equation-content=" \displaystyle g(x)=x - 12 " /> , find <img class="equation_image" title=" \displaystyle \left(\frac{f}{g}\right)(x) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%5Cfrac%7Bf%7D%7Bg%7D%5Cright%29%28x%29%20" alt="LaTeX: \displaystyle \left(\frac{f}{g}\right)(x) " data-equation-content=" \displaystyle \left(\frac{f}{g}\right)(x) " /> and the domain of <img class="equation_image" title=" \displaystyle \left(\frac{f}{g}\right)(x) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%5Cfrac%7Bf%7D%7Bg%7D%5Cright%29%28x%29%20" alt="LaTeX: \displaystyle \left(\frac{f}{g}\right)(x) " data-equation-content=" \displaystyle \left(\frac{f}{g}\right)(x) " /> </p> </p><p> <p> <img class="equation_image" title=" \displaystyle \frac{f}{g}\left(x\right)=\frac{\sqrt{x + 7}}{x - 12} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7Bf%7D%7Bg%7D%5Cleft%28x%5Cright%29%3D%5Cfrac%7B%5Csqrt%7Bx%20%2B%207%7D%7D%7Bx%20-%2012%7D%20" alt="LaTeX: \displaystyle \frac{f}{g}\left(x\right)=\frac{\sqrt{x + 7}}{x - 12} " data-equation-content=" \displaystyle \frac{f}{g}\left(x\right)=\frac{\sqrt{x + 7}}{x - 12} " /> . The domain of <img class="equation_image" title=" \displaystyle f " src="/equation_images/%20%5Cdisplaystyle%20f%20" alt="LaTeX: \displaystyle f " data-equation-content=" \displaystyle f " /> is the solution to <img class="equation_image" title=" \displaystyle x + 7\geq 0 " src="/equation_images/%20%5Cdisplaystyle%20x%20%2B%207%5Cgeq%200%20" alt="LaTeX: \displaystyle x + 7\geq 0 " data-equation-content=" \displaystyle x + 7\geq 0 " /> . Solving gives <img class="equation_image" title=" \displaystyle [-7,\infty) " src="/equation_images/%20%5Cdisplaystyle%20%5B-7%2C%5Cinfty%29%20" alt="LaTeX: \displaystyle [-7,\infty) " data-equation-content=" \displaystyle [-7,\infty) " /> . The domain of <img class="equation_image" title=" \displaystyle g " src="/equation_images/%20%5Cdisplaystyle%20g%20" alt="LaTeX: \displaystyle g " data-equation-content=" \displaystyle g " /> is all real numbers and the zero is <img class="equation_image" title=" \displaystyle x - 12=0 \iff x= 12 " src="/equation_images/%20%5Cdisplaystyle%20x%20-%2012%3D0%20%20%5Ciff%20x%3D%2012%20" alt="LaTeX: \displaystyle x - 12=0 \iff x= 12 " data-equation-content=" \displaystyle x - 12=0 \iff x= 12 " /> . The domain is the intersection of the domains of <img class="equation_image" title=" \displaystyle f " src="/equation_images/%20%5Cdisplaystyle%20f%20" alt="LaTeX: \displaystyle f " data-equation-content=" \displaystyle f " /> and <img class="equation_image" title=" \displaystyle g " src="/equation_images/%20%5Cdisplaystyle%20g%20" alt="LaTeX: \displaystyle g " data-equation-content=" \displaystyle g " /> excluding the zeros of <img class="equation_image" title=" \displaystyle g " src="/equation_images/%20%5Cdisplaystyle%20g%20" alt="LaTeX: \displaystyle g " data-equation-content=" \displaystyle g " /> . This gives <img class="equation_image" title=" \displaystyle [-7,12)\cup (12,\infty) " src="/equation_images/%20%5Cdisplaystyle%20%5B-7%2C12%29%5Ccup%20%2812%2C%5Cinfty%29%20" alt="LaTeX: \displaystyle [-7,12)\cup (12,\infty) " data-equation-content=" \displaystyle [-7,12)\cup (12,\infty) " /> </p> </p>