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+\documentclass{article}
+\usepackage[fleqn]{amsmath}
+\usepackage[pdf,cfg=quiz,forpaper,pointsonleft,
+% compile with exactly one of the following three
+ nosolutions
+% answerkey
+% vspacewithsolns
+]{eqexam}
+
+\examNum{2}\numVersions{2}\forVersion{a}
+\longTitleText
+ {Quiz~\nExam--003}
+ {Quiz~\nExam--007}
+\endlongTitleText
+\shortTitleText
+ {Q{\nExam}s3}
+ {Q{\nExam}s7}
+\endshortTitleText
+
+
+\title[\sExam]{\bfseries\Exam}
+\author{D. P. Story}
+\subject[C1]{Calculus I}
+\date{Spring \the\year}
+\keywords{Test~\nExam, Section \vA{003}\vB{007}}
+\email{dpstory@uakron.edu}
+
+\vspacewithkeyOn
+\solAtEndFormatting{\eqequesitemsep{3pt}}
+\everymath{\displaystyle}
+
+\begin{document}
+
+\maketitle
+
+\begin{exam}{qz02}
+
+\begin{instructions}[Global Instructions:]
+Solve each of the following problems without error. \textit{Show all details.} Box in your
+$\boxed{\text{answers.}}$ Use good notation, you \emph{will} be marked off for bad notation.
+\end{instructions}
+
+\begin{problem}[3]
+Identify all numbers $x$ at which the function $ f(x) = \frac{x+2}{\sqrt{\vA{x-1}\vB{2-x}}} $ is continuous.
+
+\begin{solution}[.75in]
+We require $ \vA{x - 1}\vB{2-x} >0 $ or $ \vA{x > 1}\vB{x<2} $. In
+interval notation, the set of all numbers at which $f$ is continuous is
+$\boxed{\vA{( 1, \infty )}\vB{(-\infty, 2)} }$.
+\end{solution}
+\end{problem}
+
+\begin{problem}[3]
+Given $ f(x) = \begin{cases}
+ 3x^2 - 2x & x < -1 \\
+ 6x^2 + x \vB{+1} & x \ge -1
+\end{cases}$. Is this function (a) continuous at $ x = -1 $;, (b)~discontinuous with a removable discontinuity
+at $ x = -1 $; or (c)~discontinuous with a jump discontinuity at $ x = -1 $? Justify your response.
+
+\begin{solution}[2in]
+Look at the left and right limits:
+\begin{align*}
+ \lim_{x\to-1^-}f(x) &= \lim_{x\to-1^-} 3x^2 - 2x = 5\\
+ \lim_{x\to-1^+}f(x) &= \lim_{x\to-1^+} 6x^2 + x \vB{+1} = \vA{5}\vB{6} \vA{=}\vB{\neq} f(-1)
+\end{align*}
+Thus, $\lim_{x\to-1^-}f(x) \vA{=}\vB{\neq}
+\lim_{x\to-1^+}f(x)\vA{=f(-1)}$. The two sided limit \vA{exists}\vB{does
+not exist}\vA{ and $\lim_{x\to-1}f(x)=f(-1)$}. This function \vA{is}\vB{is
+not} continuous at $x=-1$, \vB{it has a jump discontinuity, since
+$\lim_{x\to-1^-}f(x) \neq \lim_{x\to-1^+}f(x)$}; as a result, the answer
+is \vA{(a)}\vB{(c)}.
+\end{solution}
+\end{problem}
+
+\begin{problem}[4]
+Define the function $ f(x) = 3x^2 - 2x $. Use one of the formulas:
+\[
+ m = \lim_{x\to a} \frac{f(x) - f(a)}{x-a}\quad\text{or}\quad
+ m = \lim_{h\to 0} \frac{f(a+h) - f(a)}{h}
+\]
+Then the slope of the line tangent to the graph of $f$ at the point $
+\vA{( 1, 1 )}\vB{(-1,5)} $.
+
+\renameSolnAfterTo{}
+\begin{solution}[2in]\ifkeyalt We make the following calculations:\fi
+\begin{multicols}{2}
+\noindent\textbf{Calculations}
+\begin{verA}
+\begin{alignat*}{2}
+ m &= \lim_{x\to1} \frac{f(x)-f(1)}{x-1}\\&
+ = \lim_{x\to1} 3x+1&&\quad\text{from side calc}\\&
+ = \boxed4
+\end{alignat*}
+\end{verA}
+\begin{verB}
+\begin{alignat*}{2}
+ m &= \lim_{x\to-1} \frac{f(x)-f(-1)}{x+1}\\&
+ = \lim_{x\to-1} 3x-5&&\quad\text{from side calc}\\&
+ = \boxed{-8}
+\end{alignat*}
+\end{verB}
+
+\columnbreak
+\noindent\textbf{Side Calculations}
+\begin{verA}
+\begin{align*}
+ f(x)-f(1) &= 3x^2 - 2x - 1\\&
+ = (x-1)(3x+1)
+\intertext{thus, the difference quotient is}
+ \frac{f(x)-f(1)}{x-1} &= 3x+1
+\end{align*}
+\end{verA}
+\begin{verB}
+\begin{align*}
+ f(x)-f(-1) &= 3x^2 - 2x - 5\\&
+ = (x+1)(3x-5)
+\intertext{thus, the difference quotient is}
+ \frac{f(x)-f(-1)}{x+1} &= 3x-5
+\end{align*}
+\end{verB}
+
+
+\vfill
+\vspace*{\sameVspace}
+\vfill
+\end{multicols}
+\end{solution}
+\begin{workarea}{\sameVspace}\parindent0pt\bfseries
+\begin{multicols}{2}
+\textbf{Calculations}
+
+\vfil\vspace*{\sameVspace}\vfil
+
+
+\columnbreak
+\textbf{Side Calculations}
+
+\vfil\vspace*{1.9in}\vfil
+
+\end{multicols}
+\end{workarea}
+
+\end{problem}
+
+\end{exam}
+\end{document}