\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}