
\documentclass[12pt]{article}
\usepackage{latexsym,amsmath,epsfig}
\newcommand{\ds}{\displaystyle}
\newcommand{\vp}{$\vphantom{\big|}$}

\newcommand{\R}{\textbf{R}}
\newcommand{\Item}{\item{$\Box$}\quad}

\newcommand{\Answer}[1]{$\phantom{.}$\hfill
        #1\quad\fbox{$\phantom{\Big|Answer}$}} 

\newcommand{\MyAnswer}[2]{$\phantom{.}$\hfill
        #1\quad\fbox{$\vphantom{\Big|}#2$}}

\newcommand{\soln}{\paragraph{Solution}}
\newcommand{\sep}{{\medskip\hrule\bigskip}}
\newcommand{\blank}{\underline{$\phantom{A blank}$}}

\setlength{\textwidth}{6.5in}
\setlength{\topmargin}{-0.5in}
\setlength{\oddsidemargin}{0in}
\setlength{\textheight}{9.0in}
\begin{document}
\begin{center}

\textbf{College of the Holy Cross, Fall 2013} \\
\textbf{Math 135, Section 1, Midterm 1 Solutions}\\
\textbf{Friday, September 20}
\end{center}

\vskip 10pt
\noindent
I.  Match the plots below with the following formulas. Note that there is an extra plot.
\begin{itemize}
\item[(5) A)] $y = 3 - (x - 1)^2$ is Plot: $\underline{3}$  (the $x - 1$ shifts the parabola to the right, not the left)
\item[(5) B)] $y = \sin(x/2)$ is Plot: $\underline{2}$  (the $x/2$ makes the period equal to $4\pi \doteq 12.6$, so the
usual sine graph is stretched horizontally)
\item[(5) C)] $y = 1 - e^{-x}$ is Plot: $\underline{1}$  (think:  $y = e^x$ reflected across the $x$- and $y$-axes, then
shifted up)
\item[(5) D)] $y = \sin(2x)$ is Plot: $\underline{5}$ (the $2x$ makes the period equal to $\pi \doteq 3.14$, so the usual
sine graph is compressed horizontally).
\end{itemize}
\vskip 0.5in
\begin{center}
Plot 1: \includegraphics[width=1.3truein,height=1.3truein]{realplot1.ps} 
Plot 2: \includegraphics[width=1.3truein,height=1.3truein]{realplot2.ps}
\end{center}
\begin{center}
Plot 3:
\includegraphics[width=1.3truein,height=1.3truein]{realplot3.ps}
Plot 4:
\includegraphics[width=1.3truein,height=1.3truein]{realplot4.ps}
\end{center}
\begin{center}
Plot 5: \includegraphics[width=2truein,height=2truein]{realplot5.ps}
\end{center}
\vfil\eject

\noindent
II.  The manager of a furniture factory has collected the following data for the cost of manufacturing chairs.
\bigskip
\begin{center}
\begin{tabular}{|c|c|}
\hline \# Chairs (per day) $x$ & Cost (in dollars) $y$ \\

\hline 100 & 2400 \\
\hline 150 & 3100 \\
\hline 250 & 4500 \\
\hline 300 & 5200 \\
\hline
\end{tabular}
\end{center}
\begin{itemize}
\item[(10) A)]  Given that $y$ is a linear function of $x$, determine a formula for it.  
\vskip 10pt
The slope is $m = \frac{3100 - 2400}{150 - 100} = 14$ so by the point slope form, we get
$y - 2400 = 14(x - 100)$, or $y = 14x + 1000$.
\vskip 10pt
\MyAnswer{Cost function:}{\ y - 2400 = 14(x - 100)\ {\rm or}\  y = 14x + 1000}%
\item[(5) B)] What does the slope represent in real-world terms?
\vskip 10pt
The slope represents the cost of manufacturing one additional chair per day, or $C(x+1) - C(x)$.  
\vskip 10pt
\item[(5) C)] What does the $y$-intercept represent in terms of cost?
\vskip 10pt
The $y$-intercept of $1000$ represents the cost per day if no chairs are actually manufactured ($x = 0$).
These are often called \emph{fixed costs} -- things like the maintenance costs of the factory, taxes, labor costs, etc.
\vskip 10pt 
\item[(5) D)]  Using your model, determine how much it will cost to produce $350$ chairs per day.
\vskip 10pt
\MyAnswer{Cost:}{$(14)(350) + 1000 = \$ 5900$}%
\end{itemize}


%%%%%%%%%% Combining functions and finding domains %%%%%%%%%%%%

\vskip 10pt
\noindent
III. Given $f(x)=4-x^2$ and $g(x)=\sqrt{3x-2}$, answer the following questions.

\begin{itemize}
\item[(10) A)] Find the domain of $f(x)$ and the domain of $g(x)$.

\vskip 10pt
The domains here are the sets of all real $x$ that can be substituted into the formulas to yield
a well-defined result.  For $f$ there are no restrictions.  For $g$, we must have $3x - 2 \ge 0$, so 
$x \ge \frac{2}{3}$.
\vskip 10pt
\MyAnswer{Domain of $f$:}{{\rm all\ real\ } x,\ {\rm or}\ (-\infty,+\infty)}
\vskip 10pt
\MyAnswer{Domain of $g$:}{{\rm all\ real\ } x \ge \frac{2}{3},\ {\rm or}\ \left[\frac{2}{3},+\infty\right)}%
\item[(5) B)] What is the domain of the function $g(x)/f(x)$?
\vskip 10pt
Now we must be able to substitute an $x$ that makes sense for $g$, and that also avoids making $f(x) = 0$.  
\vskip 10pt
\MyAnswer{Domain of $g(x)/f(x)$:}{\left[\frac{2}{3},2\right) \cup (2,+\infty),\ {\rm or\ something\ equivalent}}%
\item[(5) C)] Find the function $g\circ f$.
\MyAnswer{}{(g\circ f)(x) = g(f(x)) = \sqrt{3(4 - x^2) - 2} = \sqrt{10 - 3x^2}}%
\end{itemize}

\vskip 10pt
\noindent
IV.  Answer the following questions. 

\begin{itemize}
\item[(5) A)] Find all values of $x$ in 
$\displaystyle \left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ for which 
$\displaystyle \left| \tan x \right| > 1$.
\vskip 10pt
This is true if $\tan(x) > 1$ or $\tan(x) < -1$.  The first occurs for $x$ between $\frac{\pi}{4}$ and $\frac{\pi}{2}$;
the second occurs for $x$ between $-\frac{\pi}{2}$ and $-\frac{\pi}{4}$.
\vskip 10pt
\MyAnswer{Values of $x$:}{\left(-\frac{\pi}{2},-\frac{\pi}{4}\right) \cup \left(\frac{\pi}{4},\frac{\pi}{2}\right)}%
\vskip 10pt
\noindent
Note:  Equivalent answers like: all $x$ with $-\frac{\pi}{4} < x < -\frac{\pi}{4}$ or $\frac{\pi}{4} < x < \frac{\pi}{2}$
are also OK.  
\vskip 10pt

\item[(5) B)] If $\displaystyle \sin \theta =\frac{2}{3}$ and $\displaystyle \frac{\pi}{2}<\theta<\pi$, give the exact value of 
$\cos \theta$. 

\vskip 10pt
We can use the basic trig identity $\sin^2(\theta) + \cos^2(\theta) = 1$ for this: 
$(\frac{2}{3})^2 + \cos^2(\theta) = 1$ so $\cos^2(\theta) = \frac{5}{9}$ and
$\cos(\theta) = \pm \frac{\sqrt{5}}{3}$.  Since $\theta$ is between $\frac{\pi}{2}$ and $\pi$, the 
cosine must be negative, so the correct answer is:

\vskip 10pt
\MyAnswer{$\cos \theta$:}{-\frac{\sqrt{5}}{3}}%
\item[(5) C)] Express as a single logarithm: $\displaystyle \frac{1}{2} \ln 3 - 3 \ln 2 + \ln 6.$
\vskip 10pt
Use the properties of logarithms: $\ln(A) + \ln(B) = \ln(AB)$, $\ln(A) - \ln(B) = \ln(A/B)$,
and $p \ln(A) = \ln(A^p)$.  Then 
$$\frac{1}{2} \ln 3 - 3 \ln 2 + \ln 6 = \ln\left(\frac{3^{1/2}\cdot 6}{2^3}\right).$$
\vskip 10pt
\MyAnswer{Single logarithm:}{\ln\left(\frac{3^{1/2}\cdot 6}{2^3}\right) = \ln\left(\frac{3\sqrt{3}}{4} \right)}%

\end{itemize}
\vskip 10pt

\noindent
V.  Consider the function $\displaystyle f(x)= \frac{1}{2} e^{x+1} + 1$. 
\begin{itemize}
\item[(15) A)] Given that $f$ is one-to-one, find a formula for the inverse function of $f$. 
\vskip 10pt
Set up $y = \frac{1}{2} e^{x+1} + 1$ and solve for $x$:
\begin{eqnarray*}
2(y - 1) &=& e^{x+1}, \text{so after taking natural log of both sides} \\
\ln(2(y - 1)) &=& x+1\\  
x &=& \ln(2(y-1)) - 1.
\end{eqnarray*}
 We can swap the variables to write the inverse function as a function of $x$:
\vskip 10pt
\MyAnswer{$f^{-1}(x) = $}{\ln(2(x-1)) - 1}%
\item[(10) B)] In the space below, plot the graphs 
of the functions $f$ and $f^{-1}$ on the same set of axes.
Label one point on each graph with its coordinates.
\vskip 10pt
Here are the graphs:
\begin{center}
\includegraphics[width=2truein,height=2truein]{Five.ps}
\end{center}
Note that $f(x) > 1$ for all $x$.  This means that $y = f(x)$ should be approaching the horizontal line 
$y = 1$ as $x \to -\infty$.  Because of this, the graph $y = f^{-1}(x)$ has a vertical 
asymptote at $x = 1$.  It is obtained by reflecting $y = f(x)$ across the line $y = x$.
The top (red) curve is $y = f(x)$; it contains the point $\left(0,\frac{e}{2} + 1\right) \doteq (0,2.36)$.  The 
bottom (blue) curve is $y = f^{-1}(x)$, obtained by reflecting $y = f(x)$ across the line $y = x$; 
it contains the point $\left(\frac{e}{2} + 1,0\right) \doteq (2.36,0)$. 
\end{itemize}
\end{document}
