%% MATH 135-01 Final Exam A, Fall 2019
\documentclass[12pt]{article}
\usepackage{latexsym,amsmath,amsfonts,mathrsfs,graphicx}
\usepackage{float}
\usepackage{subfig}

\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{0in}
%% \setlength{\topmargin}{-.5in}
\setlength{\textheight}{8.25in}

\newcommand{\ds}{\displaystyle}
\newcommand{\ts}{\textstyle}
\DeclareMathOperator{\lcm}{lcm}

\newcommand{\Neg}{\phantom{-}}
\newcommand{\vp}{$\vphantom{\big|}$}
\newcommand{\Vp}{$\vphantom{\Big|}$}

\newcommand{\by}{\times}

\newcommand{\Number}[1]{\mathbf{#1}}
\newcommand{\R}{\Number{R}}

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

\newcommand{\be}{\begin{enumerate}}
\newcommand{\ee}{\end{enumerate}}

\newcommand{\bd}{\begin{description}}
\newcommand{\ed}{\end{description}}

\newcommand{\bi}{\begin{enumerate}}
\newcommand{\ei}{\end{enumerate}}

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

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

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

%\usepackage{math131}

\pagestyle{myheadings}
%\markright{MATH 135-01, Final Exam}

\setlength{\oddsidemargin}{0in}
\setlength{\textheight}{8.5in}

\def\Answer#1{$\phantom{.}$\hfill
        #1\quad\fbox{$\phantom{\Big|Put longgggggggggggggggggggg answer here}$}}
\def\ShortAnswer#1{$\phantom{.}$\hfill
        #1\quad\fbox{$\phantom{\Big|answer}$}}

\def\Name{$\underline{\phantom{Please write your name in this
        space}}$} 
\def\Strut{$\vphantom{\Big|}$}

\begin{document}

\begin{center}

\textbf{Holy Cross College, Fall Semester, 2019} \\
\textbf{MATH 135, Section 01, Final Exam A Solutions} \\
\textbf{Saturday, December~21, 8:00~AM} \\
\end{center}


\noindent 
1. [20~points] One of the functions given in the following table is linear
and the other is exponential.  Find a formula for the \emph{linear} one and place it
in the appropriate box.  In the box for the other one, write ``Exponential.''

\[
\begin{array}{|c||r|r|r|r|r|}
\hline
x          & 1 & 2 & 3& 4& 5\Strut \\
\hline
f(x)       & 1.2 & 2.4 & 4.8 & 9.6 & 19.2 \Strut \\
\hline
g(x)       & -3.3& -4.4& -5.5 & -6.6 & -7.7\Strut \\
\hline
\end{array}
\]

\vskip 10pt
\noindent
{\it Answer:}
\begin{eqnarray*}
f(x) &=& \text{Exponential}\\
g(x) &=& -1.1 (x - 1) - 3.3 = -1.1 x - 2.2
\end{eqnarray*}
\vskip 10pt
\noindent
2.  Let $f(x) = x^3 + 3x + 1$ and $g(x) = \sqrt{x^2 + 3}$.
\begin{enumerate}
\item[(a)] [10~points]  Find a formula for the composition $f(g(x))$.
\item[] {\it Answer:} $f(g(x)) = (x^2 + 3)^{3/2} + 3(x^2 + 3)^{1/2} + 3$.
\item[(b)] [10~points] Write $g(x) = h(k(x))$ for two other functions $h$ and $k$, where 
$h(x) \ne x$ and $k(x) \ne x$.  
\item[] {\it Answer:} $h(x) = \sqrt{x}$ and $k(x) = x^2 + 3$ is one possibility.
\item[(c)] [10~points]  Find the average rate of change of $f(x)$ per unit change in $x$
on the interval $[1,3]$.
\item[] {\it Answer:}  The average rate of change is
$$\frac{f(3) - f(1)}{3 - 1} = \frac{37 - 5}{2} = 16.$$
\end{enumerate}
\vskip 10pt
\noindent
3.  Compute the following limits [10~points each]. \emph{Any legal method is OK.}
\begin{enumerate}
\item[(a)] $\ds \lim_{x\to 0} \frac{\sin(4x)}{x}$
\item[] {\it Answer:} This limit is 
$$\lim_{x \to 0} 4\cdot \frac{\sin(4x)}{4x} = 4 \cdot 1 = 4.$$
\item[(b)] $\ds \lim_{x\to 2} \frac{x^2 - 8x + 12}{x^2 - 2x}$
\item[] {\it Answer:} This is a $0/0$ indeterminate form.  It can be
done either by factoring the top and the bottom and canceling a factor of $x - 2$ get
$$\lim_{x \to 2} \frac{(x - 2)(x - 6)}{x(x - 2)} = \lim_{x \to 2} \frac{x - 6}{x} = -2$$
or by using L'Hopital's Rule.  
\item[(c)] (*) $\ds \lim_{x\to \infty} \left(1 + \frac{3}{x}\right)^x$
\item[] {\it Answer:} This is a $1^\infty$ indeterminate form.  Take logarithms, rearrange to $0/0$ form, then apply L'Hopital's Rule.  The answer is $e^3$.
\end{enumerate}
\vskip 10pt

\noindent
4. 
 \begin{enumerate}
\item[(a)] [10~points] State the limit definition of the derivative:
\item[] {\it Answer:}
$$f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h},$$
provided that the limit exists.
\item[(b)] [10~points] \emph{Use the definition} to compute $f'(x)$ for
$f(x) = \ds \frac{1}{x + 1}$.
\item[] {\it Answer:} We have 
\begin{eqnarray*}
f'(x) &=& \lim_{h\to 0} \frac{\frac{1}{x + h + 1} - \frac{1}{x + 1}}{h}\\
&=& \lim_{h\to 0} \frac{(x + 1) - (x + h + 1)}{h(x + h + 1)(x + 1)}\\
&=& \lim_{h\to 0} \frac{-h}{h(x+h+1)(x+1)}\\
&=& \lim_{h\to 0} \frac{-1}{(x+h+1)(x+1)}\\
&=& \frac{-1}{(x+1)^2}
\end{eqnarray*}
(Note that this agrees with the result obtained from the chain rule
for the derivative of $f(x) = (x + 1)^{-1}$.)  
\item[(c)] [10~points] Find the equation of the tangent line
to the graph $y = \ds \\frac{1}{x + 1}$ at the point $(2,1/3)$.  Note: you can do this one even
if you were not able to complete part b above.
\item[] {\it Answer:} The slope is $f'(2) = \frac{-1}{9}$ so the equation is
$$y - \frac{1}{3} = \frac{-1}{9}(x - 2)$$
by the point-slope form.
\end{enumerate}
\vskip 10pt
\noindent
5.  Compute the following derivatives using the derivative rules.
You need not simplify. 
\begin{enumerate} 
\item[(a)] [10 points] $\ds f(x) = x^{4/5} - \frac {1}{\sqrt[3]{x}} + e^x$.
\item[] {\it Answer:} 
$$f'(x) = \frac{4}{5} x^{-1/5} + \frac{1}{3} x^{-4/3} + e^x$$
\item[(b)] [10 points]  $\ds g(x) = \frac{x^2 + 1}{x^4 + 1}$
\item[] {\it Answer:} By the quotient rule,
$$g'(x) = \frac{(x^4+1)(2x) - (x^2+1)(4x^3)}{(x^4 + 1)^2}$$
\item[(c)] (*) [10 points] $\ds h(s) = \ln(4s^2 + 2)\tan^{-1}(s)$
\item[] {\it Answer:} By the product and chain rules, 
$$h'(s) = \ln(4s^2 + 2)\cdot \frac{1}{s^2 + 1} + \tan^{-1}(s)\cdot \frac{8s}{4s^2 + 2}.$$
\item[(d)] (*) [10 points] Find $y' = \frac{dy}{dx}$ if $4 x^2 y^3 - 2y^2 + \sin(x) = x^4$.
\item[] {\it Answer:} Differentiating implicitly,
$$\frac{dy}{dx} = \frac{-\cos(x) + 4x^3 - 8xy^3}{12x^2y^2 - 4y}$$
\end{enumerate}

\begin{figure}
\begin{center}
\includegraphics[width = .5 \linewidth]{135F6.png}
\caption{Plot for question 6.}
\end{center}
\end{figure}
\sep
\noindent
6. (*) [5~points each] All parts of this question refer to the function $f(x)$ whose
\emph{derivative} $f'(x)$ is plotted in Figure 1 above.  Assume the domain of $f$
contains only the interval shown in the plot.

\begin{enumerate}
\item[(a)]  On which interval(s) is $f(x)$ decreasing?
\item[] \emph{Answer:} On the interval $(4,7)$.
\item[(b)] On which interval(s) is $y = f(x)$ concave up?
\item[] \emph{Answer:} On $(1,2) \cup (6,7)$
\item[(c)] How many inflection points does $f$ have?  
\item[] \emph{Answer:} Two inflection points at $x = 2,6$
\end{enumerate}
\vskip 10pt
\noindent
7. (*). [20~points] If the ticket price for the upcoming ``Cats'' movie is set at
$\$10$, then 1000 tickets for the first showing will be sold.  However, for each
$\$0.25$ increase in the ticket price, the number of tickets sold will go down by
10 tickets.  Each person who buys a ticket will also purchase $\$10$ of nachos and 
drinks at the concession stand.  How should the ticket price be set to maximize
the revenue earned by the theater from the ticket and concession sales?
\vskip 10pt
\noindent
{\it Answer:} Let $x$ be the number of $\$0.25$ increases over the $\$10$ level.
Then with the ticket price at $10 + 0.25x$, $1000 - 10x$ tickets will be sold,
yielding revenue of $(1000 - 10x)(10 + 0.25x)$.  In addition, $10(1000 - x)$
dollars in nachos and drinks will be sold, yielding total revenue of 
$$R(x) = (1000 - 10x)(10 + 0.25x) + 10(1000 - 10x) = 20000 + 50x - 2.5x^2$$
We differentiate and set equal to zero to find critical points:
$$R'(x) = 50 - 5x = 0$$
When $x = 10$, and $R''(x) = -5 < 0$, so this is a local and global maximum
for the revenue.  The ticket price should be $\$(10 + 0.25\cdot 10) = \$ 12.50$.

\begin{figure}
\centering
\subfloat[Plot I]{\includegraphics[width=1.2in]{135F8I.png}}
\subfloat[Plot II]{\includegraphics[width=1.2in]{135F8II.png}}\\
\subfloat[Plot III]{\includegraphics[width=1.2in]{135F8III.png}}
\subfloat[Plot IV]{\includegraphics[width=1.2in]{135F8IV.png}}\\
\subfloat[Plot V]{\includegraphics[width=1.2in]{135F8V.png}}
\end{figure}

\noindent
8. (*) [3~points each]\quad By considering the location of vertical and 
Horizontal asymptotes, $x$- and $y$-axis intercepts, etc. determine which of 
the following functions matches each graph.  Circle the number of the graph
   showing each of the following functions. 


\begin{enumerate}
  \item[(a)] $\displaystyle{f(x) =\frac{x}{x^2 - 1}}$ \hfill
  \item[] \emph{Answer:} III -- vertical asymptotes at $x = \pm 1$
and horizontal asymptote at $y = 0$. 
  \item[(b)] $\displaystyle{f(x) = \frac{x}{x^2 + 1}}$\hfill
  \item[] \emph{Answer:} I -- no vertical asymptotes and horizontal
asymptote at $y = 0$.
  \item[(c)] $\displaystyle{f(x) = \frac{x^2 - 1}{x^2 + 1}}$\hfill 
\item[] \emph{Answer:} IV -- no vertical asymptotes and horizontal 
asymptote at $y = 1$.
\item[(d)] $\displaystyle{f(x) = \frac{1}{x^2 - 16}}$\hfill
\item[] \emph{Answer:} V -- vertical asymptotes at $x = \pm 4$.
\item[(e)] $\displaystyle{f(x) = \frac{3x + 1}{x - 1}}$\hfill
\item[] \emph{Answer:} II -- vertical asymptote at $x = 1$ (and not $-1$).
\end{enumerate}


\end{document}
