%% MATH 135-01 Final Exam B, Fall 2019
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\begin{document}

\begin{center}

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

\vskip 10pt
\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{exponential} one and place it
in the appropriate box.  In the box for the other one, write ``Linear.''

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

\noindent
{\it Answer:}
\begin{eqnarray*}
f(x) &=& 1.2 \cdot 2^{x - 1} = 0.6 \cdot 2^x\\
g(x) &=& \text{Linear}
\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 the formula for the composition $g(f(x))$
\item[] {\it Answer:}
$$g(f(x)) = \sqrt{(x^3 + 3x + 1)^2 + 3}$$
\item[(b)] [10~points] Write $g(x) = h(k(x))$ for two other functions $h$ and $k$, where 
\item[] {\it Answer:} One possibility is $h(x) = \sqrt{x}$ and $k(x) = x^2 + 3$.
\item[(c)] [10~points]  Find the average rate of change of $g(x)$ per unit change in $x$ on the interval $[0,2]$.
\item[] {\it Answer:} The average rate of change is 
$$\frac{g(2) - g(0)}{2 - 0} = \frac{\sqrt{7} - \sqrt{3}}{2} \doteq .457.$$
\end{enumerate}
\vskip 10pt
\noindent
3.  Compute the following limits [10~points each].  \emph{Any legal method is OK}.
\begin{enumerate}
\item[(a)] (*) $\ds \lim_{x\to 0} (1 + 5x)^{1/x}$
\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 $e^5$.  
\item[(b)]  $\ds \lim_{x\to 3} \frac{x^2 - 6x + 9}{x^2 - 5x + 6}$
\item[] This is a $0/0$ indeterminate form.  It can be evaluated by 
factoring the top and the bottom and cancelling a factor of $x - 3$: 
$$\lim_{x\to 3} \frac{(x-3)^2}{(x - 3)(x - 2)} = \lim_{x\to 3}\frac{x-3}{x-2} = 0.$$
Alternatively, one can also use L'Hopital's Rule.  
\item[(c)]  $\ds \lim_{x\to 0} \frac{\sin(3x)}{x}$
\item[] {\it Answer:}
$$=\lim_{x\to 0} 3\cdot \frac{\sin(3x)}{3x} = 3\cdot 1 = 3.$$
\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] Use the definition to compute $f'(x)$ for
$f(x) = \ds \frac{1}{x^2}$.
\item[] {\it Answer:} We have
\begin{eqnarray*}
f'(x) &=& \lim_{h\to 0} \frac{\frac{1}{(x + h)^2} - \frac{1}{x^2}}{h}\\
&=& \lim_{h\to 0} \frac{(x^2 - (x + h)^2}{h(x+h)^2x^2}\\
&=& \lim_{h\to 0} \frac{-2xh - h^2}{h(x+h)^2 x^2}\\
&=& \lim_{h\to 0} \frac{-2x - h^2}{(x+h)^2 x^2}\\
&=& \frac{-2x}{x^4} = \frac{-2}{x^3} 
\end{eqnarray*}
(Note that this agrees with the shortcut rule for differentiating
$f(x) = x^{-2}$.
\item[(c)] [10~points] Find the equation of the tangent line
to the graph $y = \ds \frac{1}{x^2}$ at the point $(2,1/4)$.  Note: you can do this
one even if you were not able to complete part b above.
\item[] {\it Answer:} The slope is $\frac{-2}{2^3} = \frac{-1}{4}$.  So by the 
point-slope form, the equation of the tangent line is 
$$y - 1/4 = \frac{-1}{4}(x - 2)$$
\end{enumerate}
\vskip 10pt
\noindent
5.  Compute the following derivatives using the derivative rules. You need not simplify. 
\begin{enumerate}
\item[(a)] [10 points] $\ds f(z) = z^5 - \frac {1}{\sqrt[6]{z}} + e^z$.
\item[] {\it Answer}
$$f'(z) = 5z^4 + \frac{1}{6} x^{-7/6} + e^z.$$
\item[(b)] [10 points] $\ds g(x) = (x^3 + 1)(x^2 - 1)$
\item[] {\it Answer:} By the product rule, 
$$g'(x) = (x^3 + 1)(2x) + (x^2 - 1)(3x^2).$$
This can also be done by multiplying out the product, then 
differentiating.
\item[(c)] (*) [10 points] $\ds h(t) = \frac{\ln(t^2 + 3)}{\sin^{-1}(t)}$
\item[] {\it Answer:} By the quotient rule and the rules for the natural 
Log and inverse sine,
$$h'(t) = \frac{\sin^{-1}(t)\cdot \frac{2t}{t^2 + 3} - \ln(t^2+3)\cdot \frac{1}{\sqrt{1 - t^2}}}{(\sin^{-1}(t))^2}.$$
\item[(d)] (*) [10 points] Find $y' = \frac{dy}{dx}$ if $3 x^3 y^2 - 7y^3 + \tan(x) = y$.
\item[] {\it Answer:} Differentiating implicitly,
$$\frac{dy}{dx} = \frac{-\sec^2(x) - 9x^2 y^2}{6x^3 y - 21 y^2 - 1}$$ 
\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 $f(x)$ increasing?
\item[] {\it Answer:} Increasing on $(1,4)$.

\item[(b)] On which interval(s) is $y = f(x)$ concave down?
\item[] {\it Answer:} On $(2,6)$.

\item[(c)] How many critical points does $f$ have?  Classify them as 
local maxima, local minima or neither.
\item[] {\it Answer:} One critical point at $x = 4$.  This is a local
maximum for $f(x)$ by the first derivative test.
\end{enumerate}
\vskip 10pt
\noindent
\noindent
7. (*) [20~points] If the ticket price for the upcoming ``Star Wars'' 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 $\$12$ of candy, popcorn and 
drinks at the concession stand.  What is the maximum
revenue that can be earned 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, $12(1000 - x)$
dollars in candy, popcorn, and drinks will be sold, yielding total revenue of 
$$R(x) = (1000 - 10x)(10 + 0.25x) + 12(1000 - 10x) = 22000 + 30x - 2.5x^2$$
We differentiate and set equal to zero to find critical points:
$$R'(x) = 30 - 5x = 0$$
When $x = 6$, and $R''(x) = -5 < 0$, so this is a local and global maximum
for the revenue.  The ticket price should be $\$(10 + 0.25\cdot 6) = \$11.50$
and the maximum revenue is $R(6) = \$22090$.  

\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)] $\displaystyle{f(x) = \frac{1}{x^2 - 16}}$\hfill
 \item[] {\it Answer:} V (vertical asymptotes at $x = \pm 4$, horizontal asymptote at $y = 0$)

  \item[(b)] $\displaystyle{f(x) = \frac{x^2 - 1}{x^2 + 1}}$\hfill 
  \item[] {\it Answer:} IV (no vertical asymptote and horizontal asymptote at $y = 1$)

  \item[(c)] $\displaystyle{f(x) = \frac{x}{x^2 + 1}}$\hfill
  \item[] {\it Answer:} I (no vertical asymptotes, $x$-axis intercept at $x = 0$,
  horizontal asymptote at $y = 0$.)

 

  \item[(d)] $\displaystyle{f(x) =\frac{x}{x^2 - 1}}$ \hfill
  \item[] {\it Answer:} III (vertical asymptotes at $x = \pm 1$, $x$-axis intercept at 
$x = 0$, horizontal asymptote at $y = 0$.)

  \item[(e)] $\displaystyle{f(x) = \frac{3x + 1}{x - 1}}$\hfill
  \item[] {\it Answer:} II (only one vertical asymptote at $x = 1$)

\end{enumerate}


\end{document}
