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\begin{document}
\centerline{MATH 135 -- Calculus 1}
\centerline{Exam 3 Practice Problems}
\centerline{November 20, 2019}
\vskip 10pt
\noindent
{\it Disclaimers}
\vskip 10pt
The actual exam problems may be worded differently and might
combine topics shown here.  Also, the actual exam will be
significantly shorter than this list of practice problems.
The purpose of this list is to show the range of topics that
might be covered.  
\begin{enumerate}
\item[I.]  Find the indicated derivative(s) and simplify.
\vskip 10pt
\begin{enumerate}
\item[(A)]  $y'$ for 
$$y = \ln(x)\left(x^7 - \frac{4}{\sqrt{x}}\right)$$
\item[(B)]  $y'$ for 
$$y = \sin^{-1}(e^{2x} + 2)$$
\item[(C)] $y'$ for 
 $$y = \frac{\ln(x + 1)}{3x^4 - 1}$$
\item[(D)] $y'$ for 
 $$y = \frac{\sin(x)}{1 + \cos(x)}$$
\item[(E)]  $y'$ for 
$$y = \tan^{-1}(x^2+x)$$
\item[(F)]  Find $\frac{dy}{dx}$ using implicit differentiation: $$xy^2 - 3y^3 + 2x^4 - 4xy = 2$$
\item[(G)]  Find the equation of the line tangent to the curve from (F) at $(x,y) = (1,0)$
\item[(H)]  Find
$$\frac{d}{dx}\left(5x\sqrt{x}-\frac{2}{x^3}+ 11x - 4\right)$$
\item[(I)]
$$\frac{d}{dt}\left(\frac{t^2e^{3t}}{t^4 + 1}\right)$$
\item[(J)]
$$\frac{d^2}{dz^2}\left( \frac{z^2 - 2z + 4}{z^2 + 1}\right)$$
\item[(L)] 
$$\frac{d}{dx}\left( \sin(x)\left(x^7 - \frac{4}{\sqrt{x}}\right)\right)$$
\item[(M)] Find $y'$ (note this is just another way of asking the same question!)
$$y = (e^{2x} + 2)^3$$
\item[(N)]  Find $y'$ and $y''$ 
$$y = \frac{x + 1}{3x^4 - 1}$$
\item[(O)] Find $y'$
$$y = \frac{\sin(x)}{1 + \cos(x)} + x^2 \cos(x^3 + 3)$$
\end{enumerate}
\item[II.] The total cost (in $ \$ $) of repaying a car loan at 
interest rate of $r \%$ per year is $C = f(r)$.  
\begin{enumerate}
\item[A)] What is the meaning of the statement $f(7) = 20000$?
\item[B)] What is the meaning of the statement $f'(7) = 3000$? What are the units of $f'(7)$? 
\end{enumerate}
\item[III.] The quantity of a reagent present in a chemical 
reaction is given by $Q(t) = t^3 - 3t^2 + t + 30$ grams at time $t$ seconds for all $t \ge 0$.
(Note:  For a question like this, I could also give you the plot of the function and ask
questions like those below.  In this case you need to start from the formula and compute
$Q'(t)$; if you were given the graph, you need to make the connection between slopes of 
tangent lines and signs of $Q'(t)$ visually.)
\begin{enumerate}
\item[(A)]  Over which intervals with $t \ge 0$ 
is the amount increasing? (i.e. $Q'(t) > 0$) decreasing (i.e. $Q'(t) < 0$)?
\item[(B)]  Over which intervals is the rate of change of $Q$ increasing? decreasing?
decreasing? 
\end{enumerate}
\item[IV.]  Compute the following limits using L'Hopital's Rule or other methods,
as appropriate.
\begin{enumerate}
\item[(A)] $\displaystyle \lim_{x\to 0^+} x^{1/3}\ln(x)$
\item[(B)] $\displaystyle \lim_{x\to \infty} \frac{x^3}{e^{2x}}$
\item[(C)] $\displaystyle \lim_{x\to 1} \frac{x^2 - 5x + 4}{x^2 - 3x + 2}$
\item[(D)] $\displaystyle \lim_{x\to \infty} \left(1 + \frac{4}{x}\right)^x$
\end{enumerate}
\vskip 10pt
\item[V.]   All parts of this question refer to $f(x) = 4x^3 - x^4$.
\vskip 10pt
\begin{enumerate}
\item[(A)]  Find and classify all the critical points of $f$
using the First Derivative Test.
\item[(B)]  Over which intervals is the  graph $y = f(x)$ concave
up?  concave down?
\item[(C)]  Sketch the graph $y = f(x)$.  
\item[(D)]  Find the absolute maximum and minimum of $f(x)$
on the interval $[1,4]$.  
\end{enumerate}
\vskip 10pt
\item[VI.]  All three parts of this question refer to the function $f(x)$
\emph{whose derivative is plotted in Figure 1.  NOTE:  This is the graph $y = f'(x)$
not $y = f(x)$.}
\begin{figure}
\begin{center}
\includegraphics[width=100mm]{Ex4SamplePlot.png}
\caption{Plot of $y = f'(x)$ for Problem VI}
\end{center}
\end{figure}
\begin{enumerate}
\item[(A)]  Give approximate values for all the critical points of $f(x)$
in the interval shown, and say whether $f$ has a
local maximum, a local minimum, or neither at each.
\item[(B)]  Find approximate values for all the inflection points of $f(x)$.
\item[(C)]  Over which intervals is $y = f(x)$ concave up? concave down?
\end{enumerate}
\vskip 10pt
\item[VII.]   A town wants to build a pipeline from a water station on a small
island 2 miles off the shore of its water reservoir to the town. 
One possible route is shown dotted in red. The town is 6 miles along the shore
from the point nearest the island.  It costs \$3 million   
per mile to lay pipe under the water and \$2 million per mile
to lay pipe along the shoreline.
\begin{figure}
\begin{center}
\includegraphics[width=2.5truein,height=2.5truein]{FVI.png} 
\end{center} 
\caption{Figure for problem VII.}
\end{figure}

\begin{enumerate} 
\item[A)] Give the cost $C(x)$ of constructing the pipeline
as a function of $x$.
%\item[] {\it Solution:}  By the Pythagorean theorem and the given 
%information about cost per mile, we have 
%$$C(x) = 3\sqrt{4 + x^2} + 2(6 - x)$$
\item[B)] Where along the 
shoreline should the pipeline hit land to minimize the costs
of construction? 
%\item[] {\it Solution:} To find the minimum of $C(x)$, we
%can restrict to $x$ in the closed interval $[0,6]$, since it clearly does no
%good to take $x < 0$ or $x > 6$.  The function $C(x)$ has
%a critical number for $x > 0$ at the positive solution of $C'(x) = 0$:
%\begin{eqnarray*}
%0 &=& \frac{3x}{\sqrt{4 + x^2}} - 2, \text{or}\\
%3x &=& 2 \sqrt{4 + x^2} \\
%9x^2 &=& 16 + 4x^2 \\
%5 x^2 &=& 16 \\
%x &=& \frac{4}{\sqrt{5}} \doteq 1.79.
%\end{eqnarray*}
%We have $C(0) = 18$, $C(6) = 3\sqrt{40} \doteq 19.0$, and
%$C\left(\frac{4}{\sqrt{5}}\right) \doteq 16.47$.  So 
%the minimum cost is attained at $x = \frac{4}{\sqrt{5}}\doteq 1.79$ miles.
\end{enumerate}
\end{enumerate}
\end{document}
