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\centerline{\textbf{\large Exam \#1 solutions $\cdot$ Friday, February 2, 2007}}
\vspace*{.2in}
\centerline{MATH 124 $\cdot$ Calculus I $\cdot$ Section 8 $\cdot$ Spring 2007}
\vspace*{0.2in}

\emph{Note:  Some of my solutions are wordy, for the sake of explanation.  All
I expect of you is computation, except for questions which specifically ask for
verbal responses.}

John Kerl (\texttt{kerl.john.r@gmail.com}).

\vspace*{0.1in}

%% ----------------------------------------------------------------
\textbf{Problem 1.}  Let $f(x) = A(x-B)^2(x+C)$ with $A>0$, $B>0$, and $C>0$.
Using the following space, sketch a graph of $f(x)$.  Include and label the
$x$-intercept(s), if any, and the $y$-intercept.  (The $x$-intercepts and
$y$-intercept should be in terms of $A$, $B$, and $C$ --- please do not plug in
specific numerical values for $A$, $B$, and $C$.)

\vspace*{0.2in}
\emph{Solution}:
The function has a double root at $x=B$ and a single root at $x=-C$.  The
$y$-intercept is found by setting $x=0$, namely, $y=AB^2C$.  The leading
coefficient is positive and the function is a cubic polynomial, so $f(x) \to
\infty$ as $x \to \infty$ and $f(x) \to -\infty$ as $x \to -\infty$.
Writing down these things, we have

\begin{center}
\includegraphics{e1_1.eps}
\end{center}

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\textbf{Problem 2.}  Sally's track coach tells her that to avoid injury,
she should increase her weekly mileage by 10\% per week.

\vspace*{0.2in}

\textbf{Part (a).} If she currently runs 2 miles per week, and if she follows
her coach's advice, by how many weeks from now will she be running 25 miles per
week?

\vspace*{0.2in}
\emph{Solution}:
Sally's current mileage is 2 miles per week.  Next week it will be
$2\cdot 1.1$; the week after it will be $2\cdot 1.1^2$, and so on.
After $t$ weeks it will be $2 \cdot 1.1^t$.  When will this be 25?  Solve
for $t$:
\begin{eqnarray*}
	2 \cdot 1.1^t &=& 25 \\
	1.1^t &=& 12.5 \\
	\ln(1.1^t) &=& \ln(12.5) \\
	t \ln(1.1) &=& \ln(12.5) \\
	t &=& \frac{\ln(12.5)}{\ln(1.1)} \approx 26.50.
\end{eqnarray*}
So, her mileage will surpass 25 miles per week after 27 weeks.

\vspace*{0.2in}

\textbf{Part (b).} If she currently runs 2 miles per week, and if she instead
follows her friend's advice and increases her mileage by 1.5 miles per week, by
how many weeks from now will she be running 25 miles per week?

\vspace*{0.2in}
\emph{Solution}:
Sally's current mileage is 2 miles per week.  Next week it will be $2 + 1.5$;
the week after it will be $2 + 1.5\cdot 2$, and so on.  After $t$ weeks it will
be $2 + 1.5t$.  When will this be 25?  Solve for $t$:
\begin{eqnarray*}
	2 + 1.5t &=& 25 \\
	1.5t &=& 23 \\
	t &=& \frac{23}{1.5} \approx 15.33.\\
\end{eqnarray*}
So, her mileage will surpass 25 miles per week after 16 weeks.

%% ----------------------------------------------------------------
\newpage
\textbf{Problem 3.}  Let
$$
	f(z) = \frac{(z-1)(z-2)}{(z-3)(z-4)}.
$$

\vspace*{0.2in}

\textbf{Part (a).}  Find the domain of $f(z)$.

\vspace*{0.2in}
\emph{Solution}:
We can evaluate this function for any input unless a division by zero
happens.  Division by zero happens when the denominator is zero, so we
need to solve
$$
	(z-3)(z-4) = 0
$$
to find all $z$ which are not in the domain of $f(z)$.  Remember that the
product of two real numbers can be zero only when one (or both) of them are
zero.  So, $(z-3)(z-4)=0$ when either $z-3=0$ or $z-4=0$, i.e. $z=3$ or $4$.
The domain of $f(z)$ is all other $z$'s, namely, $z \ne 3,4$.  (You may write
this in interval notation as $(-\infty,3) \cup (3,4) \cup (4, +\infty)$ if you
wish.  The notation is less important to me than the computation.)
\vspace*{0.2in}

\textbf{Part (b).}  Find the zeroes of $f(z)$.

\vspace*{0.2in}
\emph{Solution}:
The zeroes of any function are the input(s), if any, which give zero as
output.  We find these by setting $f(z)=0$ and solving for $z$.  We have
\begin{eqnarray*}
	\frac{(z-1)(z-2)}{(z-3)(z-4)} &=& 0 \\
	(z-1)(z-2) &=& (z-3)(z-4) \cdot 0 = 0.
\end{eqnarray*}
Again, the product of two real numbers is zero when at least one of them is
zero, so either $z-1=0$ or $z-2=0$, i.e. the zeroes of $f(z)$ are $z=1$ and
$2$.
\vspace*{0.2in}

\textbf{Part (c).}  Find the vertical asymptote(s), if any, of $f(z)$.

\vspace*{0.2in}
\emph{Solution}:
Vertical asymptotes occur when the numerator is not zero but the denominator is
zero\footnote{In fact, they occur when the multiplicity of a zero of the
denominator exceeds the multiplicitiy of a zero in the numerator.
For example,
$(z-1)^2/(z-1)^3$ has a vertical asymptote at $z=1$;
$(z-1)^3/(z-1)^2$ has a hole there.
However,
I've not discussed this generality in class.}.
In part (a) we found the zeroes of the denominator, namely, $z=3$ and $4$.
Neither of these are zeroes of the numerator, so $f(z)$ has
vertical asymptotes at $z=3$ and $4$.

\vspace*{0.2in}

\textbf{Part (d).}  Find the horizontal asymptote(s), if any, of $f(z)$.

\vspace*{0.2in}
\emph{Solution}:
The horizontal asymptote of a rational function is the ratio of its leading
coefficients, namely, $y=1/1=1$.  Alternatively, graph the rational function
and keeping zooming out in the horizontal direction.  Alternatively, evaluate
the function for very large positive $z$ and very large negative $z$ and see
what the function values approach.
\vspace*{0.2in}

\textbf{Part (e).}  Find the hole(s), if any, of $f(z)$.

\vspace*{0.2in}
\emph{Solution}:
By the discussion in part (c), $f(z)$ has no holes.  We would need the
numerator and the denominator to have a common zero; they do not.
\vspace*{0.2in}

%% ----------------------------------------------------------------
\newpage
\textbf{Problem 4.}  In your physics class, you attach an oscillosope to a
piece of electronic equipment and discover that the voltage you observe looks
like a sine wave.  In particular, you notice that
\begin{itemize}
\item The maximum voltage you observe is 1.8 volts.
\item The minimum voltage you observe is 0.0 volts.
\item This maximum voltage is found at the start of your data (i.e. $t=0.0$
	picoseconds); the first time it reoccurs is 2.0 picoseconds later (i.e.
	$t=2.0$ picoseconds).
\end{itemize}
Find an equation of the form $a+b\cos(ct)$ that describes the voltage $V$ as a
function of time $t$ in picoseconds.  (You don't need to know how many
picoseconds make a second.) (Hint:  You might want to sketch a graph.)

\vspace*{0.2in}
\emph{Solution}:
Sketching a graph as suggested, we have

\includegraphics{e1_4.eps}

From the graph we can read off the following data:

\begin{itemize}
\item Amplitude = 0.9V.
\item Vertical shift = 0.9V.
\item Period = 2.0 ps.
\item Horizontal shift = 0.
\end{itemize}

From data, we can write down an equation:
\begin{eqnarray*}
	V &=& \textrm{vertical shift} + \textrm{amplitude} \cdot
		\cos\left(
		{\frac{2\pi}{\textrm{period}}(t - \textrm{horizontal shift)}}\right) \\
	&=& 0.9 + 0.9 \cos\left({\frac{2\pi}{\textrm{2.0}}(t - 0)}\right) \\
	&=& 0.9 + 0.9 \cos(\pi t).
\end{eqnarray*}
You can (and should) graph this equation in your calculator to make sure
it looks like the given graph.

\vspace*{0.2in}

%% ----------------------------------------------------------------
\newpage
\textbf{Problem 5.}  In a lab experiment, you have microorganisms living in a
petri dish.  Assume that the microorganisms are dying at an exponential rate.
You have the following data:
\begin{itemize}
\item You forgot to count the initial population.
\item You count a population of 2500 organisms on day one.
\item You count a population of 1500 organisms on day two.
\end{itemize}
Find an exponential model of the form $y=Ce^{rt}$ for these data, where $y$ is
the number of microorganisms in the dish after $t$ days.  That is, find $C$ and
$r$ so this model matches the given data.

\vspace*{0.2in}
\emph{Solution}:
We are given $y=2500$ at $t=1$ and $y=1500$ at $t=2$.
Writing these down, we have
\begin{eqnarray*}
	\left\{ \begin{array}{lll}
	2500 &=& Ce^r\\
	1500 &=& Ce^{2r}.
	\end{array}\right.
\end{eqnarray*}
This is two equations in two unknowns, so we need to eliminate a variable.  We
can solve for $C$ in one equation and plug it into the other.  (Some students
prefer to divide one equation by the other.  This works too.) From the first
equation,
$$C=2500e^{-r}.$$
Plugging into the second, we have
\begin{eqnarray*}
	1500 &=& 2500 e^{-r}e^{2r} \\
	e^{r} &=& \frac{1500}{2500} = 0.6 \\
	\ln(e^{r}) &=& \ln(0.6) \\
	r &=& \ln(0.6) \approx -0.5108.
\end{eqnarray*}
Since we know $C=2500e^{-r}$ and $r = -0.5108$, we have
$$
	C=2500e^{0.5108} \approx 4166.
$$
In summary, we have found
$$
	y = 4166e^{-0.5108t}.
$$

\vspace*{0.2in}

%% ----------------------------------------------------------------
\newpage
\textbf{Problem 6.}  You leave on a car trip for San Diego.
The following events happen:
\begin{itemize}
\item[(i)] You drive at medium speed (in town) for the first 15 minutes.
\item[(ii)] Once you get out of town, you accelerate almost immediately
	to the speed limit and drive at that speed for 45 minutes.
\item[(iii)] Then, you hit a traffic jam.  You come to a stop almost
	immediately and sit at a stop for 30 minutes.
\item[(iv)] Once past the accident, you accelerate almost immediately back
	to the speed limit and drive at that speed until you reach the coast 4
	hours later.
\end{itemize}
Sketch a graph of distance $D$ from your house vs.
time $t$ in hours.  Label all the above events on the $t$ axis.

\vspace*{0.2in}
\emph{Solution}:
\vspace*{0.2in}

\includegraphics{e1_6.eps}
\vspace*{0.2in}

%% ----------------------------------------------------------------
\newpage
\textbf{Problem 7.}  The following graph shows the altitude $y$ of a rocket
(in miles) as a function of time $t$ since launch (in seconds):

\includegraphics{rocketfig.eps}

\textbf{Part (a).}  Estimate $f(18)$, and describe verbally what it means to
you.  Include units in your response.

\vspace*{0.2in}
\emph{Solution}:
From the graph, $f(18) \approx 18$.  This means that the rocket is 18 miles
high after 18 seconds.
\vspace*{0.2in}

\textbf{Part (b).}  Estimate $f'(18)$, and describe verbally what it means to
you.  Include units in your response.

\vspace*{0.2in}
\emph{Solution}:
Sketching a tangent line to $t=18$ in red, I see something like

\includegraphics{rocketfig2.eps}

The slope of this line looks like about 1.2; let's call it 1.  This must be
miles per second; it's the (vertical) speed of the rocket after 18 seconds.

\vspace*{0.2in}

\textbf{Part (c).}  Estimate $f^{-1}(40)$, and describe verbally what it means
to you.  Include units in your response.

\vspace*{0.2in}
\emph{Solution}:
From the graph, it looks like $f^{-1}(40)=26$, i.e. the rocket is 40 miles
high after about 26 seconds of flight.
\vspace*{0.2in}

%% ----------------------------------------------------------------
\newpage
\textbf{Problem 8.}  Let $f(x) = 2x^2+1$.
Evaluate
$$
	\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}.
$$
using properties of limits.

\vspace*{0.2in}
\emph{Solution}:
\begin{eqnarray*}
	\lim_{h\to 0}\frac{f(a+h)-f(a)}{h} &=&
		\lim_{h\to 0}\frac{[2(a+h)^2+1]-[2a^2+1]}{h} \\
	&=& \lim_{h\to 0}\frac{[2(a^2+2ah+h^2)+1]-[2a^2+1]}{h} \\
	&=& \lim_{h\to 0}\frac{[2a^2+4ah+2h^2+1]-[2a^2+1]}{h} \\
	&=& \lim_{h\to 0}\frac{2a^2+4ah+2h^2+1-2a^2-1}{h} \\
	&=& \lim_{h\to 0}\frac{4ah+2h^2}{h} \\
	&=& \lim_{h\to 0} (4a+2h) \\
	&=& 4a.
\end{eqnarray*}

\vspace*{0.2in}

%% ----------------------------------------------------------------
\newpage
\textbf{Problem 9.}  The following are selected values of a function $f(x)$.
$$
	f(3) = \textrm{ is undefined} \qquad \qquad
	f(4) = -6 \qquad \qquad
	f(6) = 9
$$
$$
\begin{tabular}{|r|r|r|r|}
\hline \multicolumn{2}{|c|}{Table A} \\
\hline $x$ & $f(x)$ \\
\hline
\hline $     2.9    $ & $    -8.14286$ \\
\hline $     2.99   $ & $   -98.01493$ \\
\hline $     2.999  $ & $  -998.00150$ \\
\hline $     2.9999 $ & $ -9998.00015$ \\
\hline
\end{tabular}
\hspace*{0.2in}
\begin{tabular}{|r|r|r|r|}
\hline \multicolumn{2}{|c|}{Table B} \\
\hline $x$ & $f(x)$ \\
\hline
\hline $     3.1    $ & $   -12.15789$ \\
\hline $     3.01   $ & $  -102.01508$ \\
\hline $     3.001  $ & $ -1002.00150$ \\
\hline $     3.0001 $ & $-10002.00015$ \\
\hline
\end{tabular}
\hspace*{0.2in}
\begin{tabular}{|r|r|r|r|}
\hline \multicolumn{2}{|c|}{Table C} \\
\hline $x$ & $f(x)$ \\
\hline
\hline $     3.9    $ & $    -5.56566$ \\
\hline $     3.99   $ & $    -5.95070$ \\
\hline $     3.999  $ & $    -5.99501$ \\
\hline $     3.9999 $ & $    -5.99950$ \\
\hline
\end{tabular}
\hspace*{0.2in}
\begin{tabular}{|r|r|r|r|}
\hline \multicolumn{2}{|c|}{Table D} \\
\hline $x$ & $f(x)$ \\
\hline
\hline $     4.1    $ & $     6.57576$ \\
\hline $     4.01   $ & $     6.05071$ \\
\hline $     4.001  $ & $     6.00501$ \\
\hline $     4.0001 $ & $     6.00050$ \\
\hline
\end{tabular}
$$
$$
\begin{tabular}{|r|r|r|r|}
\hline \multicolumn{2}{|c|}{Table E} \\
\hline $x$ & $f(x)$ \\
\hline
\hline $     5.9    $ & $     6.32184$ \\
\hline $     5.99   $ & $     6.49616$ \\
\hline $     5.999  $ & $     6.49956$ \\
\hline $     5.9999 $ & $     6.49999$ \\
\hline
\end{tabular}
\hspace*{0.2in}
\begin{tabular}{|r|r|r|r|}
\hline \multicolumn{2}{|c|}{Table F} \\
\hline $x$ & $f(x)$ \\
\hline
\hline $     6.1    $ & $     6.63196$ \\
\hline $     6.01   $ & $     6.59234$ \\
\hline $     6.001  $ & $     6.50034$ \\
\hline $     6.0001 $ & $     6.50002$ \\
\hline
\end{tabular}
\hspace*{0.2in}
\begin{tabular}{|r|r|r|r|}
\hline \multicolumn{2}{|c|}{Table G} \\
\hline $x$ & $f(x)$ \\
\hline
\hline $   100$ & $     1.05285$ \\
\hline $  1000$ & $     1.00503$ \\
\hline $ 10000$ & $     1.00050$ \\
\hline $100000$ & $     1.00005$ \\
\hline
\end{tabular}
\hspace*{0.2in}
\begin{tabular}{|r|r|r|r|}
\hline \multicolumn{2}{|c|}{Table H} \\
\hline $x$ & $f(x)$ \\
\hline
\hline $   -100$ & $     2.95257$ \\
\hline $  -1000$ & $     2.99503$ \\
\hline $ -10000$ & $     2.99950$ \\
\hline $-100000$ & $     2.99995$ \\
\hline
\end{tabular}
$$
\vspace*{0.2in}

\textbf{Part (a).} Find the following limits, when they exist.  If a limit does
not exist, say why not.  For each, state which table(s) (A, B, C, D, E, F, G,
H) you use to support your answer.

\vspace*{0.2in}

\begin{itemize}

\item $\lim_{x \to 3}f(x)$:  This does not exist.  As $x\to 3$ from either side
(tables A and B), $f(x)$ goes more and more negative.  This looks like a
vertical asymptote at $x=3$.  (Some people would say that the limit does exist,
and is $-\infty$.)

\item $\lim_{x \to 4}f(x)$: The left-hand limit (table C) looks like $-6$; the
right-hand limit (table D) looks like $6$.  Since those two don't agree, the
limit does not exist.

\item $\lim_{x \to 6^-}f(x)$:  This looks like $6.5$ (table E).

\item $\lim_{x \to -\infty}f(x)$:  From table $H$, it looks like $f(x) \to 3$
as $x \to -\infty$.

\end{itemize}

\vspace*{0.2in}

\textbf{Part (b).} Determine if $f(x)$ is continuous or not continuous
at the following values.  If $f(x)$ is not continuous at the point,
describe what feature occurs there.

\begin{itemize}

\item $x = 3$:  The function isn't defined there, so it can't be continuous
there.  It looks like a vertical asymptote.

\item $x = 4$:  The function value and the left-hand limit are $-6$, which
disagrees with the right-hand limit $6$.  So, the function is not continuous
there.  It has a jump.

\item $x = 6$:  The left-hand and right-hand limits are both 6.5, but the
function value at $x=6$ is 9.  The function is not continuous there.  It has a
jump.

\end{itemize}



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