%% Math 131 Midterm 1, Fall 2004
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\begin{center}

\textbf{College of the Holy Cross, Fall Semester, 2017} \\
\textbf{MATH 133, Midterm 1 Solutions} \\
\textbf{Friday, September 22}
\end{center}


\vskip 10pt
\begin{enumerate}

\item{} 
\begin{enumerate}
\item[(a)] (10) Express the set of all $x$ satisfying $|3x - 9| \le 6$ as an interval or union of intervals.
\vskip 10pt
{\it Solution:}  Algebraically, this inequality says $-6 \le 3x - 9 \le 6$, so $3 \le 3x \le 15$, so $1 \le x \le 5$.
As an interval this is $[1,5]$.  
Geometrically, we could get the same result by dividing by $3$ to get $|x - 3| \le 2$.
The numbers at distance at most $2$ along the number line from $3$ are exactly $x \in [1,5]$ as before.  
\vskip 10pt 
\item[(b)] (10) What is the domain of the function $f(x) = \displaystyle{ \frac{\sqrt{3 - x}}{x}}$?  Any correct form is OK.
\vskip 10pt
{\it Solution:}  We must have $3 - x \ge 0$, or $x \le 3$ for the square root to 
be defined.  We must also have $x\ne 0$ since we cannot divide by zero.  These conditions define
$(-\infty,0) \cup (0,3]$.  Another correct way to say this:  all real $x \le 3$, except $x = 0$.  
\end{enumerate}
\vskip 10pt
\item{} (20) The graph $y=f(x)$ and four graphs obtained by transforming it
  are shown in Figures 1 and 2. Match the given formulas with the corresponding graph.
  

 
  (a) $y=f(\frac{1}{2}x)$:$\underline{B}$ (horizonal stretching)  \hfil 
  (b) $y=\frac{1}{2}f(x)$:$\underline{D}$ (vertical compression)  \hfil 
  (c) $y=f(2x)$:$\underline{C}$ (horizontal compression) \\
  (d) Note that there is an extra graph that does not match any of the formulas.  A is the 
Graph that does not match any formula.  (It is actually $y = 2f(x)$ (vertical stretching)).


\vskip 10pt
\item{}
\begin{enumerate}
\item[(a)] (15) Complete the square:  $q(x) = 4x^2 - 16x + 24$.
\vskip 10pt
{\it Solution:}  By the usual process, 
$$q(x) = 4(x^2 - 4x + 6) = 4((x - 2)^2 + 2) = 8 + 4(x - 2)^2.$$
\item[(b)] (5) What is the minimum value of $q(x)$?
{\it Solution:}  The smallest value attained by $q(x)$ is $8$ since $4(x - 2)^2 \ge 0$ for all $x$.  
\end{enumerate}

\vskip 10pt
\item{} The temperature~$f(t)$ at a desert location varies
  sinusoidally from a low of~$40^\circ\textrm{F}$ at $t = 0$ hours (midnight) to a high
  of~$80^\circ\textrm{F}$ at $t = 12$ hours (noon).  (See graph in Figure 3, which shows
the temperature over two complete days.)
\begin{figure}
\begin{center}
\includegraphics[width=40mm]{Orig.png}
\end{center}
\caption{The graph $y = f(x)$}
\end{figure}

\begin{figure}
\centering
\begin{tabular}{cc}
(A) \includegraphics[width=39mm]{M1PA.png}&(B) \includegraphics[width=39mm]{M1PD.png}\\
(C) \includegraphics[width=39mm]{M1PC.png}&(D) \includegraphics[width=39mm]{M1PB.png}
\end{tabular}
\caption{The transformed graphs}
\end{figure}
  \begin{figure}
\begin{center}
\includegraphics[width=3.0in]{Sinus.png}
\end{center}
\caption{Graph of temperature as a function of time.  }
\end{figure} 
\vglue 10pt
\begin{enumerate}
\item[(a)] (6)  What is the period of this sinusoidal oscillation? {\it Answer:} 24 hours \hfil
\item[(b)] (6)  What is the amplitude? {\it Answer:}  $\frac{1}{2}(80 - 40) = 20$ \hfil
\item[(c)] (8) Give a possible formula for $f$ as a function of $t$.
\vskip 10pt
{\it Solution:}  Since the period starts at one of the minimum values, this looks like a cosine
graph, with horizontal and vertical scaling, plus a vertical shift and reflection 
across the $x$-axis.  The vertical shift can be 
found from the difference between the maximum and the amplitude:  $80 - 20 = 60$.  One form is
$$f(t) = -20\cos\left(\frac{2\pi t}{24}\right) + 60.$$
Another possibility (with horizontal shifting too by 6 hours to the right) would be 
$$f(t) = 20 \sin\left(\frac{2\pi (t - 6)}{24}\right) + 60.$$
\end{enumerate}
0
\item{}  You are traveling by horse along a straight line road starting from $x = 0$ (miles) 
at time $t = 0$ (hours).  For the first hour, you move in the positive $x$-direction at 5 miles per hour.  
At $t = 2$, you realize you have dropped an important item from your saddle bag.  So you turn around and 
retrace your steps to 5 miles per hour.  You find the item at $t = 3$.  Then you turn back around
and continue at 5 miles per hour for an additional 2 hours.  
\vskip 10pt
\begin{enumerate}
\item[(a)] (10) Sketch the graph of your position $x$ as a function of time $t$ for $0 \le t \le 5$.  
\item[] {\it Solution:}  The graph is shown in Figure 4 on the next page.  
\begin{figure}
\begin{center}
\includegraphics[width=.5\linewidth]{Trip.png}
\end{center}
\caption{Figure for Question 6 (a)}
\end{figure}
\item[(b)]  (10) Give your position $x$ as a piecewise-defined function of $t$.
\end{enumerate}
\vskip 10pt
{\it Solution:} Using the point-slope form for equation of lines, this is the function
$$x(t) = \begin{cases}
             5t & {\rm if}\ 0 \le t \le 2\\
             -5t +20 & {\rm if}\ 2 \le t \le 3\\
             5t - 10 & {\rm if}\ 3 \le t \le 5.
\end{cases}
$$
\item{} 
Extra Credit. (10)  A sample of a radioactive element is decaying over time.  The mass present
at time $t$ is given by 
according to $M(t) = 139 e^{-0.003 t}\ {\rm grams}$, where $t$ is in months.  When will the mass 
present reach 50 grams?
\vskip 10pt
{\it Solution:}  We want to solve the equation 
$$50 = 139 e^{-0.003 t}$$
for $t$.  Divide by $139$
first to get 
$$e^{-0.003 t} = \frac{50}{139}.$$  
Then taking natural logs of both sides gives
$$-0.003 t = \ln\left(\frac{50}{139}\right) \Rightarrow t = \frac{\ln\left(\frac{50}{139}\right)}{-0.003}
\doteq 340.8\ {\rm months}.$$
\end{enumerate}

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