%% Math 131 Midterm 1, Fall 2004
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\textbf{College of the Holy Cross, Fall Semester, 2015} \\
\textbf{MATH 133, section 1, Midterm 1 Solutions} \\
\textbf{Thursday, September 24}
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\noindent
{\it Note:} a technical issue made it impossible to reproduce all of the graphs together in 
this document.  The exam papers are also posted on the course homepage in case you
want to look at the graphs that were given.
\vskip 10pt
\begin{enumerate}

\item{} 
\begin{enumerate}
\item[(a)] (7.5) Express the set of all $x$ satisfying $|4x - 12| \le 8$ as an interval or union of intervals.
\vskip 10pt
{\it Solution:}  Algebraically, this inequality says $-8 \le 4x - 12 \le 8$, so $4 \le 4x \le 20$, so $1 \le x \le 5$.
As an interval this is $[1,5]$.  Geometrically, we could get the same result by dividing by $4$ 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)] (7.5) What is the domain of the function $f(x) = \frac{1}{x \sqrt{4 - x}}$?  Any correct form is OK.
\vskip 10pt
{\it Solution:}  We must have $x \ne 0$ and $4 - x > 0$, or $x < 4$  The $x$'s satisfying both conditions form
$(-\infty,0) \cup (0,4)$.  Another correct way to say this:  all real $x < 4$, except $x = 0$.  
\end{enumerate}
\vskip 10pt
\item{} (15) The graph $y=f(x)$ and four graphs obtained by transforming it
  are shown. Match the given formulas with the corresponding graph.
  Note that there is an extra graph that does not match any of the formulas.
\begin{enumerate}
\item[(a)] $y=f(\frac{1}{2}x)$ is graph (iv) -- a horizontal stretch by a factor of 2 
\item[(b)] $y=\frac{1}{2}f(x)$: is graph (i) -- a vertical compression by a factor of 1/2  
\item[(c)] $y=f(2x)$: is graph (iii).  
\end{enumerate}
(The remaining graph (ii) is actually $y = \frac{1}{2} - f(x)$.)

\vskip 10pt
\item{}
\begin{enumerate}
\item[(a)] (10) Complete the square:  $q(x) = 3x^2 + 18x + 36$.
\vskip 10pt
{\it Solution:}  By the usual process, 
$$q(x) = 3(x^2 + 6x + 12) = 3((x + 3)^2 + 3) = 9 + 3(x + 3)^2.$$
\item[(b)] (5) What is the minimum value of $q(x)$?
\vskip 10pt
{\it Solution:}  The smallest value attained by $q(x)$ is $9$ since $3(x + 3)^2 \ge 0$ for all $x$.  
\vskip 10pt 
\item[(c)] (5) Using your answers from parts (a) and (b), determine the \emph{range} of the function 
$r(x) = \sqrt{q(x)} = \sqrt{3x^2 + 18x + 36}$.
\vskip 10pt
{\it Solution:}  Since $q(x) \ge 9$ for all $x$, $\sqrt{q(x)} \ge 3$ for all $x$.  Moreover, all values
$\ge 3$ are obtained for some $x$, so the range is the interval $[3,\infty)$.    
\end{enumerate}

\vskip 10pt
\item{} The temperature~$H$ in a desert varies
  sinusoidally from a high of~$80^\circ\textrm{F}$ at 5~PM to a low
  of~$40^\circ\textrm{F}$ at 5~AM.  (See graph below)
  
\vglue 10pt
\begin{enumerate}
\item[(a)] (5)  What is the period of this sinusoidal oscillation? {\it Answer:} 24 hours \hfil
\item[(b)] (5)  What is the amplitude? {\it Answer:}  $\frac{1}{2}(80 - 40) = 20$ \hfil
\item[(c)] (10)  If $t = 0$ is the first 5PM time shown, give a possible formula for $H$ as a function
of $t$.
\vskip 10pt
{\it Solution:}  Since the period starts at one of the maximum values this looks like a cosine
graph, with horizontal and vertical scaling, plus a vertical shift.  The vertical shift can be 
found from the average of the minimum and maximum values: $\frac{40 + 80}{2} = 60$.  One form is
$$H(t) = 20 \cos\left(\frac{2\pi t}{24}\right) + 60.$$
Another possibility would be 
$$H(t) = 20 \sin\left(\frac{2\pi (t + 6)}{24}\right) + 60.$$
\end{enumerate}

\item{}
\begin{enumerate}
\item[(a)] (7.5)  Solve for $x$:  $4^{x + 1} = 8^{x - 3}$.
\vskip 10pt
{\it Solutions:}
The basic method:  Take $\ln$ of both sides to get 
$$(x + 1) \ln(4) = (x - 3) \ln(8).$$  
Then isolate the $x$ with basic algebra:
$$x(\ln(4) - \ln(8)) = -3\ln(8) - \ln(4).$$
So
$$x = \frac{-3\ln(8) - \ln(4)}{\ln(4) - \ln(8)}.$$  
This is 
one correct form, but it also can be simplified.  Since $4 = 2^2$ and $8 = 2^3$, this simplifies to 
$$x = \frac{-11\ln(2)}{-\ln(2)} = 11.$$
\vskip 10pt
The clever method:  Using laws of exponents, rewrite $4=2^2$ and $8 = 2^3$, so the equation becomes
$$2^{2(x + 1)} = 2^{3(x - 3)}.$$
This implies $2(x + 1) = 3(x - 3)$ or $2x + 2 = 3x - 9$, so $x = 11$.  (!) 
\vskip 10pt
\item[(b)] (7.5) 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}

\item{}  You are traveling by donkey 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)] (7.5) Sketch the graph of your position $x$ as a function of time $t$ for $0 \le t \le 5$.  
\begin{figure}
\begin{center}
\includegraphics[width=.5\linewidth]{Trip.png}
\end{center}
\caption{Figure for Question 6 (a)}
\end{figure}
\item[(b)]  (7.5) 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
$$x(t) = \begin{cases}
             5t & {\rm if}\ 0 \le t \le 2\\
             -5t +20 & {\rm if}\ 2 \le t \le 3\\
             5t - 10 & {\rm if}\ 3 \le t \le 5.
\end{cases}
$$

\vfil\eject
\end{enumerate}


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