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

\begin{center}

\textbf{College of the Holy Cross, Fall Semester, 2004} \\
\textbf{Math 131, Midterm 1 Solutions}
\end{center}

\begin{enumerate}
  
\item{} [5~points each]\quad 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.

  \begin{center}
    \input{midterm1a.eepic}
  \end{center}

  (a) $y=f(2x)$: (iii)\hfil 
  (b) $y=\frac{1}{2}f(x)$: (i)\hfil 
  (c) $y=\frac{1}{2}-f(x)$: (ii)

  Graph~(iv) has equation $y=f(\frac{1}{2}x)$.
\bigskip

\begin{center}
  \input{midterm1b.eepic}\hspace*{0.5in}\input{midterm1c.eepic}
\end{center}
\vfil

\begin{center}
  \input{midterm1d.eepic}\hspace*{0.5in}\input{midterm1e.eepic}
\end{center}

\vfil\eject

\item{} [20~points]\quad The desert temperature~$H$ varies
  sinusoidally from a high of~$80^\circ\textrm{F}$ at 5~PM to a low
  of~$40^\circ\textrm{F}$ at 5~AM. Find a formula for~$H$ as a
  function of~$t$, with $t$~measured in hours from 5~PM. You may use
  the graph below for reference; it's a good idea to start by labeling
  the vertical and horizontal scales.

  \begin{center}
    \input{midsoln1f.eepic}
  \end{center}
  
  \bigskip

  \soln Because $t=0$ at a ``crest'' (the temperature is maximum), we
  seek a function of the form $H(t)=D+A\cos(Bt)$ with $A>0$ the
  amplitude of oscillation. The maximum minus minimum is $40=2A$, so
  $A=20$. The vertical shift~$D$ is the halfways point between the
  extremes, so $D=\frac{1}{2}(80+40)=60$. Finally, the period is
  24~hours, so $B=2\pi/24=\pi/12$. In summary
  \[
  \ts H(t) = 60 + 20\cos(\frac{\pi t}{12}).
  \]

  \vfil

  \sep

\item{} [10~points each]\quad In each part, fill in the table as
  indicated.
  
  \bd

\item{(a)} Assuming~$f$ is a \textbf{linear function}. 

  \soln The line through the points $(-1,24)$~and $(1,12)$ has slope
  \[
  m=(24-12)/(-1-1)=12/-2=-6.
  \]
  The point-slope form of the equation is $(y-12)=-6(x-1)$, which
  simplifies to $y=18-6x$. To fill in the table, set $x=5$~and $x=10$:

  \[
    \begin{array}{|c||r|r|r|r|}
      \hline
      x    & \ \ -1 &  \phantom{f()}1 & 5 & 10 \Strut \\
      \hline
      f(x) & 24 & 12 & -12 & -42 \Strut \\
      \hline
    \end{array}
  \]

\vfil\eject

\item{(b)} Assuming~$f$ is an \textbf{exponential function}. 

  \soln Here we are told to assume $f(x)=P_0 a^x$. Using the values
  provided gives two equations,
  \[
  24 = P_0a^{-1},\qquad 12=P_0a.
  \]
  Dividing the first equation by the second cancels~$P_0$:
  $2=24/12=a^{-1}/a=a^{-2}$, or $a=2^{-1/2}$. Next, we plug this value
  into one of the equations (say the first) and solve for~$P_0$: Since
  $24=P_0a^{-1}$, we have $P_0=24a=24/\sqrt{2}=12\sqrt{2}$. Thus
  \[
  f(x) = P_0 a^x = (12\sqrt{2})2^{-x/2}.
  \]
  As above, we fill in the table by setting $x=5$~and $x=10$.
  \[
    \begin{array}{|c||r|r|r|r|}
      \hline
      x    & \ \ -1 &  \phantom{f()}1 & 5 & 10 \Strut \\
      \hline
      f(x) & 24 & 12 & 3 & \frac{3\sqrt{2}}{8}\simeq0.53 \Strut \\
      \hline
    \end{array}
  \]

\ed

\sep

\item{} [10~points]\quad Find the polynomial whose graph is shown;
  express your answer in \textbf{both} factored and expanded
  (multiplied out) form.
\begin{center}
  \input{midterm1g.eepic}
\end{center}

\soln We have $p(x)=k(x+1)^2(x-3)$. Since $p(0)=6$, we find that
$6=k(1)^2(-3)$, or $k=-2$. Thus,
\[
\begin{aligned}
  p(x) &= -2(x+1)^2(x-3) \qquad\text{(factored)} \\
       &= -2(x^2+2x+1)(x-3) \\
       &= -2(x^3-x^2-5x-3) \\
       &= -2x^3+2x^2+10x+6. \qquad\text{(expanded)}
\end{aligned}
\]

\vfil\eject

\item{} [10~points each]\quad Let $y=f(x)=x^2-3$ for $x\geq0$.

  \bd
  
  \item{(a)} Find the inverse function in the form $y=f^{-1}(x)$, and
    find the domain of~$f^{-1}$.
  
    \soln Formally, exchange $x$~and $y$, obtaining $x=y^2-3$, then
    solve for~$y$.  This gives $y=\pm\sqrt{x+3}$. To see whether we
    need a negative sign or not, it's easiest to consider the graph:
    \begin{center}
      \input{midsoln1h.eepic}
    \end{center}
    We conclude that $y=f^{-1}(x)=\sqrt{x+3}$.

  \item{(b)} Suppose $g(x)=x+e^x$. With $f(x)$~as above, find 
    $g\bigl(f(x)\bigr)$.

    \soln $g\bigl(f(x)\bigr)=g(x^2-3)=(x^2-3) + e^{(x^2-3)}$.

  \ed

\sep

\item{} [15~points]\quad An automobile costs \$25,000 and depreciates
  in value by 20\%~per year. How many years pass before the car is
  worth~\$5000? Give both an \textbf{exact answer} (using fractions,
  square roots, logarithms, etc.\ as needed) and a numerical answer
  rounded to two decimal places.

  \soln Let $V(t)$ be the value of the car in thousands of dollars
  after $t$~years. (Note the explicit choice of units. This should be
  your first step when creating a mathematical model!) Because
  $V(0)=25$ and $V$~decreases to $0.8$~times its current value after
  1~year, we have
  \[
  V(t) = 25(0.8)^t.
  \]
  The problem asks for the value~$T$ so that
  $V(T)=25(0.8)^T=5$. Dividing by~25 and taking natural logarithms
  gives
  \[
  (0.8)^T=0.2,\qquad T\ln(0.8)=\ln(0.2),\qquad
  T=\frac{\ln(0.2)}{\ln(0.8)}\simeq7.21~\text{years}
  \]

\end{enumerate}

\end{document}
