%% Math 131, Fall, 2003
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\begin{document}

\noindent{\bf Name}:\ \vbox{\hrule width 3in}

\begin{center}
\textbf{Holy Cross College, Fall Semester, 2003} \\
\textbf{Math 131, Quiz \#1} \\
Professor Hwang \\
\end{center}

\be
\item The general exponential function has formula $P=f(t)=P_0\,a^t$,
  with $a>0$. Suppose~$f$ is the exponential function satisfying
  $f(2)=3$~and $f(4)=12$.

  \bd
  \item{(a)} Find $P_0$~and $a$.
    \medskip

    \textbf{Solution}: The equation $f(2)=3$ says that $P_0\,a^2=3$.
    Similarly, $f(4)=12$ means $P_0\,a^4=12$. Divide the second
    equation by the first:
    \[
    4 = \frac{12}{3} = \frac{P_0\,a^4}{P_0\,a^2} 
      = \frac{a^4}{a^2} = a^2,
    \]
    or $a=2$. Since $P_0\,a^2=3$, we find that $4P_0=3$, or $P_0=3/4$.

 \vfil

    \MyAnswer{$P_0=$}{\qquad3/4\qquad}\MyAnswer{$a=$}{\qquad2\qquad}

    \begin{center}
      \hrule
    \end{center}

  \item{(b)} Find the average rate of change of~$f$ on the
    interval~$[3,4]$. Hint: First find $f(3)$.
    \medskip

    \textbf{Solution}: From part~(a), we know that
    \[
    f(t) = \textstyle\frac{3}{4}\cdot2^t,
    \]
    so $f(3) = \frac{3}{4}\cdot2^3 = \frac{3}{4}\cdot8 = 6$. The 
    average rate of change over the interval~$[3,4]$ is
    \[
    \frac{f(4) - f(3)}{4 - 3} = \frac{12 - 6}{4 - 3} 
    = \frac{6}{1} = 6.
    \]
    
    Note that incrementing~$t$ by~1 doubles the function values:
    $f(2)=3$, $f(3)=6$, $f(4)=12$, etc.

    \vfil
    \MyAnswer{Average rate of change on~$[3,4]$ is}{\qquad6\qquad}
  \ed
\ee
\eject

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