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{\LARGE{
\hfil Math 120 - Autumn 2015 \hfil \pp
\hfil Final Exam\hfil  \pp
\hfil Answers \hfil \pp
}}

\begin{enumerate}

\item $82.62$

\item (a) $f(x)=1-2e^{-3x}$.

$f^{-1}(x) = -\frac{1}{3} \ln \left( \frac{1-x}{2} \right)$.

The domain of $f$ is all real numbers, and its range is all real
numbers $<1$.

The domain of $f^{-1}$ is all real numbers $<1$, and its range is all real numbers.

(b) 

\includegraphics[width=6in]{problem2b.png}


\item (a) $x=-\frac{5}{3}t+12, y=\frac{2}{3}t$.

(b) $x=-\frac{5}{3}t+6, y=-\frac{5}{3}t+13$.

(c) $\frac{39}{7} \approx 5.57$ seconds.


\item $3$ years.

\item The trees are 
\[
\frac{100}{\tan(78^\circ) - \tan(64^\circ)} \approx 37.67 \text{ feet apart.}
\]



\item (a) 
\[
A(x) = \begin{cases}
-0.25x^2+54x & \text{ if } 0 \le x \le 48, \\
2241-(63-x)^2 & \text{ if } 48 < x \le 63.
\end{cases}
\]
(b) Cut should be made at $x = 108-2\sqrt{1795.5} \approx 23.25$ cm.

\item The populations will be the same in year $558$.

\item (a) Height is $15-6\sqrt{3}$.

(b) Distance = 
\[
\sqrt{ \left( 12-12\cos \frac{2\pi}{3} \right)^2+ 
\left( 50+12 \sin \frac{2\pi}{3} \right)^2 }.
\]

\end{enumerate}

\end{document}


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