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\begin{document}

\begin{center}
\textbf{College of the Holy Cross} \\
\textbf{MATH 135, section 1}\\
\textbf{Solutions for Final Examination -- Thursday, December 12}
\end{center}

\noindent
I.  The  graph $y = f(x)$ is given in blue.  Match each equation with one
of the numbered pink graphs.
\vskip 10pt
\begin{center}
\includegraphics[width=3truein,height=2truein]{FI.png} 
\end{center} 

\begin{itemize}
\item[(5) A)] $y = f(x - 4)$ is plot number: 3 (shift $4$ units right)
\item[]
\item[(5) B)] $y = f(x) + 3$ is plot number: 1 (shift $3$ units up)
\item[]
\item[(5) C)] $y = \frac{1}{3} f(x) $ is plot number: 4 (compress vertically by a factor of 
$\frac{1}{3}$)
\item[]
\item[(5) D)] $y = -f(x + 4)$ is plot number: 5 (shift $4$ units left and reflect across
the $x$-axis)
\item[]
\item[(5) E)] $y = 2 f(x + 6)$ is plot number: 2 (shift $6$ units lift and stretch vertically
by a factor of $2$)
\end{itemize}

\noindent
II.  A cup of hot chocolate is set out on a counter at $t = 0$.  The 
temperature of the chocolate $t$ minutes later is
$C(t) = 70 + 80e^{-t/3}$ (in degrees F).
\vskip 10pt
\begin{enumerate}
\item[A)] (5) What is the temperature of the chocolate at $t = 0$?
\item[] {\it Answer:}  $C(0) = 70 + 80 e^{-0/3} = 150$ degrees F.  
\item[B)] (10) What is the rate of change of the temperature at $t = 10$ minutes?
\item[] {\it Solution:}  The (instantaneous) rate of change at $t = 10$ is $C'(10)$.  
Since $C'(t) = \frac{-80}{3} e^{-t/3}$
by the chain rule, $C'(10) = \frac{-80}{3} e^{-10/3} \doteq -.95$ degrees F per minute.
\item[] {\it Comment:}  Since the question says ``at $t = 10$'' you should 
think: ``instantaneous rate of change.''  Quite a few people in the class
computed an average rate of change from $t = 0$ to $t = 10$, which is not the same!
\item[C)] (10) How long does it take for the temperature to 
reach $100^\circ F$?
\item[] {\it Solution:}  The time is the solution of $100 = 70 + 80 e^{-t/3}$, or 
$t = -3 \ln(30/80) \doteq 2.9$ minutes. 
\end{enumerate}

\noindent
III.  Compute the following limits.  Any legal method is OK.  
\vskip 10pt
\begin{enumerate}
\item[(A)] (10) $\displaystyle{\lim_{x \to 3} \frac{x^2 + x - 12}{x^2 - 5x + 6}}$.
\item[] {\it Solution:}  Since $x^2 + x - 12 = (x - 3)(x + 4)$ and $x^2 - 5x + 6 = (x-3)(x-2)$,
for $x \ne 3$, the function is 
$$\frac{x^2 + x - 12}{x^2 - 5x + 6} = \frac{x + 4}{x-2}.$$
Hence the limit equals 
$$\lim_{x \to 3}  \frac{x + 4}{x-2} = 7$$
by the limit quotient rule.  (Note: this could also be done with L'Hopital's Rule.)
\item[(B)] (10) $\displaystyle{\lim_{x \to 1^-} \frac{|x - 1|}{x^2 - 1}}$.
\item[] {\it Solution:}  The denominator is $x^2 - 1 = (x - 1)(x + 1)$.
The numerator is $x - 1$ if $x > 1$ and $-(x - 1)$ if $x < 1$.  
Hence the function equals 
$$\begin{cases}
           \frac{-1}{x+1} & \text{ if } x < 1\\
           \frac{1}{x + 1} & \text{ if } x > 1.
\end{cases}$$
This shows that the one-sided limit exists and equals
$$\lim_{x \to 1^-} \frac{-1}{x + 1} = \frac{-1}{2}.$$
(The overall limit does not exist since the limit from the other side
exists but equals a different value, namely $\frac{+1}{2}$.)
\item[(C)] (10) $\displaystyle{\lim_{x \to 0^+} x^2 \ln(x)}$
\item[] {\it Solution:}  This is indeterminate of the form $0\cdot \infty$.  So 
we want to rearrange, and then apply L'Hopital's Rule like this:
\begin{eqnarray*}
\lim_{x \to 0^+} x^2 \ln(x) &=& \lim_{x \to 0^+} \frac{\ln(x)}{x^{-2}}, \text{ which is an } \infty/\infty \text{ indeterminate form}\\
&=& \lim_{x \to 0^+} \frac{x^{-1}}{-2 x^{-3}}\\
&=&  \lim_{x \to 0^+} \frac{x^2}{-2}\\
&=& 0.
\end{eqnarray*}
\item[(D)] (10) $\displaystyle{\lim_{x \to 0} \frac{\tan(x)}{x^{1/2}}}$
\item[] {\it Solution:}  This is indeterminate of the form $0/0$.  We can apply 
L'Hopital directly like this:
\begin{eqnarray*} 
\lim_{x \to 0} \frac{\tan(x)}{x^{1/2}} &=& \lim_{x\to 0} \frac{\sec^2(x)}{\frac{1}{2x^{1/2}}}\\
&=& \lim_{x\to 0} 2x^{1/2} \sec^2(x) \\
&=& 0\cdot 1 = 0.
\end{eqnarray*}
\item[] {\it Comment:}  I gave some partial credit for experimentation with 
a table of values on C and D.  But you should be aware that that is not a complete 
justification for saying the limit is zero in C and D.  It's suggestive, but
it's not a complete reason.  

\end{enumerate}

\noindent
IV.  
\begin{enumerate}
\item[A)] (10) {\it Using the limit definition}, and showing all necessary steps
to justify your answer, compute $f'(x)$ for $f(x) = 5x^2 - x + 3$.\\
\vskip 10pt
\noindent
{\it Solution:}
\begin{eqnarray*}
f'(x) &=& \lim_{h\to 0} \frac{f(x + h) - f(x)}{h}\\
&=& \lim_{h\to 0} \frac{5(x + h)^2 - (x + h) + 3 - 5x^2 + x - 3}{h}\\
&=& \lim_{h\to 0} \frac{10xh + 5h^2 - h}{h}\\
&=& \lim_{h\to 0} 10x -1 + 5h\\
&=& 10x - 1.
\end{eqnarray*} 
\end{enumerate}

\noindent
IV. (continued)  Using appropriate derivative rules, compute the derivatives
of the following functions.  You do not need to simplify your answers.
\begin{enumerate}
\item[B)] (5) $\displaystyle{g(x) = 4x^3 +  \sqrt{x} + \frac{2}{\sqrt[4]{x}} + e^2}$.
\item[] {\it Solution:}  We can rewrite $g(x)$ as 
$$g(x) = 4x^3 + x^{1/2} + 2 x^{-1/4} + e^2.$$
So by the power and sum rules for derivatives
$$g'(x) = 12x^2 + \frac{1}{2}x^{-1/2} - \frac{1}{2} x^{-5/4} + 0.$$ 
\item[C)] (10) $\displaystyle{h(x) = \frac{\sin(x) + x}{\sec(x)}}$.
\item[] {\it Solution:}  By the quotient rule, 
$$h'(x) = \frac{\sec(x) (\cos(x) + 1) - (\sin(x) + x)\sec(x)\tan(x)}{\sec^2(x)}.$$
\item[D)] (10) $i(x) = \ln(x^3+3)$.
\item[] {\it Solution:}  By the chain rule, 
$$i'(x) = \frac{3x^2}{x^3 + 3}.$$ 
\item[E)] (10) $j(x) = \tan^{-1}(12x + 2) + x^x$\\
\item[] {\it Solution:}  By the derivative rules for the inverse tangent and 
the chain rule, plus logarithmic differentiation for the $x^x$,
$$j'(x) = \frac{12}{1 + (12x + 2)^2} + x^x(1 + \ln(x)).$$
\end{enumerate}
\vskip 10pt
\noindent
V.  The following graph shows the {\it derivative}
$f'(x)$ for some function $f(x)$ defined on $0\le x \le 4$.
Note:  This \emph{is not} $y = f(x)$, it is $y = f'(x)$.  
\begin{center}
\includegraphics[width=3truein,height=2truein]{FV.png} 
\end{center} 
Using the graph, {\it estimate}
\begin{enumerate}
\item[A)] (5) The interval(s) on which $f(x)$ is increasing.
\item[] {\it Solution:}  $f(x)$ is increasing on intervals where
$f'(x) > 0$.  Here that is true for $x$ in $(0,1)$ and $(2,3)$.  
\item[B)] (10) The critical numbers of $f(x)$ \emph{in the open interval} $(0,4)$.  
Say what the behavior of $f(x)$ is at each critical number (local max, local min, neither).
\item[] {\it Solution:}  The critical numbers in this interval are the places where $f'(x) = 0$, 
so $x = 1,2,3$.  By the First Derivative Test, $f$ has local maxima at $x = 1$ and $x = 3$
($f'$ goes from positive to negative), while $f$ has a local minimum at $x = 2$ ($f'$ goes
from negative to positive). 
\item[C)] (10) The interval(s) on which $y = f(x)$ is concave down. 
\item[] {\it Solution:}  $f$ is concave down on intervals where $f''(x) < 0$, 
or equivalently where $f'(x)$ is decreasing.  That is true here for $x$ in $(.4,1.3)$ 
and again for $x$ in $(2,4,3.3)$ (approximately).
\end{enumerate}

\noindent
VI. A town wants to build a pipeline from a water station on a small
island 2 miles from the shore of its water reservoir to the town. 
One possible route is shown dotted in red. The town is 6 miles along the shore
from the point nearest the island.  It costs \$3 million   
per mile to lay pipe under the water and \$2 million per mile
to lay pipe along the shoreline.
\begin{center}
\includegraphics[width=2.5truein,height=2.5truein]{FVI.png} 
\end{center} 

\begin{enumerate} 
\item[A)] (5)  Give the cost $C(x)$ of constructing the pipeline
as a function of $x$.
\item[] {\it Solution:}  By the Pythagorean theorem and the given 
information about cost per mile, we have 
$$C(x) = 3\sqrt{4 + x^2} + 2(6 - x)$$
\item{B)} (10) Where along the 
shoreline should the pipeline hit land to minimize the costs
of construction? 
\item[] {\it Solution:} To find the minimum of $C(x)$, we
can restrict to $x$ in the closed interval $[0,6]$, since it clearly does no
good to take $x < 0$ or $x > 6$.  The function $C(x)$ has
a critical number for $x > 0$ at the positive solution of $C'(x) = 0$:
\begin{eqnarray*}
0 &=& \frac{3x}{\sqrt{4 + x^2}} - 2, \text{or}\\
3x &=& 2 \sqrt{4 + x^2} \\
9x^2 &=& 16 + 4x^2 \\
5 x^2 &=& 16 \\
x &=& \frac{4}{\sqrt{5}} \doteq 1.79.
\end{eqnarray*}
We have $C(0) = 18$, $C(6) = 3\sqrt{40} \doteq 19.0$, and
$C\left(\frac{4}{\sqrt{5}}\right) \doteq 16.47$.  So 
the minimum cost is attained at $x = \frac{4}{\sqrt{5}}\doteq 1.79$ miles.
\end{enumerate}

\noindent
VII. (15)  A block of dry ice (solid $CO_2$) is 
evaporating and losing volume at the rate of $10$ cm${}^3$/min. 
It has the shape of a cube at all times.  How 
fast are the edges of cube shrinking when the block has volume 
$216$ cm${}^3$?  
\vskip 10pt
\noindent
{\it Solution:}  Call the side of the cube $x$.  Then $V = x^3$.
Taking time derivatives, we have $V' = 3x^2 x'$. From the given information,
when $V = 216$, $x = 6$ and $V' = -10$.  Therefore the rate of change of 
the side of the cube is 
$$x' = \frac{-10}{3\cdot 6^2} = \frac{-5}{54} \doteq -.093$$
(units cm/min).  The side of the cube is decreasing at
about $.09$ cm/min.

\noindent
VIII.  (10) True or false:  The graph
obtained by stretching $y = e^{-x}$ vertically by a factor of 2 can
{\it also} be obtained from $y = e^{-x}$ by a horizontal shift.  Explain 
your answer.
\vskip 10pt
\noindent
{\it Solution:} This is TRUE, because 
$$2e^{-x} = e^{\ln(2)} e^{-x} = e^{-(x - \ln(2))}.$$
So exactly the same graph is obtained if we stretch $y = e^{-x}$ vertically by 
a factor of 2, or shift $y = e^{-x}$ to the right by $\ln(2)$
units.  This seems counterintuitive, but it is a general property of exponential
functions that this sort of thing is true(!)

\end{document}




