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\begin{document}
\begin{center}

\textbf{College of the Holy Cross} \\
\textbf{MATH 133, section 1 -- Calculus with Fundamentals}\\
\textbf{Solutions for Final Exam -- December 15, 2015}
\end{center}
\vskip 10pt

\noindent
I.  A portion of the graph $y = f(x)$ is given in black (\emph{on the top of the next page}).  Match each equation 
with one of the red graphs identified with lower-case letters.
\begin{itemize}
\item[A)] (5) $y = -f(x + 5)$ is letter: {\bf d} (shifted 5 units left and reflected across $x$-axis)
\item[B)] (5) $y = f(x + 4) - 3.2$ is letter: {\bf a} (shifted 4 units left and 3.2 units down)
\item[C)] (5) $y = 3 f(x)$ is letter: {\bf e} (stretched by a factor of 3 vertically)
\item[D)] (5) $y = f(x) + 2 $ is letter: {\bf c} (shifted 2 units up)
\item[E)] (5) $y =  f(x - 4)$ is letter: {\bf b} (shifted 4 units right)
\item[F)] (5) The correct formula for $f(x)$ is $f(x) = xe^{-x}$, not $f(x) = \cos(x) - 1$.  
Note: $\cos(x) - 1$ would have a local maximum at $x = 0$ so that cannot be the correct formula.
\end{itemize}
\vskip 10pt
\noindent
II.  The power delivered by a battery to an apparatus of resistance $R$ (in ohms)
is 
$$P(R) = \frac{5 R}{(R + 0.5)^2}$$
(in watts).  
\vskip 10pt
\begin{enumerate}
\item[A)] (5) If $R = 10$ ohms, what is the power delivered by the battery?
\item[] $P(10) = \frac{5 \cdot 10}{(10 + 0.5)^2} \doteq .45$ watts.
\item[B)] (5) A power of 2 watts can be obtained with two different values of the 
resistance $R$.  What are they? 
\item[]  They are the solutions of $2 = \frac{5R}{(R + 0.5)^2}$, which 
rearranges to $2 R^2 - 3R + .5 = 0$.  By the quadratic formula, the roots
are 
$$R = \frac{3 \pm \sqrt{9 - (4)(2)(.5)}}{4} = \frac{3 \pm \sqrt{5}}{4} \doteq .19,1.3\ {\rm ohms}.$$
\item[C)] (10) What is the \emph{rate of change} of the power at $R = 10$ ohms?
\item[]  This is the derivative $P'(10)$.  By the quotient rule, 
$$P'(R) = \frac{(R+.5)^2\cdot 5 - 5R\cdot(2R + 1)}{(R+.5)^4}$$
so $P'(10) \doteq -.041$ (units are: watts per ohm).  
\end{enumerate}
\vskip 10pt
\begin{figure}
\begin{center}
\includegraphics[width=3truein]{133FI.png} 
\caption{Figure for problem I}
\end{center} 
\end{figure}
\vskip 10pt
\noindent
III.  Compute the following limits.  Any legal method is OK.  
\vskip 10pt
\begin{enumerate}
\item[(A)] (10) $\displaystyle{\lim_{x \to 3} \frac{x^2 - 6x + 9}{x^2 - 5x + 6}}$.
\item[]  This is a 0/0 form.  We can factor the top and bottom, cancel one $x - 3$, then
take the limit to obtain:
\begin{eqnarray*}
\lim_{x \to 3} \frac{x^2 - 6x + 9}{x^2 - 5x + 6} &=&
\lim_{x \to 3} \frac{(x - 3)^2}{(x - 2)(x - 3)} \\
&=&
\lim_{x \to 3} \frac{(x - 3)}{(x - 2)} \\
&=& 0.
\end{eqnarray*}
\item[(B)] (10) $\displaystyle{\lim_{x \to 2} \frac{\sqrt{x + 7} - 3}{x - 2}}$.
\item[] Another 0/0 form.  For this one, we multiply the top and bottom 
by the conjugate radical, simplify, then evaluate the resulting limit by 
continuity:
\begin{eqnarray*}
\lim_{x \to 2} \frac{\sqrt{x + 7} - 3}{x - 2} &=&
\lim_{x \to 2} \frac{(\sqrt{x+7} - 3)(\sqrt{x+7} + 3)}{(x - 2)(\sqrt{x+7}+3)}\\
&=&
\lim_{x \to 2} \frac{(x+7) - 9}{(x - 2)(\sqrt{x+7}+3)}\\
 &=&
\lim_{x \to 2} \frac{(x - 2)}{(x - 2)(\sqrt{x+7}+3)}\\
 &=&
\lim_{x \to 2} \frac{1}{(\sqrt{x+7}+3)}\\
&=& \frac{1}{6}.
\end{eqnarray*}
\item[(C)] (10) $\displaystyle{\lim_{x \to 0} x\cot(x)}$
\item[] Rewrite $\cot(x) = \frac{\cos(x)}{\sin(x)}$ and use $\lim_{x\to 0} \frac{\sin(x)}{x} = 1$,
so $\lim_{x\to 0} \frac{x}{\sin(x)} = 1$ as well by the limit quotient rule.  
Then 
$$\lim_{x\to 0} x\cot(x) = \lim_{x\to 0} \frac{x\cos(x)}{\sin(x)} = \lim_{x\to 0} \frac{x}{\sin(x)} \cdot \lim_{x\to 0}\cos(x) = 1\cdot 1 
= 1.$$ 
\end{enumerate}
\vskip 10pt
\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 $f(x) = \displaystyle{\frac{1}{x - 3}}$.
\item[]  We have
\begin{eqnarray*}
f'(x) &=& \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}\\
&=& \lim_{h\to 0} \frac{\frac{1}{x+h-3}-\frac{1}{x-3}}{h}\\
&=& \lim_{h\to 0} \frac{(x - 3) - (x + h - 3)}{h(x+h-3)(x-3)}\\
&=& \lim_{h\to 0} \frac{-h}{h(x+h-3)(x-3)}\\
&=& \lim_{h\to 0} \frac{-1}{(x+h-3)(x-3)}\\
&=& \frac{-1}{(x - 3)^2}.
\end{eqnarray*}
(Note that this agrees with the result of applying the chain rule to the function 
$f(x) = (x-3)^{-1}$.) 
\item[] Using appropriate derivative rules, compute the derivatives
of the following functions.  You do \emph{not} need to simplify your answers.
\item[B)] (10) $\displaystyle{g(x) = \pi x^e + \frac{3}{x^7} + \sqrt[3]{x} + 5}$
\item[] $g'(x) = \pi e x^{e-1} - \frac{21}{x^8} + \frac{1}{3} x^{-2/3}$.  
\item[C)] (10) $\displaystyle{h(x) = \frac{x^2 - 4x + 1}{x^3 + 4}}$
\item[] (quotient rule)
$$h'(x) = \frac{(x^3 + 4)(2x - 4) - (x^2 - 4x + 1)(3x^2)}{(x^3+4)^2}.$$
\item[D)] (10) $i(x) = x \ln(\cos(x)) + e^{\cos(x)}$
\item[]  (product and chain rules)
$$i'(x) = x\cdot\left(\frac{-\sin(x)}{\cos(x)}\right) + \ln(\cos(x)) + e^{\cos(x)}\cdot (-\sin(x))$$
\item[E)] (10) $j(x) = \sin^{-1}(3x - 4)$
\item[] (derivative rule for inverse sine and chain rule)
$$j'(x) = \frac{1}{\sqrt{1 - (3x - 4)^2}}\cdot 3$$
\end{enumerate}
\vskip 10pt
\noindent
V.  The graph in Figure 2 shows the {\it derivative}
$f'(x)$ for some function $f(x)$ defined on $-1.5\le x \le 2.5$.
In particular you should assume $f(1), f(2)$ are defined and finite.
Note:  This graph \emph{is not $y = f(x)$, it is $y = f'(x)$}. 
\begin{figure} 
\begin{center}
\includegraphics[width=3truein]{133FV.png} 
\caption{$y = f'(x)$ for problem V.}
\end{center} 
\end{figure}
Using the graph, answer these questions:
\begin{enumerate}
\item[A)] (5) What are the critical points of $f$ in the interval $[-1.5,2.5]$? 
\item[] Answer: $x = 0,1,2$ are all critical points.  The ``outside ones'' at $x = 0,2$
have  $f'(x) = 0$, the ``middle one'' $f'(1)$ does not exist because of the 
vertical asymptote to the graph $y = f'(x)$.
\item[B)] (5)  Classify each of the points you found in part A) as a local 
maximum, local minimum, or neither.
\item[] $x = 0$ is a local minimum of $f$; $x = 1,2$ are neither.
\item[C)] (5) Explain briefly how you know your answer in B) is correct.
\item[] Answer:  From the given graph,
$f'(x)$ changes sign from negative to positive at $x = 0$, 
so that is a local minimum by the First Derivative Test.  At the other two critical points, 
$f'$ does not change sign (it's positive on both sides of $x = 1,2$). So they neither 
local maxima nor local minima.
\item[D)] (5)  Is $x = 0$ a point of inflection of $f$?  Why or why not?
\item[] Answer: No -- the concavity of $y = f(x)$ does not change at $x = 0$
because $f'(x)$ is increasing on both sides of $0$.
\item[E)] (5)  Does it appear that $f''(2)$ exists? (Yes/No):  
\item[] Answer: No -- $y = f'(x)$ appears to have a cusp (or maybe a ``corner'') at
$x = 2$, so the derivative $f''(2) = (f')'(2)$ doesn't exist.  
\item[F)] (5)  Over which interval(s) contained in $[-1.5,2.5]$ is the graph $y = f(x)$
concave down.
\item[] Answer: Where $f'$ is decreasing, so $(1,2)$ only.
\end{enumerate}
\vskip 10pt
\noindent
VI.  A Spanish factory can produce $P = 2 L K^2$ million lightbulbs
if labor costing $L$ million euros is hired and equipment costing $K$ million euros 
is obtained.  If an order for $1.7$ million lightbulbs is received, which 
combination of $L$ and $K$ will \emph{minimize} the total cost of labor plus equipment
to fill the order?
\begin{enumerate} 
\item[A)] (5) Express the total cost of producing $1.7$ million lightbulbs
as a function of one of the two variables $K,L$.  (Your choice, but choose wisely!)
\item[] Setting $P = 1.7$, we have
$$1.7 = 2 L K^2 \Rightarrow L = \frac{.85}{K^2}.$$
So the total cost (labor, plus equipment) is
$$L + K = \frac{.85}{K^2} + K.$$
The function we want to minimize is 
$$C(K) = \frac{.85}{K^2} + K.$$
(Note that you want to differentiate this, {\it not} the formula for 
$P$ as a function of $L$ and $K$.)
\item[B)] (10) Find a critical point of your function from part A which is 
a realistic solution of this problem; solve for the other variable.
\item[] We have
$$C'(K) = \frac{-1.7}{K^3} + 1.$$
This is $0$ when 
$$K = (1.7)^{1/3} \doteq 1.19\ {\rm million\ euros}.$$
Then $L = \frac{.85}{(1.19)^2} \doteq .60$\ {\rm\ million euros}.
\item[C)] (5) How do you know that your solution \emph{minimizes} the total cost?
\item[] One way to tell is to compute the second derivative of the total cost 
function (as a function of $K$):
$$C''(K) = \frac{5.1}{K^4} > 0$$
at $K = 1.19$ (in fact for all $K$.)  This implies $K = 1.19$ is a local minimum
by the Second Derivative Test.
\end{enumerate}
\vskip 10pt
\noindent
VII. (15) A moth ball is 
evaporating and losing volume at the rate of $.1$ cm${}^3$/week. 
It has the shape of a sphere at all times.  How 
fast is the surface area of the moth ball shrinking when the
radius of the ball is $1$cm?  (Note: A sphere of radius $r$
has volume $V = \frac{4\pi r^3}{3}$ and surface area $A = 4\pi r^2$.) 
\vskip 10pt
\noindent
Solution:  We are given that the volume of the moth ball is decreasing
at $.1$ cubic centimeters per week.  This says 
$$-0.1 = \frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}.$$  
Hence when $r = 1$, 
$$\frac{dr}{dt} = \frac{-0.1}{4\pi} \doteq -.008$$ 
cm/week.  Then 
$$\frac{dA}{dt} = 8\pi r \frac{dr}{dt} = \frac{8\pi\cdot(-0.1)}{4\pi} = -0.2$$
(the units here are (square cm)/week). 
\vskip 10pt 
\noindent
VIII. (5) True/False:  The graph
obtained by shifting $y = \ln(x)$ vertically by 2 units can
{\it also} be obtained from $y = \ln(x)$ by a horizontal compression.  Explain 
your answer.
\vskip 5pt\noindent
This is {\bf True} because of properties of the logarithm function:  Shifting $y = \ln(x)$
vertically by 2 units gives $y = \ln(x) + 2$.  But
$$\ln(x) + 2 = \ln(x) + \ln(e^2) = \ln(e^2\cdot x).$$
Multiplying $x$ by a constant greater than 1 ``inside'' the function compresses
the graph horizontally.   Since $e^2 > 1$, the graph of this function
is a horizontal compression of the graph $y = \ln(x)$.
\end{document}




