%% MATH 135-01 Final Exam, Fall 2016
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\begin{document}

\begin{center}

\textbf{Holy Cross College, Fall Semester, 2016} \\
\textbf{MATH 135, Section 01, Final Exam Solutions} \\
\textbf{Friday, December~16, 8:00~AM} \\
\end{center}

\noindent
1. [5~points each]\quad Circle the number of the graph (on next page)
   showing each of the following functions.  Note that there is one ``extra''
graph that does not match any of these functions.


\begin{enumerate}
  \item[(a)] $f(x) = e^{-x} + 2$ is Plot III


  \item[(b)] $f(x) = x^3 - 4x$ is Plot V


  \item[(c)] $f(x) = 2\cos(2\pi x)$ is Plot I


  \item[(d)] $f(x) = \ds \frac{1}{x^2 - 9}$ is Plot IV

  \item[(e)] Give a formula of a function that matches the graph you 
     did not circle.  {\it Answer:}  Plot II is the plot of $f(x) = \sin(x)$.

\end{enumerate}


\noindent 
2. [20~points] One of the functions given in the following table is linear
and the other is exponential.  Find a formula for the linear one and place it
in the appropriate box.  In the box for the other one, write ``Exponential.''

\[
\begin{array}{|c||r|r|r|r|r|}
\hline
x          & 1 & 2 & 3& 4& 5\Strut \\
\hline
f(x)       & 1.2 & 0.6 & 0.3 & 0.15 & 0.075 \Strut \\
\hline
g(x)       & -2.3& -0.6& 1.1 & 2.8 & 4.5\Strut \\
\hline
\end{array}
\]

\begin{figure}
\begin{center}
\subfloat[Plot I]{\includegraphics[width=1.2in]{135FIc.png}}
\subfloat[Plot II]{\includegraphics[width=1.2in]{135FIe.png}}\\
\subfloat[Plot III]{\includegraphics[width=1.2in]{135FIa.png}}
\subfloat[Plot IV]{\includegraphics[width=1.2in]{135FId.png}}\\
\subfloat[Plot V]{\includegraphics[width=1.2in]{135FIb.png}}\\
\caption{Plots for problem 1.}
\end{center}
\end{figure}

\noindent
{\it Solution:}  $g(x)$ is the linear function $g(x) = 1.7 (x - 1) - 2.3 = 1.7 x - 4$.
$f(x)$ is the exponential function (whose formula is $f(x) = 2.4\left(\frac{1}{2}\right)^x$, but that was not asked for).
\vskip 10pt 
\noindent
3.  
\begin{enumerate}
\item[(a)] [15~points]  The depth of water in a tank oscillates sinusoidally
once every 4 hours according to $d(t) = 2\cos\left(\frac{\pi t}{2}\right) + 4$.
Sketch the graph of the depth versus time.
\item[] {\it Solution:} This is a cosine graph with amplitude $2$, shifted up $4$, and
period $4$.  See plot on next page.
\item[(b)] [10~points]  Find the average rate of change of the depth on the interval $[1,1.1]$.
\item[] {\it Solution:}  The average rate of change is 
$$\frac{2\cos(1.1\pi/2) + 4 - (2\cos(\pi/2) + 4)}{1.1 - 1} \doteq -3.13$$
(If the units of $d$ were feet and $t$ was in hours, this would have the 
units of feet/hour, but that was not asked for.)
\end{enumerate}

\begin{figure}
\begin{center}
\includegraphics[width=2.5in]{Sinusoid.png}
\caption{The plot $y = 2 \cos\left(\frac{\pi t}{2}\right) + 4$ in question 3.}
\end{center}
\end{figure} 

\vskip 10pt
\noindent
4.  Compute the following limits [5~points each].  Any legal method is OK.
\begin{enumerate}
\item[(a)]  $\ds \lim_{x\to 2} \frac{x^3 + 2x}{x - 4}$
\item[] {\it Solution:}  This rational function is continuous at $x = 2$, 
so the limit is equal to the value $12/(-2) = -6$.  
\item[(b)]  $\ds \lim_{x\to 2} \frac{x^2 - 7x + 10}{x^2 - 5x + 6}$
\item[] {\it Soluiton:}  This is a 0/0 limit, so we factor the top and 
bottom, cancel and then evaluate:
$$= \lim_{x\to 2} \frac{(x - 2)(x - 5)}{(x - 2)(x - 3)} = \lim_{x\to 2} \frac{x - 5}{x - 3} = 3.$$
\item[(c)]  $\ds \lim_{x\to \infty} \frac{5x^2 - x + 21}{8x^2 - 9x + 1}$
\item[] {\it Solution:}  Either divide the top and bottom by $x^2$ and take 
the limit, or use L'Hopital's Rule.  The value is $\frac{5}{8}$.  
\end{enumerate}
\vskip 10pt
\noindent
5. 
 \begin{enumerate}
\item[(a)] [5~points] State the limit definition of the derivative:
\item[] {\it Solution:} 
$$f'(x) = \lim_{h\to 0} \frac{f(x + h) - f(x)}{h}$$
if the limit exists.
\item[(b)] [10~points] Use the definition to compute $f'(x)$ for
$f(x) = 3\sqrt{x + 2}$.
\item[] {\it Solution:}
\begin{eqnarray*}
f'(x) &=& \lim_{h\to 0} \frac{3\sqrt{x + h + 2} - 3\sqrt{x + 2}}{h}\\
&=& \lim_{h\to 0} \frac{3\sqrt{x + h + 2} - 3\sqrt{x + 2}}{h} \cdot \frac{3\sqrt{x + h + 2} + 3\sqrt{x + 2}}{3\sqrt{x + h + 2} + 3\sqrt{x + 2}}\\
&=& \lim_{h\to 0} \frac{9(x + 2 + h) - 9(x + 2)}{h (3\sqrt{x + h + 2} + 3\sqrt{x + 2})}\\
&=& \lim_{h\to 0} \frac{9}{3\sqrt{x + h + 2} + 3\sqrt{x + 2}}\\
&=& \frac{3}{2\sqrt{x+2}}.
\end{eqnarray*}
\item[(c)] [10~points] Find the equation of the tangent line
to the graph $y = 3\sqrt{x + 2}$ at the point $(2,6)$.
\item[] {\it Solution:} Since $f(2) = 3\sqrt{4} = 6$ and $f'(2) = \frac{3}{4}$, the tangent line is 
$$y - 6 = \frac{3}{4}(x - 2) \quad {\rm or} \quad y = \frac{3}{4} x + \frac{9}{2}.$$
\end{enumerate}
\vskip 10pt
\noindent
6.  Compute the following derivatives using the derivative rules.
You need not simplify. [5~points each]
\begin{enumerate} 
\item[(a)]  $\ds f(t) = t^4 - \frac {1}{\sqrt[5]{t}} + e^t = t^4 - t^{-1/5} + e^t$.
\item[] {\it Solution:}  $f'(t) = 4t^3 + \frac{1}{5} t^{-6/5} + e^t = 4t^3 + \frac{1}{5 t^{6/5}} + e^t$.

\item[(b)]  $\ds g(x) = \frac{x^2 - 2}{\cos(x) + 1}$

\item[] {\it Solution:}  By the quotient rule:
$$g'(x) = \frac{(\cos(x) + 1)(2x) - (x^2 - 2)(-\sin(x))}{(\cos(x)+1)^2}.$$

\item[(c)] $\ds h(z) = \ln(4z^2 + 2 \tan^{-1}(z))$

\item[] {\it Solution:}  By the chain rule
$$h'(z) = \frac{1}{4z^2 + 2\tan^{-1}(z)}\cdot \left(8z + \frac{1}{1 + z^2}\right).$$


\item[(d)] Find $\frac{dy}{dx}$ if $5 x^2 y^2 - 2y^5 + x = 1$.

\item[] {\it Solution:} Using implicit differentiation:
$$10 x^2 y \frac{dy}{dx} + 10x y^2 - 10 y^4 \frac{dy}{dx} + 1 = 0,$$
so 
$$\frac{dy}{dx} = \frac{-1 - 10 xy^2}{10 x^2y - 10y^4}.$$
\end{enumerate}

\vskip 10pt
\noindent
7.  All parts of this question refer to the functions defined by 
$f(x) = x^4 + 2ax^2$, where $a$ is any fixed real number.  

\begin{enumerate}
\item[(a)]  [10~points]  Assuming $a < 0$, find the {\it critical points} of $f$, 
and construct a sign diagram for $f'(x)$.  Which of your 
critical points are local maxima and which are local minima?

\item[] {\it Solution:} Since $a < 0$, The derivative of $f$ factors as
$$f'(x) = 4x^3 + 4ax = 4x(x^2 + a) = 4x(x + \sqrt{-a})(x - \sqrt{-a})$$
This
is equal to zero at $x = 0, \pm \sqrt{-a}$.  The derivative is negative on 
$(-\infty,-\sqrt{-a})$ and again on $(0,\sqrt{-a})$.  It is positive on
$(-\sqrt{-a},0)$ and $(\sqrt{-a},+\infty)$.  By the first derivative test,
this says $x = \pm\sqrt{-a}$ are local minima and $x = 0$ is a local maximum. 
 
\item[(b)]  [10~points] Repeat part a, but assume now that $a > 0$.  

\item[] {\it Solution:}  We still have $f'(x) = 4x(x^2 + a)$, but when 
$a > 0$, the quadratic $x^2 + a = 0$
has no real roots, so $x = 0$ is the only critical point.  Moreover, 
$x^2 + a > 0$ for all $x$, so the derivative
is $< 0$ when $x < 0$ and $> 0$ when $x > 0$.  The function
has a local minimum at $x = 0$. 

\item[(c)] [10~points] How many different {\it inflection points} does
the graph $y = f(x)$ have if $a < 0$?  Explain.

\item[] {\it Solution:}  $f''(x) = 12x^2 + 4a$ is zero at $x = \pm \sqrt{-a/3}$
when $a < 0$.  The second derivative changes sign at each of those points.
So there are \emph{two inflection points}.                                                                                                                  
\end{enumerate}

\vskip 10pt

\noindent
8. [20~points] The radius and the height of a circular cone increase at
a rate of 2 cm/sec.  How fast is the volume of the cone increasing when $r = 10$
and $h = 20$?
\vskip 10pt
\noindent
{\it Solution:} From the volume formula for a cone, $V = \frac{\pi r^2 h}{3}$.
Hence
$$\frac{dV}{dt} = \frac{\pi}{3}\left(r^2 \frac{dh}{dt} + 2rh \frac{dh}{dt}\right).$$
Substituting $r = 10, h = 20$, $\frac{dr}{dt} = \frac{dh}{dt} = 2$, we get
$$\frac{dV}{dt} = \frac{\pi}{3}\left(10^2 \cdot 2 + 2\cdot 10 \cdot 20 \cdot 2\right) = 
\frac{1000\pi}{3}$$
(cubic cm per second).

\vskip 10pt
\noindent
9.  [20~points] A rectangular poster is to have total area 600 square inches, including
blank 1 inch wide margins on all four sides of a central printed area.  What overall
dimensions will maximize the printed area?  
\vskip 10pt 
\noindent
{\it Solution:}  Call the overall dimensions $x,y$. Then the dimensions of the
printed area are $x-2$ by $y-2$ inches because of the blank margins.
 We have $xy = 600$ and we want to maximize 
$$A = (x - 2)(y - 2) = (x - 2)\left(\frac{600}{x} - 2\right) = 600 - \frac{1200}{x} - 2x + 4$$
We differentiate this and set it equal to zero to find the critical points:
$$A'(x) = \frac{1200}{x^2} - 2 = 0$$
so $x = \pm \sqrt{600}$. We discard the negative root because
$x$ is supposed to represent a length.  So $x = 10\sqrt{6}$.  Then $y = 10\sqrt{6}$ also.  This is a 
maximum because $A''(x) = \frac{-2400}{x^3} < 0$ for all $x > 0$. 
\end{document}
