%% MATH 131-01 Final Exam Solutions, Fall 2007
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\begin{center}

\textbf{College of the Holy Cross, Fall Semester, 2007} \\
\textbf{MATH 131, Section 01, Final Exam Solutions} \\
\textbf{December~10} \\
\end{center}


\vskip 10pt

\noindent
1. [5~points each]\quad Circle the number of the graph
showing each of the following functions.  Note that there is one more
graph than there are formulas.  

\soln

\begin{itemize}
\item[(a)] $\ds f(x) = 2 - 2\sin(\pi x)$ is graph III.

\item[(b)] $\ds f(x) = \frac{x^2}{x^2-1}$ is graph V.  (Note the vertical asymptotes at $x = \pm 1$.) 

\item[(c)] $\ds f(x) = \ln(\sqrt{x})$ is graph I.  
   
\item[(d)] $\ds f(x) = e^x + e^{-x}$ is graph II.  (This is not graph IV which has a somewhat
similar shape, since $f(0) = e^0 + e^0 = 2$.  This matches II but not IV.)
\end{itemize}

\begin{center}
I: \includegraphics[width=1.5truein,height=1.5truein]{1I.ps} 
II: \includegraphics[width=1.5truein,height=1.5truein]{1II.ps}
\end{center}
\begin{center}
III: \includegraphics[width=1.5truein,height=1.5truein]{1III.ps} 
IV: \includegraphics[width=1.5truein,height=1.5truein]{1IV.ps}
\end{center}
\begin{center}
V: \includegraphics[width=1.5truein,height=1.5truein]{1V.ps} 
\end{center}

\vskip 10pt
\noindent 
2.  
\begin{itemize}
\item[(a)] [5~points] What is the domain of the function $f(x) = \sqrt{x^2 - 4}$?

\soln The domain is the set of all real $x$ with $x^2 - 4 \ge 0$, or $x$ in 
$(-\infty,2] \cup [2,+\infty)$.

\item[(b)] [10~points] Find a formula for the inverse function of $f(x) = \frac{x}{3} - 1$ and
plot $y = f(x)$ and $y = f^{-1}(x)$ on the same coordinate axes below.

\soln To find the formula of $f^{-1}$, set $y = \frac{x}{3} - 1$ and solve for $x$:
$x = 3(y + 1) = 3y + 3$.  Switch variables to get $y = f^{-1}(x) = 3x + 3$.
The graph $y = f(x)$ is a straight line with slope $\frac{1}{3}$ and $y$-intercept at $-1$.
The graph $y = f^{-1}(x)$ is a straight line with slope $3$ and $y$-intercept at $y = 3$:

\begin{center}
\includegraphics[width=2.0truein,height=2.0truein]{2b.ps} 
\end{center}

\item[(c)] [5~points] Find the interval of all real 
$x$ satisfying $-\pi/4 < \tan^{-1}(x) < \pi/3$. 

\soln Since the tangent function is increasing, this set of inequalities is
the same as $\tan(-\pi/4) < x < \tan(\pi/3)$,  or $-1 < x < \sqrt{3}$.

\item[(d)] [5~points] Express as a single logarithm: $\ln(6) - \ln(12) + 5\ln(2)$. 
  
\soln Using properties of logarithms, this equals $\ds \ln\left(\frac{6\cdot 2^5}{12}\right)
 = \ln(16).$
\end{itemize}

\vskip 10pt
\noindent
3.  Compute the following limits [5~points each].  Any legal method is OK.
\begin{itemize}
\item[(a)]  $\ds \lim_{x\to 3^-} \frac{|x-3|}{x^2 - 2x - 3}$

\soln The denominator factors as $x^2 - 2x - 3 = (x - 3)(x + 1)$.  But
$\ds \frac{|x - 3|}{x - 3} = -1$ when $x < 3$ (which is what we want for
the left-hand limit).  Hence 
$$\ds \lim_{x\to 3^-} \frac{|x-3|}{x^2 - 2x - 3} = \lim_{x\to 3^{-1}} \frac{-1}{x+1} = \frac{-1}{4}.$$

\item[(b)]  $\ds \lim_{x\to -1} \frac{x^3-x}{x^2 + 5x + 4}$

\soln This one is indeterminate of the form $0/0$ and
can be done via ``preliminary algebra'' or L'Hopital's Rule.  If
we use L'Hopital, 

$$\lim_{x\to -1} \frac{x^3-x}{x^2 + 5x + 4} = \lim_{x\to -1}\frac{3x^2 - 1}{2x + 5} = \frac{2}{3}.$$

\item[(c)]  $\ds \lim_{x\to +\infty} \frac{\ln(x^2)}{x^{3/2}}$

\soln Indeterminate of the form $\infty/\infty$, so we use L'Hopital, then
simplify:

$$\lim_{x\to +\infty} \frac{\ln(x^2)}{x^{3/2}} = \lim_{x\to +\infty} \frac{\frac{2x}{x^2}}{(3/2)x^{1/2}}
=  \lim_{x\to +\infty} \frac{4}{3 x^{3/2}} = 0.$$
\end{itemize}

\vskip 10pt
\noindent
4. 
\begin{itemize}
\item[(a)] [5~points] State the limit definition of the derivative.

\soln The derivative is
$$f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$$
provided that the limit exists.

\item[(b)] [10~points] Use the definition to compute $f'(x)$ for
$\ds f(x) = \frac{1}{x-1}$.

\soln We compute
\begin{eqnarray*}
\lim_{h\to 0} \frac{f(x+h) - f(x)}{h} &=&  \lim_{h\to 0} \frac{\frac{1}{x+h-1} - \frac{1}{x-1}}{h}\\
&=& \lim_{h\to 0} \frac{(x-1) - (x + h - 1)}{h(x-1)(x+h-1)}\\
&=& \lim_{h\to 0} \frac{-h}{h(x-1)(x+h-1)}\\
&=& \lim_{h\to 0} \frac{-1}{(x-1)(x+h-1)}\\
&=& \frac{-1}{(x-1)^2}
\end{eqnarray*}

\item[(c)] [10~points] Find the equation of the tangent line
to the graph $y = \frac{1}{x-1}$ at the point $(5,1/4)$.

\soln The tangent line has slope $f'(5) = \frac{-1}{16}$ and passes through the 
point $(5,1/4)$ on the graph of $f$.  So the equation is
$$y - 1/4 = \frac{-1}{16}(x - 5).$$ 
\end{itemize}


\noindent
5.  The following graph shows $y = g'(x)$, the derivative
of some function $g(x)$.  
\begin{center}
\includegraphics[width=2.0truein,height=2.0truein]{7.ps} 
\end{center}
\centerline{\it Note: this is $y=g'(x)$, not $y = g(x)$.}
\begin{itemize}
\item[(a)] [5~points] At which of the points $x = 0,a,b,c,d,e,f$
does the function $g(x)$ have local maxima? 

\soln By the first derivative test, $g(x)$ has local maxima where
$g'(x)$ changes from positive to negative: at $x = c,e$.

\item[(b)] [5~points] At which of the points $x = 0,a,b,c,d,e,f$
does the function $g(x)$ have local minima? 

\soln By the first derivative test, $g(x)$ has local minima where
$g'(x)$ changes from negative to positive: at $x = a,d,f$.

\item[(c)] [5~points] On the interval $(a,b)$, is $y = g(x)$ concave up or down?

\soln On the interval $(a,b)$, $g'(x)$ is increasing, so $(g')'(x) = g''(x) > 0$.
This implies $y = g(x)$ is concave {\it up}.

\end{itemize}
\vskip 10pt

\noindent
6.  Compute the following derivatives using the derivative rules.
You need {\it not} simplify. [5~points each]
\begin{itemize} 
\item[(a)]  $f(x) = 3x^5 + \frac {5}{x^3} + e^x$.

\soln $f'(x) = 15x^4 - 15 x^{-4} + e^x$.

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

\soln By the quotient rule, 
$$g'(x) = \frac{(\sin^{-1}(x) + 1)\cos(x) - (\sin(x) + 1)\frac{1}{\sqrt{1 - x^2}}}{(\sin^{-1}(x) + 1)^2}.$$

\item[(c)] $\ds h(x) = \sqrt{\ln(x) + 4\cos(\tan(x))}$.

\soln By the chain rule and the derivative rules for $\ln, \cos, \tan$:
$$h'(x) = \frac{1}{2}\left(\ln(x) + 4\cos(\tan(x))\right)^{-1/2}\cdot \left(\frac{1}{x} - 4\sin(\tan(x))\sec^2(x)\right).$$

\item[(d)] $\ds j(x) = (x^2 + 3)^{(x + 2)}$.   

\soln Using logarithmic differentiation, $\ln(j(x)) = (x + 2)\ln(x^2 + 3)$. So then 
by the product rule
$$\frac{j'(x)}{j(x)} = \frac{2x(x+2)}{x^2+2} + \ln(x^2 + 3)$$
Hence
$$j'(x) = (x^2 + 3)^{(x+2)}\left(\frac{2x(x+2)}{x^2+2} + \ln(x^2 + 3)\right).$$

\item[(e)] Find $\ds \frac{dy}{dx}$ if $x^3 + y^3 = 18xy$.

\soln Differentiating implicitly,
$$3x^2 + 3y^2 y' = 18y + 18xy'.$$
Then we solve algebraically for $y' = \frac{dy}{dx}$:
$$y' = \frac{18y - 3x^2}{3y^2 - 18x} = \frac{6y - x^2}{y^2 - 6x}.$$

\end{itemize}
\vskip 10pt
\noindent
7.  All parts of this question refer to the function 
$$f(x) = \frac{(x+1)^2}{x^2 + 1}.$$
\begin{itemize}
\item[(a)] [5~points] Compute $f'(x)$ and simplify.

\soln By the quotient and chain rules:
$$f'(x) = \frac{(x^2 + 1)\cdot 2(x+1) - (x + 1)^2\cdot 2x}{(x^2+1)^2} = 
\frac{-2(x^2 - 1)}{(x^2 + 1)^2}.$$

\item[(b)] [10~points] Find all critical numbers of $f$ and determine the
intervals on which $f$ is increasing and decreasing.

\soln $f'(x) = 0$ at $x = \pm 1$.  These are the critical numbers. 
$f'(x) > 0$ so $f(x)$ is increasing on $(-1,1)$.  $f'(x) < 0$
so $f(x)$ is decreasing on $(-\infty,-1)\cup (1,+\infty)$.

\item[(c)] [5~points] {\bf Given:} the second derivative in simplified
form is
$$f''(x) = \frac{4x(x^2 - 3)}{(x^2+1)^3}.$$ 
Determine all points of inflection and the intervals where $y = f(x)$ is concave
up and concave down.

\soln From the form of the numerator, $f''(x) = 0$ at $x = 0, \pm \sqrt{3}$.  
$f''(x) > 0$ so $f(x)$ is concave up on $(-\sqrt{3},0) \cup (\sqrt{3},+\infty)$.
$f''(x) < 0$ so $f(x)$ is concave down on $(-\infty,-\sqrt{3}) \cup (0,\sqrt{3})$.
The concavity changes at $-\sqrt{3},0,\sqrt{3}$, so all three are inflections.

\item[(d)] [5~points] Find the absolute maximum and minimum of $f(x)$ on the interval $[0,5]$.   

\soln We compare $f(0) = 1$, $f(1) = 2$ (the value at the critical number in the interval), 
and $1 < f(5) = 18/13 < 2$.  Hence the absolute minimum {\it on this interval} is $f(0) = 1$,
and the absolute maximum is $f(1) = 2$.  
\end{itemize}

\vskip 10pt
\noindent
8. [20~points]  A potter forms 20 cm${}^3$ of clay into 
a cylindrical tube.  As she rolls the tube on a flat surface, 
its length $L$ increases and its radius $r$ decreases but the quantity
of clay stays constant.  If the length is
increasing at the rate of $.1$ cm/sec, find the rate at which
the radius is changing when the radius is $1$cm.  ({\it Note:} the 
volume of a cylinder of radius $r$ and height $h$ is $V = \pi r^2 h$.)   

\soln The volume of the clay is $20 = \pi r^2 L$ at all times.  
Taking derivatives with respect to time, and using the product rule,
$$0 = 2\pi r L \frac{dr}{dt} + \pi r^2 \frac{dL}{dt}.$$
When $r = 1$, $\frac{dL}{dt} = +0.1$ and from the volume equation $L = \frac{20}{\pi}$.
Hence the radius is changing at the rate:
$$\frac{dr}{dt} = \frac{-\pi r^2 \frac{dL}{dt}}{2\pi r L} = \frac{-\pi (1)^2 (0.1)}{2\pi (1)\frac{20}{\pi}} = \frac{-\pi}{400} \doteq -0.0079 {\rm\ cm/sec}.$$
 

\vskip 10pt
\noindent
9.  [20~points] A rectangular swimming pool with sides $x,y$
is to be built with an area of 1800 square feet.  The owner
wants 5-foot wide decks along either side and 10-foot wide
decks at the two ends.  Find the dimensions of the rectangular
lot of smallest area on which such a swimming pool can be built.
Be sure you say how you know your answer gives a minimum of the area of the lot.

\soln The area of the pool is $xy = 1800$, so $y = 1800/x$.  If we
let $x$ be width of the pool, then the width of pool plus side decks
is $x + 10$.  The length of the pool plus the end decks is then $y + 20$.
The total area of the rectangle containing the pool and the decks is
$$A = (x + 10)(y + 20) = xy + 10y + 20x + 200 = 2000 + \frac{18000}{x} + 20x.$$
Taking derivative with respect to $x$, and setting equal zero to find
critical numbers, we find
$$0 = A'(x) = 20 - \frac{18000}{x^2}$$
so $x = \pm \sqrt{18000/20} = \pm \sqrt{900} = \pm 30$.  The negative
root is irrelevant here.  So $x = 30$ is the only critical number
we need to consider.  By the equation $y = 1800/x$ we get $y = 60$.
Now, at $x = 30$, $A''(x) = \frac{36000}{30^3} > 0$. Hence this gives
a minimum of the area of the pool plus the decks.  The final answer
is that the lot should be $x + 10 = 40$ by $y + 20 = 80$ feet to 
fit the pool and the decks.  

\vskip 10pt
\noindent
10.  On and near the surface of the moon, the acceleration of gravity is $-1.66$ meters
per second per second.

\begin{itemize}
\item[(a)] [5~points]  An astronaut throws a moon rock upward
from a height of 1.5 meters above the surface 
with an initial velocity of $3$ meters per second.
Find the height of the rock above the surface as a function of time $t$
in seconds. 

\soln We antidifferentiate once to find the velocity:
$$v(t) = -1.66 t + c.$$
To get $v(0) = 3$, $c = 3$.  Then we antidifferentiate again
to get the height:
$$h(t) = -.83 t^2 + 3t + d.$$
To get $h(0) = 1.5$, $d = 1.5$.  So $h(t) = -.83 t^2 + 3 t + 1.5$.

\item[(b)] [3~points]  How long does it take for the rock from part 
a to fall back to the surface of the moon?

\soln We need to solve the equation $h(t) = -.83 t^2 + 3t + 1.5 = 0$.
The only way to do this is via the quadratic formula:
$$t = \frac{-3 \pm \sqrt{9 - 4(-.83)(1.5)}}{-1.66}.$$
The positive root is about $t = 4.06$ sec.

\item[(c)] [2~points]  What is the velocity of the rock from part a when 
it strikes the surface of the moon?  

\soln Substitute $t = 4.06$ into the formula for $v(t)$ above:
$v(4.06) \doteq -3.74$ m/sec.

\end{itemize}

\end{document}
