\documentclass[11pt]{article}
%
%  You can add additional definitions
%  to the following preamble to create
%  new "theorem-type" environments,
%  set up new abbreviations for commonly
%  used symbols, and so forth.
%
\newcommand\bC{\mathbb C}
\newcommand\bZ{\mathbb Z}
\renewcommand\a{\alpha}
\newtheorem{proposition}{Proposition}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
\usepackage{amssymb,epsfig,amscd}
\usepackage{graphicx}
%
%  The actual document starts here
%  the first lines generate a title and
%  lines for each author's name.  A line
%  with the date the file is typeset 
%  by LaTeX is generated each time you
%  build the file.
%
\begin{document}
\title{Solutions for Problem Set 2}
\author{MATH 131, Fall 2007}
\maketitle

\section*{Section 1.4}

\noindent
4.  The graphs $y = e^x$ and $y = 8^x$ are increasing exponentials;
$y = e^{-x}$ is the reflection of $y = e^x$ across the $y$-axis and
$y = 8^{-x}$ is the reflection of $y = 8^x$ across the $y$-axis.
Those two are decreasing exponentials.
\vskip 15pt
\noindent
8.  Shift $y = 4^x$ 3 units to the right ($y$-axis intercept
becomes $(0,4^{-3}) = \left(0,\frac{1}{64}\right)$).
\vskip 15pt
\noindent
10.  Take $y = 2e^x$, stretch vertically by a factor of $2$, then 
shift up one unit.
\vskip 15pt
13.
\begin{enumerate}
\item[a.]  $y = e^x - 2$
\item[b.]  $y = e^{x-2}$
\item[c.]  $y = -e^x$
\item[d.]  $y = e^{-x}$
\item[e.]  $y = -e^{-x}$
\end{enumerate}
\vskip 15pt
14.  
\begin{enumerate}
\item[a.]  $y = 8 - e^x$
\item[b.]  $y = 2

\section*{Section 1.2}

\noindent
2.  
\begin{enumerate}
\item[a.]  This is a quotient of polynomials, so {\it a rational function}.
\item[b.]  This is an {\it algebraic function} because of the square root.
\item[c.]  This is an {\it exponential function}.
\item[d.]  This is a {\it power function}.
\item[e.]  This is a {\it polynomial function}.
\item[f.]  This is a {\it trigonometric function}.
\end{enumerate}
\vskip15pt
\noindent
4.
\begin{enumerate}  
\item[a] $y = 3x$ is the graph $y = G(x)$ (yellow) because that 
is the only straight line.
\item[b] $y = 3^x$ is the graph $y = f(x)$ (blue) because
it has the correct shape for an exponential function (and it is 
the only one that takes only positive values).
\item[c] $y = x^3$ is the graph $y = F(x)$ because it has the 
correct shape for an odd integer power function
\item[d] By process of elimination, $y = x^{1/3}$ is the graph $y = g(x)$ (black).
\end{enumerate}
\vskip15pt
\noindent
5. 
\begin{enumerate}
\item[a] The family of lines of slope $m = 2$ is given by $y = 2x + b$,
where $b$ can be any real number.  The graph should show several parallel
lines with slope $2$.
\item[b] $f(2) = 1$ means that the line contains the point $(x,y) = (2,1)$.
By the point-slope form, $y - 1 = m(x - 2)$, so $y = mx - 2m + 1$.  The 
graph should show several lines passing through the point $(2,1)$, with different slopes.
\item[c] The line in both families should have slope $m = 2$ and pass through $(2,1)$,
so by the point-slope form, $y - 1 = 2(x - 2)$, or $y = 2x - 3$.
\end{enumerate}
\vskip15pt
\noindent
6.  From the given equation $f(-3) = 1$ for all these $f$.  The graphs are
all straight lines passing through the point $(x,y) = (-3,1)$.  
\vskip15pt
\noindent
8.  The first graph is the parabola $y = 2x^2$ shifted $3$ units to the 
right along the $x$-axis, so $y = 2(x - 3)^2$.  The second graph is 
$y = f(x) = Ax^2 + Bx + C$ where from the graph $f(0) = 1$, $f(-2) = 2$, and
$f(1) = -2.5$.  The first equation says $1 = f(0) = A0^2 + B0 + C$, so $C = 1$.
The second and third equations then give:
\begin{eqnarray*}
2 = f(-2) & = & A(-2)^2 + B(-2) + 1 = 4A - 2B + 1 \\
-2.5 = f(1) & = & A(1)^2 + B(1) + 1 = A + B + 1
\end{eqnarray*}
or
\begin{eqnarray*}
4A - 2B &=& 1\\
A + B &=& -\frac{7}{2}
\end{eqnarray*}
Solving simultaneously yields $A = -1$, $B = - \frac{5}{2}$, so the function
is $f(x) = -x^2 - \frac{5}{2} x + 1$.
\vskip15pt
\noindent
12.  
\begin{enumerate}
\item[a.]  The graph should be the portion of a
line sloping down to the right {\it in the first quadrant}, with vertical
axis intercept at $y = 200$, and horizontal axis intercept at $x = 50$.
\item[b.]  The slope of $-4$ means that each dollar increase in the price
will reduce the number of spaces that will be rented by $4$.  The $y$-intercept
gives the number of spaces that could be occupied if the rent was free.
The $x$-intercept $x = 50$ gives the price at which no spaces will be rented.
\end{enumerate}
\vskip15pt
\noindent
15.
\begin{enumerate}
\item[a.] $P = 15 + .434 d$, where $d$ is the depth in feet and $P$ is the pressure
in pounds per square inch at depth $d$.  
\item[b.] $100 = 15 + .434 d$, when $d = \frac{100 - 15}{.434} \doteq 195.9$ feet.
\end{enumerate}
\vskip15pt
\noindent
18.
\begin{enumerate}
\item[a.]  The slope is $\frac{460 - 380}{800 - 480} = \frac{1}{4}$, so the 
equation is $C - 460 = \frac{1}{4}(d - 800)$, or $C = \frac{1}{4}d + 260$.  
\item[b.]  When $d = 1500$, the model in part a predicts
$C = \frac{1}{4}(1500) + 260 = 375 + 260 = 635$ dollars.
\item[c.]  The slope represents the increase in cost for each additional 
mile driven.
\item[d.]  The $y$-intercept represents the {\it fixed costs} (car payment, 
insurance, etc.) that are incurred even if the car is not driven at all.
\item[e.]  A linear model is reasonable here because there are 
some costs that are fixed (car payment, insurance, etc.) that are
independent of the mileage driven, and others (gas, oil, etc.) that
are variable, but proportional to the amount driven.  The total cost is the sum
of the fixed and variable costs.
\end{enumerate}

\section*{Section 1.3}

\noindent
2.  
\begin{enumerate}
\item[a.]  The graph is stretched vertically by a factor of 5.
\item[b.]  The graph is shifted to the right by 5 units along the $x$-axis.
\item[c.]  The graph is reflected across the $x$-axis.
\item[d.]  The is stretched vertically by a factor of 5 and reflected across the $x$-axis.
\item[e.]  The graph is compressed horizontally by a factor of $5$.
\item[f.]  The graph is stretched vertically by a factor of 5, then shifted down 
by $3$ units.  
\end{enumerate}
\vskip15pt
\noindent
3. 
\begin{enumerate}
\item[a.] This is graph 3, because it is a horizontal shift of $y = f(x)$ by 4 units to the right.
\item[b.] This is graph 1, because it is vertical shift by 1 unit up.
\item[c.] This is graph 4, because it is a vertical shrinking of $y = f(x)$.
\item[d.] This is graph 5, because it is a reflection of $y = f(x)$ across the 
$x$-axis, shifted to the left by 4 units.
\item[e] This is graph 2, because it is stretched vertically and shifted 6 units
to the left.
\end{enumerate}
\vskip15pt
\noindent
4. a)  Shift 4 units left.  b)  Shift 4 units up.  c)  Stretch vertically
by 2.  d) Compress vertically by 2, reflect across $x$-axis, then shift up by 3.
\vskip15pt
\noindent
6.  The graph is stretched vertically by a factor of 2 and shifted 2 units to the 
right, so we must have 
\[
y = 2f(x - 3) = 2\sqrt{3(x - 2) - (x - 2)^2} = 2\sqrt{-10 + 7x - x^2}.
\] 
\vskip15pt
\noindent
10.  The graph is obtained by reflecting the parabola $y = x^2$ across the 
$x$-axis, then shifting the resulting graph up by 1 unit.
\vskip15pt
\noindent
16.  The graph is obtained by shifting the graph $y = \frac{1}{x}$ to the 
right by 4 units.  This puts the vertical asymptote at $x = 4$.
\vskip15pt
\noindent
30.  For each $x$, the $y$-coordinate on $y = f(x) + g(x)$ is obtained by 
adding $f(x)$ and $g(x)$.  For example, since it appears $f(-1) = 1$
and $g(-1) = -1$, $f(-1) + g(-1) = 0$.  The domain of $f+g$ is the 
set of $x$ where {\it both} $f,g$ are defined, so $[-2,3]$.  
\vskip15pt
\noindent
32.  The domain of $f$ is $[-1,+\infty)$.  The domain of $g$ is $(-\infty,1]$.
\begin{itemize}
\item $(f + g)(x) = \sqrt{1 + x} + \sqrt{1 - x}$.  The domain is the 
set of $x$ that are in both the domain of $f$ and the domain of $g$, 
or $[-1,1]$ (that is all $x$ with $-1\le x \le 1$).
\item  $(f - g)(x) = \sqrt{1 + x} - \sqrt{1 - x}$.  The domain is the 
set of $x$ that are in both the domain of $f$ and the domain of $g$, 
or $[-1,1]$ (that is all $x$ with $-1\le x \le 1$).
\item $(f \cdot g)(x) = \sqrt{1 + x} \cdot \sqrt{1 - x} = \sqrt{1 - x^2}$.  The domain is the 
set of $x$ that are in both the domain of $f$ and the domain of $g$, 
or $[-1,1]$ (that is all $x$ with $-1\le x \le 1$).
\item $(f/g)(x) = \frac{\sqrt{1 + x}}{\sqrt{1 - x}}$.  The domain is the 
set of $x$ that are in both the domain of $f$ and the domain of $g$, and 
where $g(x) \ne 0$: $[-1,1)$ (that is all $x$ with $-1\le x < 1$).
\end{itemize}
\vskip15pt
\noindent
32.  The domains of $f$ and $g$ are $(-\infty,+\infty)$.  The domain
of a composition is the set of $x$ in the domain of the ``inside''
function such that the value is in the domain of the ``outside'' function.
\begin{itemize}
\item $(f\circ f)(x) = 1 - (1 - x^3)^3$.  The domain is the set of all real $x$.
\item $(g\circ g)(x) = \frac{1}{1/x} = x$.  The domain is the set of all $x \ne 0$
(since we cannot substitute $x = 0$ into the ``inside'' $g(x) = \frac{1}{x}$.
\item $(f \circ g)(x) =  1 - \frac{1}{x^3}$.  The domain is the 
set of $x\ne 0$.
\item $(g\circ f)(x) = \frac{1}{1 - x^3}$.  The domain is the set of all $x \ne 1$
in the real numbers.
\end{itemize}
\vskip15pt
\noindent
\vskip15pt
\noindent
40.   The domain
of a composition is the set of $x$ in the domain of the ``inside''
function such that the value is in the domain of the ``outside'' function.
\begin{itemize}
\item $(f\circ f)(x) = \sqrt{2\sqrt{2x + 3} + 3}$.  The domain is the set of all 
real $x \ge -3/2$.
\item $(g\circ g)(x) = (x^2 + 1)^2 + 1$.  The domain is the set of all real $x$.
\item $(f \circ g)(x) = \sqrt{2(x^2+1) + 3} = \sqrt{2x^2 + 5}$.  The domain is the 
set of all real $x$.
\item $(g\circ f)(x) = (\sqrt{2x + 3})^2 + 1 = 2x + 4$.  The domain is the set of 
all $x \ge -3/2$ in the real numbers, since we must be able to substitute
$x$ into the inside $f(x)$.  
\end{itemize}
\vskip15pt
\noindent
42.  $$(f\circ g\circ h)(x) = \frac{2}{\cos(\sqrt{x+3}) + 1}$$
\vskip15pt
\noindent
44.  $F(x) = \sin(\sqrt{x}) = (f \circ g)(x)$ for $f(x) = \sin(x)$ and $g(x) =\sqrt{x}$.
\vskip15pt
\noindent
50.  a)  $f(g(1)) = f(6) = 5$.  b) $g(f(1)) = g(3) = 2$. c) $f(f(1)) = f(3) = 4$.
d)  $g(g(1)) = g(6) = 3$.  e) $(g\circ f)(3) = g(f(3)) = g(4) = 1$.  f)  
$(f\circ g)(6) = f(g(6)) = f(3) = 4$.
\vskip15pt
\noindent
54.  a)  The radius is increasing at the constant rate of 2 cm/sec.  After
$t$ seconds, it is given by $r = 2t$.  b)  From the volumn formula
for a sphere, $V = \frac{4}{3}\pi r^3$, so $(V\circ r)(t) = \frac{4}{3}\pi (2t)^3 = 
\frac{32}{3}\pi t^3$.  This gives the volume of the balloon as a function of time.
\vskip15pt
\noindent
55. a) The positions of the ship when it is closest to the lighthouse,
the position at a later time, and the lighthouse form a right triangle with
sides $d,6$ and hypotenuse $s$.   
By the Pythagorean Theorem, $s = \sqrt{d^2 + 36}$.  b)  $d = 30t$.  c)
$s = \sqrt{(30t)^2 + 36} = \sqrt{900t^2 + 36}$.  This function
gives the distance to the lighthouse as a function of time.
\end{document}
