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\title{Solutions for Problem Set 2}
\author{MATH 131, Fall 2007}
\maketitle

\section*{Section 1.5}

\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 $y = e^x$, stretch vertically by a factor of $2$, then 
shift up one unit.
\vskip 15pt
\noindent
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
\noindent
14.  
\begin{enumerate}
\item[a.]  $y = 8 - e^x$ (same as reflecting across the $x$-axis, then shifting
up by 8 units).
\item[b.]  $y = e^{4 - x} = e^{-(x - 4)}$ (same as shifting right by 4 units, then 
reflect across the $y$-axis).
\end{enumerate}
\vskip 15pt
17.  From the given points, $6 = Ca$ and $24 = Ca^3$.  Hence
$4 = \frac{Ca^3}{Ca} = a^2$, so $a = 2$ and $C = 3$.  $f(x) = 3\cdot 2^x$.
\vskip 15pt
\noindent
18.  Same method as 17:  $2 = Ca^0 = C$ and $2/9 = C a^2 = 2a^2$.  So
$a^2 = 1/9$, and $a = 1/3$.  $f(x) = 2\cdot \left(\frac{1}{3}\right)^x$.
\vskip 15pt
\noindent
20.  Definitely take option II.  The reason is that since each 
month has at least 28 days, on the last day you will get
at least $2^{27}$ pennies.  This is $\$(.01) 134,217,728 = \$1,342,177.28$,
which is more than a million dollars, just on the last day.  
\vskip 15pt
\noindent
26.
\begin{enumerate}
\item[a.] There are 6 doublings in the first three hours, so 
$500 \cdot 2^6 = 32000$.
\item[b.] After $t$ hours, there are $500 \cdot 2^{2t}$ bacteria.
\item[c.] 40 minutes $ = \frac{2}{3}$ hour, so 
$500 \cdot 2^{4/3} \doteq 1260$ bacteria.
\item[d.] The graph is a standard increasing exponential, with
vertical axis intercept at $500$.  The population reaches
$10000$ when $100000 = 500 \cdot 2^{2t}$, so $200 = 2^{2t}$, so 
$t = \frac{1}{2} \log_2(200)$ hours (about $3.82$ hours).  
\end{enumerate}

\section*{Section 1.6}

\noindent
4.  Apparently one-to-one, since no duplicate $f(x)$-values appear in table.
\vskip 15pt
\noindent
6.  Not one-to-one.  Note that $f(x) = 0$ for three different $x$-values
in the picture.  The graph does not pass the horizontal line test.
\vskip 15pt
\noindent
10.  The graph $y = 10 - 3x$ is a line with slope $-3$, so it crosses
horizontal lines only once.  This is a one-to-one function.
\vskip 15pt
\noindent
12.  Not one-to-one, since, for instance $\cos(0) = 1 = \cos(2\pi)$
\vskip 15pt
\noindent
14.  Not one-to-one, since peoples' heights typically stay roughly 
constant for extended periods (e.g. between growth spurts as children,
or once they reach maturity). 
\vskip 15pt
\noindent
15. $f^{-1}(9) = 2$.
\vskip 15pt
\noindent
16 b.  For any function $f$ that is one-to-one, if $5$ is in
the range of $f$, $f(f^{-1}(5)) = 5$.  (This has nothing to do
with the given formula for $f(x)$.)
\vskip 15pt
\noindent
21.  $y = \sqrt{10 - 3x}$ implies $x = \frac{1}{-3}(y^2 - 10) = \frac{10}{3} - \frac{y^2}{3}$.
So $f^{-1}(x) = \frac{10}{3} - \frac{x^2}{3}$.
\vskip 15pt
\noindent
24.  $y = 2x^3 + 3$ implies $x = \left(\frac{y - 3}{2}\right)^{1/3} = 
\root{3}\of{\frac{y - 3}{2}}$, so $f^{-1}(x) = \root{3}\of{\frac{x - 3}{2}}$.
\vskip 15pt
\noindent
26.  $y = \frac{1 + e^x}{1 - e^x}$ implies $y - ye^x = 1 + e^x$, so 
$e^x = \frac{y - 1}{y + 1}$, and $x = \ln\left(\frac{y - 1}{y + 1}\right)$.
The inverse function is $f^{-1}(x) = \ln\left(\frac{x - 1}{x + 1}\right)$.
\vskip 15pt
\noindent
30.  Reflect the graph across the line $y = x$.  The inverse function
graph should be decreasing from about $(-1,4)$, through $(0,2)$, to $(1,0)$.
\vskip 15pt
\noindent
34.  a. $\log_8(2) = 1/3$, since $8^{1/3} = 2$. b. $\ln(e^{\sqrt{2}}) = \sqrt{2}$.
\vskip 15pt
\noindent
36. a)  Using properties of logs and exponentials, 
$2^{\log_2(3) + \log_2(5)} = 2^{\log_2(15)} = 15$.  
\vskip 15pt
\noindent
48.
\begin{enumerate}
\item[a.]  Take $\ln$ of both sides in $e^{2x + 3} = 7$.  We get $2x + 3 = \ln(7)$,
so $x = \frac{\ln(7) - 3}{2}$.
\item[b.]  Take exponential of both sides in $\ln(5 - 2x) = -3$ to get
$5 - 2x = e^{-3}$.  Then $x = \frac{5 - e^{-3}}{2}$.
\end{enumerate}
\vskip 15pt
\noindent
50 a.  Take exponentials twice, $\ln(\ln(x)) = 1$ implies $\ln(x) = e$, so $x = e^e$.
\vskip 15pt
\noindent
54.  Let $f(x) = \ln(2 + \ln(x))$.  The domain of this composition is the 
set of $x$ in the domain of $\ln(x)$ such that $2 + \ln(x)$ is in the domain of 
the ``outside'' $\ln$.  Recall that $\ln(x)$ has domain the set of all positive 
reals.  But $2 + \ln(x) > 0$ only for $x > e^{-2}$.  So the domain is the 
set of all real $x$ with $x > e^{-2}$.  
The inverse function is found by the usual process: $y = \ln(2 + \ln(x))$
when $e^y = 2 + \ln(x)$, so $e^{e^y - 2} = x$.  Thus $f^{-1}(x) = e^{e^x - 2}$.
The domain of this function is the same as the range of $f(x)$, which is
the set of all real $x$.  

\section*{Appendix C}

\noindent
2. a. $-315^\circ = -\frac{7\pi}{4}$ radians.  (The negative
angle means measure clockwise, not counterclockwise as usual.)
b.  $36^\circ = \frac{\pi}{5}$ radians.
\vskip 15pt
\noindent
4.  a.  $-\frac{7\pi}{2} = -630^\circ$.  b.  $\frac{8\pi}{3} = 480^\circ$.
\vskip 15pt
\noindent
16.  $x = 25\cos(40^\circ) \doteq 19.15111$ (using Maple).
\vskip 15pt
\noindent
24.  Square out the left:
$$(\sin(x) + \cos(x))^2 = \sin^2(x) + 2\sin(x)\cos(x) + \cos^2(x).$$
Regrouping the terms, we have
$$= (\sin^2(x)  + \cos^2(x)) + 2\sin(x)\cos(x) = 1 + \sin(2x),$$
by the basic identity 7, and the double angle formula 15a.
\vskip 15pt
\noindent
30.  $2\sin^2(x) = 1$ when $\sin(x) = \pm \frac{1}{\sqrt{2}}$, so 
$x = \frac{\pi}{4},  \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$
(the first two give $+ \frac{1}{\sqrt{2}}$ the other two give $- \frac{1}{\sqrt{2}}$).
\vskip 15pt
\noindent
32.  $|\tan(x)| = 1$ when $\tan(x) = \pm 1$, so $\sin(x) = \pm \cos(x)$.  This 
happens at the same $x$-values given in problem 30:
$x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.
\vskip 15pt
\noindent
37.  Shift the graph $y = \cos(x)$ by $\frac{\pi}{3}$ units to the right.
\vskip 15pt
\noindent
38.  Compress the graph $y = \tan(x)$ by a factor of 2 horizontally.
This puts vertical asymptotes at $x = \pm \frac{\pi}{4}, \pm \frac{3\pi}{4}$, etc.
\vskip 15pt
\noindent
41. a.  This is the unique angle $\theta$ in the interval 
$\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$
with $\sin(\theta) = .5$, so $\theta = \frac{\pi}{6}$.
b.  $\arctan(-1) = \tan^{-1}(-1) = -\frac{\pi}{4}$.
\vskip 15pt
\noindent
42.  a.  $\tan^{-1}\sqrt{3} = \frac{\pi}{3}$.  b.  $\arcsin(1) = \sin^{-1}(1) = \frac{\pi}{2}$.
\vskip 15pt
\noindent
43. a. $\sin(\sin^{-1}(0.7)) = 0.7$ (see problem 16 b in Section 1.6 above).  
b. $\sin\left(\frac{5\pi}{4}\right) = -\frac{1}{\sqrt{2}}$, so 
$\arcsin\left(-\frac{1}{\sqrt{2}}\right) = -\frac{\pi}{4}$.
(Values of $\arcsin = \sin^{-1}$ are all in the interval 
$\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$.)

\vskip 15pt
\noindent
44.  a)  If $\theta = \arctan(2)$, then $\tan(\theta) = 2$, so 
we can get $\theta$ in a right triangle with opposite side $2$ and adjacent side
1 (draw the picture!).  Then the hypotenuse is $\sqrt{5}$, so 
$\sec(\theta) = \sqrt{5}$ (hypotenuse/adjacent).  

b)  First use the double angle formula:
$$\sin(2\sin^{-1}(3/5)) = 2\sin(\sin^{-1}(3/5))\cos(\sin^{-1}(3/5)) = 
2\cdot \frac{3}{5}\cdot \sqrt{1 - \left(\frac{3}{5}\right)^2} = 
\frac{24}{25}.$$
Note: the value $\frac{4}{5} = \cos(\sin^{-1}(3/5))$ is obtained
by a calculation similar to what we did in part a.  If 
$\theta = \sin^{-1}(3/5)$, then $\theta$ is an angle in a right
triangle with opposite side 3 and hypotenuse 5.  The adjacent side
is $\sqrt{25 - 9} = 4$, so $\cos(\theta) = 4/5$ (adjacent/hypotenuse).

\end{document}
