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\usepackage{enumerate}
\usepackage{calculus}
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\newcommand{\problem}[1]{\bigskip\textbf{Problem #1}\ }

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\title{Math 165: Honors Calculus I\\
Assignment 20 Solutions}

\begin{document}

\maketitle

\thispagestyle{empty}

\bigskip
\problem{6} Suppose $f$ and $g$ are functions such that $g(f(x)) = x$
for all $x$ in the domain of $f$. Prove that $f$ is one-to-one.

\medskip
\begin{proof}
We must show that if $x_1 \ne x_2$ then $f(x_1) \ne f(x_2)$. Suppose this
statement were not true. Then there would be numbers $x_1 \ne x_2$ such
that $f(x_1) = f(x_2)$. But then $x_1 = g(f(x_1)) = g(f(x_2)) = x_2$,
a contradiction.  Therefore, the statement must be true.
\end{proof}

\bigskip
\problem{7}  Let $f(x) = \int_1^x \frac{1}{t}\,dt$ for $x > 0$.

  \begin{enumerate}[a)]

  \item Prove that $f$ is strictly increasing and $f^{-1}(0) = 1$.

\begin{proof}
Let $0 < x_1 < x_2$. We must show that $f(x_1) < f(x_2)$. But this follows
from:
\[
f(x_2) = \int_1^{x_2} \frac{1}{t}\,dt 
       = \int_1^{x_1} \frac{1}{t}\,dt  +  \int_{x_1}^{x_2} \frac{1}{t}\,dt 
       = f(x_1) + \int_{x_1}^{x_2} \frac{1}{t}\,dt 
\]
Since $\frac{1}{t} \ge \frac{1}{x_2}$ on the interval $[x_1,x_2]$,
$\int_{x_1}^{x_2} \frac{1}{t}\,dt \ge \frac{1}{x_2}(x_2 - x_1) > 0$, and so
$f(x_2) > f(x_1)$. Finally, $f^{-1}(0) = 1$ because $f(1) = \int_{1}^1
\frac{1}{t}\,dt = 0$.
\end{proof}


  \item Prove that $f(ab) = f(a) + f(b)$.

\begin{proof}
$f(ab) = \int_1^{ab}\frac{1}{t}\,dt =\int_1^{a}\frac{1}{t}\,dt +
\int_a^{ab}\frac{1}{t}\,dt = f(a) + \int_a^{ab}\frac{1}{t}\,dt$. By
Theorem~1.19, $\int_a^{ab}\frac{1}{t}\,dt = a \int_{1}^{b}
\frac{1}{at}\,dt =  \int_{1}^{b} \frac{1}{t}\,dt = f(b)$.
\end{proof}

  \item Prove that $f^{-1}(a+b) = f^{-1}(a)f^{-1}(b)$.

\begin{proof}
Let $c = f^{-1}(a)$ and $d = f^{-1}(b)$. Then $f(c) = a$ and $f(d) = b$.
Using part b) we find that
$f^{-1}(a + b) = f^{-1}(f(c) + f(d)) = f^{-1}(f(cd)) = cd =
f^{-1}(a)f^{-1}(b)$.
\end{proof}

  \end{enumerate}

\bigskip
\problem{8}  Let $f$ be strictly decreasing on $[a,b]$. Prove that the
inverse $f^{-1}$  is strictly decreasing on $[f(b),f(a)]$.

\begin{proof}
Let $y_1 < y_2$ in $[f(b),f(a)]$. We must show that $f^{-1}(y_1) >
f^{-1}(y_2)$. Let $x_1 = f^{-1}(y_1)$ and $x_2 = f^{-1}(y_2)$ so
$f(x_1) = y_1$ and $f(x_2) = y_2$. Since $f$ is strictly decreasing, $x_1 >
x_2$ (if $x_1 \le x_2$ we would have $y_1 = f(x_1) \ge f(x_2) = y_2$, a
contradiction). Therefore, $f^{-1}(y_1) = x_1 > x_2 = f^{-1}(y_2)$.
\end{proof}

\bigskip
\problem{9}  Let $f(x) = x^4 - 8 x^2 + 8$. Determine the intervals on
which $f(x)$ is strictly increasing and strictly decreasing and find
the inverse of $f(x)$ on each of those intervals. Plot $f(x)$ and each
of the inverses as functions of $x$.

\medskip
\emph{Solution.}
It is clear from the graph that $f(x) = (x^2 - 4)^2 - 8$ is strictly
decreasing on the intervals $(-\infty,-2]$ and $[0,2]$, and strictly
increasing on the intervals $[-2,0]$ and $[2,\infty)$. To prove this
analytically: if $x_1 < x_2 \le -2$, then $x_1^2 > x_2^2 \ge 4$
so $x_1^2-4 > x_2^2-4 \ge 0$. Therefore, $(x_1^2-4)^2 > (x_2^2-4)^2$
and $f(x_1) > f(x_2)$. if $-2 \le x_1 < x_2 \le 0$, then $4 \ge x_1^2 >
x_2^2$ so $0 \ge x_1^2-4 > x_2^2-4$. Hence $(x_1^2-4)^2 < (x_2^2 -4)^2$
and $f(x_1) < f(x_2)$. If $0 \le x_1 < x_2 \le 2$, then $x_1^2 < x_2^2
\le 4$ so $x_1^2 - 4 < x_2^2 - 4 \le 0$. It follows that $(x_1^2-4)^2 >
(x_2^2-4)^2$ and $f(x_1) > f(x_2)$. Finally, if $2 \le x_1 < x_2$, then
$4 \le x_1^2 < x_2^2$ so $0 \le x_1^2-4 < x_2^2-4$.  Thus, $(x_1^2-4)^2 <
(x_2^2-4)^2$ and $f(x_1) < f(x_2)$.

To find the inverses of $f$ on these intervals we solve the equation $ y =
(x^2-4)^2 - 8$ for $x$ to get $x = \pm \sqrt{4 \pm \sqrt{y + 8}}$.
\[
\resizebox{2in}{!}{\includegraphics{assign20.1.eps}}
\]

\begin{enumerate}

\item If $x < -2$, then $x$ is negative and $|x| > 2$, forcing the outside
sign to be negative and the inside sign to be positive (otherwise the
square root would be less than 2): $x = - \sqrt{4 + \sqrt{y+8}}$.  As a
function of $x$ the inverse is: $f^{-1}(x) = -\sqrt{4 + \sqrt{x+8}}$
for $x \ge -8$ (the domain of the inverse is the range of the original
function $f$ on the interval $(-\infty,-2)$).
\[
\resizebox{2in}{!}{\includegraphics{assign20.p1.eps}}\qquad
\]
\item If $-2\le x \le 0$, then $x$ is still negative, but we need the
square root to be less than 2, so $x = -\sqrt{4 - \sqrt{y+8}}$.  As a
function of $x$ the inverse is: $f^{-1}(x) = -\sqrt{4 - \sqrt{x+8}}$
for $-8 \le x \le 8$.
\[
\resizebox{2in}{!}{\includegraphics{assign20.p2.eps}}
\]

\item If  If $0 \le x \le 2$, then $x$ is positive so the outside sign is
positive, but we still need the square root to be less than 2 so $x = +
\sqrt{4 - \sqrt{y + 8}}$.  As a function of $x$ the inverse is: $f^{-1}(x)
= +\sqrt{4 - \sqrt{x+8}}$ for $-8 \le x \le 8$
\[
\resizebox{2in}{!}{\includegraphics{assign20.p3.eps}}\qquad
\]

\item Finally, if
$x \ge 2$, the both signs should be positive so the square root will be
greater than 2, $x = + \sqrt{4 + \sqrt{y + 8}}$
As a function of $x$ the inverse is:
$f^{-1}(x) = +\sqrt{4 + \sqrt{x+8}}$ for $x \ge -8$
\end{enumerate}
\[
\resizebox{2in}{!}{\includegraphics{assign20.p4.eps}}
\]


\bigskip
\problem{10}  Determine the largest interval containing~$0$ on which the
functions $\sin(x)$ and $\tan(x)$ have inverses and plot the inverses on
those intervals. Explain why there is no interval containing~$0$ (i.e.,
no interval of the form $[a,b]$ where $a<0$ and $b>0$) on which $\cos(x)$
has an inverse. Then find an interval on which $\cos(x)$ \emph{does}
have an inverse and plot the inverse there.

\medskip
\emph{Solution.}
$\sin(x)$ and $\tan(x)$ are strictly increasing on $[-\pi/2,\pi/2]$,
but on no larger interval comtaining~0. The range of $\sin(x)$ on this
intevral is $[-1,1]$ which becomes the domain of its inverse function.
\[
\resizebox{2in}{!}{\includegraphics{assign20.s1.eps}}\qquad
\resizebox{2in}{!}{\includegraphics{assign20.s2.eps}}
\]

The range of $\tan(x)$ on this interval is $(-\infty,\infty)$, which means
its inverse is defined for all real numbers.
\[
\resizebox{1.5in}{!}{\includegraphics{assign20.t1.eps}}\qquad
\raisebox{.75in}{\resizebox{3in}{!}{\includegraphics{assign20.t2.eps}}}
\]


$\cos(x)$ is increasing on $[-\pi,0]$ and decreasing on $[0,\pi]$, so is
cannot be monotonic on any interval containing 0 as an interior point. The
natural interval to restrict to is $[0,\pi]$. The range of $\cos(x)$ on
this interval is $[-1,1]$, so becomes the domain of its inverse function.
\[
\resizebox{2in}{!}{\includegraphics{assign20.c1.eps}}\qquad
\resizebox{2in}{!}{\includegraphics{assign20.c2.eps}}
\]


\end{document}

