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\markright{Chapter 1: Worksheet 1}

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\Bf{Chapter 1: Answers 1 \hfill Jack K. Cohen \hfill Colorado School of Mines}

\begin{enumerate}
\item (1.56)  $A = P^2 / 16$.
\begin{figure}[htb]
\epsfysize 70pt
\centerline{\epsffile{ans1p1.eps}}
\caption{Problem 1: Area as a function of perimeter .}
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\item (1.58) $V =$ {\Large $\frac{S^{2/3}}{6 \sqrt{\pi}}$}.

\begin{figure}[htb]
\epsfysize 75pt
\centerline{\epsffile{ans1p2.eps}}
\caption{Problem 2: Volume as a function of surface area .}
\end{figure}



\item  For $a  > 0$,  $| a| = a$.  Moreover  $|-a| = a$ too.  Note that $|-a| = |a|$ \Em{all} the time!

\item  For $a < 0$,  $| a| = -a$;  $|-a| = -a$.

\item  For $x \geq a$, the inequality reads $b \leq x - a \leq c$.  Adding $a$ everywhere gives $a + b \leq x \leq a + c$.  Since $c \geq b$, this inequality can be satisfied by a non-empty set and since $b \geq 0$ the lower limit is compatible with the assumption $x \geq a$.  Thus part of the solution set is the closed interval,  $[a + b, a + c]$.  Similar analysis of the $x < a$ range gives the other part of the solution set.

\item
        \begin{enumerate}
        \item The solution set is empty.
        \item The solution set is the same as for $b = 0$.
        \end{enumerate}

\item
        \begin{enumerate}
        \item The solution set of $b \leq |x - a| < c$ is $(a - c, a - b] \cup [a + b, a + c)$.
        \item The solution set of $b < |x - a| \leq c$ is $[a - c, a - b) \cup (a + b, a + c]$.
        \item The solution set of $b <|x - a| \ <c$ is $(a - c, a - b) \cup (a + b, a + c)$.
        \end{enumerate}

\item (1.19) After dividing by 5, we see that $a = 4/5, b = 2/5, c = 4/5$.  Thus the solution set $[a - c, a -b] \cup [a + b, a + c]$ becomes $[0, 2/5] \cup [6/5, 8/5]$.

\item (1.17) If $c \rightarrow +\infty$, then the solution set becomes $(-\infty, a -b] \cup [a + b, +\infty)$.  Inequality $|1 - 3x| > 2$ can be rewritten as $2/3 < |x - 1/3|$, so $a = 1/3, b = 2/3$.  Noting the strict inequalities in this problem, the solution set becomes $(-\infty, -1/3) \cup(1, +\infty)$.

\item (1.15)  If $b = 0$, then the solution set $[a - c, a - b] \cup [a + b, a + c]$ becomes $[a - c, a ] \cup [a, a + c]$  or simply $[a - c,  a + c]$.  Rewrite  $|3 - 2x| < 5$ as $|x - 3/2| < 5/2$ to identify $a = 3/2, c = 5/2$.   The solution set is then $(-1, 4)$.   If we have \Em{strict} inequality at the lower end ($0 < |3 - 2x| < 5$), then the point $a = 3/2$ is ``missing'' from the solution set: $(-1, 3/2) \cup (3/2, 4)$.  Such ``punctured'' intervals play a key role in the study of limits.

\end{enumerate}
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