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

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\Cb{Symbol Problems, Inequalities}
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\noindent \Bf{Suggested Problems}
Section 1.1: 1, 5, 11, 15, 17, 19

\begin{enumerate}

\item \Star\ (1.56) Express the area $A$ of a square as a function of its perimeter $P$.  Sketch a graph of this function---label your axes.

\item \Star\ (1.58) Express the volume $V$ of a sphere as a function of its surface area $S$.  Sketch a graph of this function---label your axes.

\item  If $a  > 0$, correctly remove the absolute value signs from $| a|$.  Same for  $|-a|$.

\item  If $a < 0$, correctly remove the absolute value signs from $| a|$.  Same for $|-a|$.

\item True confession time.  Problems like 1.19 drive me crazy!  One day, the Voice said, ``Use the symbols, Luke''.  Huh?  Symbols?   Oh yeah, symbols \ldots algebra \ldots do a general case!  Our inequality looks like $B \leq | D x + E| \leq C$.  By dividing by a positive constant, we can simplify to $b \leq | x - a| \leq c$ (convince yourself!).  The ``interesting'' case is $0 \leq b \leq c$, so assume this.

\noindent By doing the cases $x \leq a$ and $x > a$ separately,
show that the solution to the simplified inequality, $b \leq | x - a| \leq c$, ($0 \leq b \leq c$) is the union of the closed intervals, $[a - c, a - b]$ and $[a + b, a + c]$.

\item Why are these cases not ``interesting''?
        \begin{enumerate}
        \item $b \leq | x - a| \leq c$ with $b > c$.
        \item $b \leq | x - a| \leq c$ with $b < 0$.
        \end{enumerate}
        
\item Write down the solution of the closely related inequalities:
        \begin{enumerate}
        \item $b \leq |x - a| < c$
        \item $b < |x - a| \leq c$
        \item $b <|x - a| \ <c$
        \end{enumerate}

\item (1.19) Use the general inequality solution of the previous problems to check up on $2 \leq |4 - 5x| \leq 4$.   Don't forget to put the problem into the right form by ``dividing by a positive constant''.

\item (1.17)  What happens to the general inequality solution if $c \rightarrow +\infty$?  Check up on $|1 - 3x| > 2$.  (Hint: explain why this is equivalent to $2 < |1 - 3x|  < +\infty$.)

\item (1.15)  What happens to the general inequality solution if $b = 0$?  Check up on $|3 - 2x| < 5$.  Is the solution of  $0 < |3 - 2x| < 5$ the same as for the given inequality?

\end{enumerate}
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