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# 4.E: Rational Functions (Exercises)

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## 4.1: Introduction to Rational Functions

\subsection{Exercises}

In Exercises \ref{alltheasympfirst} - \ref{alltheasymplast}, for the given rational function $f$:

\begin{itemize}

\item Find the domain of $f$.

\item Identify any vertical asymptotes of the graph of $y = f(x)$.

\item Identify any holes in the graph.

\item Find the horizontal asymptote, if it exists.

\item Find the slant asymptote, if it exists.

\item Graph the function using a graphing utility and describe the behavior near the asymptotes.

\end{itemize}

\begin{multicols}{3}

\begin{enumerate}

\item $f(x) = \dfrac{x}{3x - 6}$ \label{alltheasympfirst}

\item $f(x) = \dfrac{3 + 7x}{5 - 2x}$

\item $f(x) = \dfrac{x}{x^{2} + x - 12}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $f(x) = \dfrac{x}{x^{2} + 1}$

\item $f(x) = \dfrac{x + 7}{(x + 3)^{2}}$

\item $f(x) = \dfrac{x^{3} + 1}{x^{2} - 1}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $f(x) = \dfrac{4x}{x^2+4}$

\item $f(x) = \dfrac{4x}{x^2-4}$

\item $f(x) = \dfrac{x^2-x-12}{x^2+x-6}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $f(x) = \dfrac{3x^2-5x-2}{x^2-9}$

\item $f(x) = \dfrac{x^3+2x^2+x}{x^2-x-2}$

\item $f(x) = \dfrac{x^{3} - 3x + 1}{x^{2} + 1}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $f(x) = \dfrac{2x^{2} + 5x - 3}{3x + 2}$

\item $f(x) = \dfrac{-x^{3} + 4x}{x^{2} - 9}$

\item \small $f(x) = \dfrac{-5x^{4} - 3x^{3} + x^{2} - 10}{x^{3} - 3x^{2} + 3x - 1}$ \normalsize

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $f(x) = \dfrac{x^3}{1-x}$

\item $f(x) = \dfrac{18-2x^2}{x^2-9}$

\item $f(x) = \dfrac{x^3-4x^2-4x-5}{x^2+x+1}$ \label{alltheasymplast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item The cost $C$ in dollars to remove $p$\% of the invasive species of Ippizuti fish from Sasquatch Pond is given by $C(p) = \frac{1770p}{100 - p}, \quad 0 \leq p < 100$

\begin{enumerate}

\item Find and interpret $C(25)$ and $C(95)$.

\item What does the vertical asymptote at $x = 100$ mean within the context of the problem?

\item What percentage of the Ippizuti fish can you remove for \$40000? \end{enumerate} \item In Exercise \ref{Sasquatchfunc1} in Section \ref{FunctionNotation}, the population of Sasquatch in Portage County was modeled by the function $P(t) = \frac{150t}{t + 15},$ where$t = 0$represents the year 1803. Find the horizontal asymptote of the graph of$y = P(t)$and explain what it means. \item Recall from Example \ref{costrevenueprofitex1} that the cost$C$(in dollars) to make$x$dOpi media players is$C(x) = 100x+2000$,$x \geq 0$. \begin{enumerate} \item Find a formula for the average cost$\overline{C}(x)$. Recall:$\overline{C}(x) = \frac{C(x)}{x}$. \item Find and interpret$\overline{C}(1)$and$\overline{C}(100)$. \item How many dOpis need to be produced so that the average cost per dOpi is$\$200$?

\item Interpret the behavior of $\overline{C}(x)$ as $x \rightarrow 0^{+}$. (HINT: You may want to find the fixed cost $C(0)$ to help in your interpretation.)

\item Interpret the behavior of $\overline{C}(x)$ as $x \rightarrow \infty$. (HINT: You may want to find the variable cost (defined in Example \ref{PortaBoyCost} in Section \ref{LinearFunctions}) to help in your interpretation.)

\end{enumerate}

\item In Exercise \ref{circuitexercisepoly} in Section \ref{GraphsofPolynomials}, we fit a few polynomial models to the following electric circuit data. (The circuit was built with a variable resistor. For each of the following resistance values (measured in kilo-ohms, $k \Omega$), the corresponding power to the load (measured in milliwatts, $mW$) is given in the table below.)\footnote{The authors wish to thank Don Anthan and Ken White of Lakeland Community College for devising this problem and generating the accompanying data set.}

\smallskip

\noindent \begin{tabular}{|l|r|r|r|r|r|r|} \hline

Resistance: ($k \Omega$) & 1.012 & 2.199 & 3.275 & 4.676 & 6.805 & 9.975 \\ \hline

Power: ($mW$) & 1.063 & 1.496 & 1.610 & 1.613 & 1.505 & 1.314 \\ \hline

\end{tabular}

\smallskip

\noindent Using some fundamental laws of circuit analysis mixed with a healthy dose of algebra, we can derive the actual formula relating power to resistance. For this circuit, it is $P(x) = \frac{25x}{(x + 3.9)^2}$, where $x$ is the resistance value, $x \geq 0$.

\begin{enumerate}

\item Graph the data along with the function $y = P(x)$ on your calculator.

\item Use your calculator to approximate the maximum power that can be delivered to the load. What is the corresponding resistance value?

\item Find and interpret the end behavior of $P(x)$ as $x \rightarrow \infty$.

\end{enumerate}

\item \index{Learning Curve Equation} \index{Thurstone, Louis Leon} In his now famous 1919 dissertation \underline{The Learning Curve Equation}, Louis Leon Thurstone presents a rational function which models the number of words a person can type in four minutes as a function of the number of pages of practice one has completed. (This paper, which is now in the public domain and can be found \href{http://books.google.com/books?id=pb5...derline{here}}, is from a bygone era when students at business schools took typing classes on manual typewriters.) Using his original notation and original language, we have $Y = \frac{L(X + P)}{(X + P) + R}$ where $L$ is the predicted practice limit in terms of speed units, $X$ is pages written, $Y$ is writing speed in terms of words in four minutes, $P$ is equivalent previous practice in terms of pages and $R$ is the rate of learning. In Figure 5 of the paper, he graphs a scatter plot and the curve $Y = \frac{216(X + 19)}{X + 148}$. Discuss this equation with your classmates. How would you update the notation? Explain what the horizontal asymptote of the graph means. You should take some time to look at the original paper. Skip over the computations you don't understand yet and try to get a sense of the time and place in which the study was conducted.

\end{enumerate}

\newpage

\begin{multicols}{2}

\begin{enumerate}

\item $f(x) = \dfrac{x}{3x - 6}$ \vphantom{$\dfrac{7x}{7x}$}\\

Domain: $(-\infty, 2) \cup (2, \infty)$\\

Vertical asymptote: $x = 2$\\

As $x \rightarrow 2^{-}, f(x) \rightarrow -\infty$\\

As $x \rightarrow 2^{+}, f(x) \rightarrow \infty$\\

No holes in the graph\\

Horizontal asymptote: $y = \frac{1}{3}$ \\

As $x \rightarrow -\infty, f(x) \rightarrow \frac{1}{3}^{-}$\\

As $x \rightarrow \infty, f(x) \rightarrow \frac{1}{3}^{+}$\\

\vfill

\columnbreak

\item $f(x) = \dfrac{3 + 7x}{5 - 2x}$\\

Domain: $(-\infty, \frac{5}{2}) \cup (\frac{5}{2}, \infty)$\\

Vertical asymptote: $x = \frac{5}{2}$\\

As $x \rightarrow \frac{5}{2}^{-}, f(x) \rightarrow \infty$\\

As $x \rightarrow \frac{5}{2}^{+}, f(x) \rightarrow -\infty$\\

No holes in the graph\\

Horizontal asymptote: $y = -\frac{7}{2}$ \\

As $x \rightarrow -\infty, f(x) \rightarrow -\frac{7}{2}^{+}$\\

As $x \rightarrow \infty, f(x) \rightarrow -\frac{7}{2}^{-}$\\

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $f(x) = \dfrac{x}{x^{2} + x - 12} = \dfrac{x}{(x + 4)(x - 3)}$\\

Domain: $(-\infty, -4) \cup (-4, 3) \cup (3, \infty)$\\

Vertical asymptotes: $x = -4, x = 3$\\

As $x \rightarrow -4^{-}, f(x) \rightarrow -\infty$\\

As $x \rightarrow -4^{+}, f(x) \rightarrow \infty$\\

As $x \rightarrow 3^{-}, f(x) \rightarrow -\infty$\\

As $x \rightarrow 3^{+}, f(x) \rightarrow \infty$\\

No holes in the graph\\

Horizontal asymptote: $y = 0$ \\

As $x \rightarrow -\infty, f(x) \rightarrow 0^{-}$\\

As $x \rightarrow \infty, f(x) \rightarrow 0^{+}$\\

\vfill

\columnbreak

\item $f(x) = \dfrac{x}{x^{2} + 1}$\\

Domain: $(-\infty, \infty)$\\

No vertical asymptotes\\

No holes in the graph\\

Horizontal asymptote: $y = 0$ \\

As $x \rightarrow -\infty, f(x) \rightarrow 0^{-}$\\

As $x \rightarrow \infty, f(x) \rightarrow 0^{+}$\\

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $f(x) = \dfrac{x + 7}{(x + 3)^{2}}$ \vphantom{$\dfrac{x^{3}}{x^{2}}$}\\

Domain: $(-\infty, -3) \cup (-3, \infty)$\\

Vertical asymptote: $x = -3$\\

As $x \rightarrow -3^{-}, f(x) \rightarrow \infty$\\

As $x \rightarrow -3^{+}, f(x) \rightarrow \infty$\\

No holes in the graph\\

Horizontal asymptote: $y = 0$ \\

\footnote{This is hard to see on the calculator, but trust me, the graph is below the $x$-axis to the left of $x = -7$.}As $x \rightarrow -\infty, f(x) \rightarrow 0^{-}$\\

As $x \rightarrow \infty, f(x) \rightarrow 0^{+}$\\

\vfill

\columnbreak

\item $f(x) = \dfrac{x^{3} + 1}{x^{2} - 1} = \dfrac{x^{2} - x+ 1}{x-1}$\\

Domain: $(-\infty, -1) \cup (-1, 1) \cup (1, \infty)$\\

Vertical asymptote: $x = 1$\\

As $x \rightarrow 1^{-}, f(x) \rightarrow -\infty$\\

As $x \rightarrow 1^{+}, f(x) \rightarrow \infty$\\

Hole at $(-1, -\frac{3}{2})$\\

Slant asymptote: $y=x$ \\

As $x \rightarrow -\infty$, the graph is below $y=x$\\

As $x \rightarrow \infty$, the graph is above $y=x$\\

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $f(x) = \dfrac{4x}{x^{2} + 4}$\\

Domain: $(-\infty, \infty)$\\

No vertical asymptotes \\

No holes in the graph\\

Horizontal asymptote: $y = 0$ \\

As $x \rightarrow -\infty, f(x) \rightarrow 0^{-}$\\

As $x \rightarrow \infty, f(x) \rightarrow 0^{+}$\\

\vfill

\columnbreak

\item $f(x) = \dfrac{4x}{x^{2} -4} = \dfrac{4x}{(x + 2)(x - 2)}$\\

Domain: $(-\infty, -2) \cup (-2, 2) \cup (2, \infty)$\\

Vertical asymptotes: $x = -2, x = 2$\\

As $x \rightarrow -2^{-}, f(x) \rightarrow -\infty$\\

As $x \rightarrow -2^{+}, f(x) \rightarrow \infty$\\

As $x \rightarrow 2^{-}, f(x) \rightarrow -\infty$\\

As $x \rightarrow 2^{+}, f(x) \rightarrow \infty$\\

No holes in the graph\\

Horizontal asymptote: $y = 0$ \\

As $x \rightarrow -\infty, f(x) \rightarrow 0^{-}$\\

As $x \rightarrow \infty, f(x) \rightarrow 0^{+}$\\

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $f(x) = \dfrac{x^2-x-12}{x^{2} +x - 6} = \dfrac{x-4}{x - 2}$\\

Domain: $(-\infty, -3) \cup (-3, 2) \cup (2, \infty)$\\

Vertical asymptote: $x = 2$\\

As $x \rightarrow 2^{-}, f(x) \rightarrow \infty$\\

As $x \rightarrow 2^{+}, f(x) \rightarrow -\infty$\\

Hole at $\left(-3, \frac{7}{5} \right)$ \\

Horizontal asymptote: $y = 1$ \\

As $x \rightarrow -\infty, f(x) \rightarrow 1^{+}$\\

As $x \rightarrow \infty, f(x) \rightarrow 1^{-}$\\

\vfill

\columnbreak

\item $f(x) = \dfrac{3x^2-5x-2}{x^{2} -9} = \dfrac{(3x+1)(x-2)}{(x + 3)(x - 3)}$\\

Domain: $(-\infty, -3) \cup (-3, 3) \cup (3, \infty)$\\

Vertical asymptotes: $x = -3, x = 3$\\

As $x \rightarrow -3^{-}, f(x) \rightarrow \infty$\\

As $x \rightarrow -3^{+}, f(x) \rightarrow -\infty$\\

As $x \rightarrow 3^{-}, f(x) \rightarrow -\infty$\\

As $x \rightarrow 3^{+}, f(x) \rightarrow \infty$\\

No holes in the graph\\

Horizontal asymptote: $y = 3$ \\

As $x \rightarrow -\infty, f(x) \rightarrow 3^{+}$\\

As $x \rightarrow \infty, f(x) \rightarrow 3^{-}$\\

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $f(x) = \dfrac{x^3+2x^2+x}{x^{2} -x-2} = \dfrac{x(x+1)}{x - 2}$\\

Domain: $(-\infty, -1) \cup (-1, 2) \cup (2, \infty)$\\

Vertical asymptote: $x = 2$\\

As $x \rightarrow 2^{-}, f(x) \rightarrow -\infty$\\

As $x \rightarrow 2^{+}, f(x) \rightarrow \infty$\\

Hole at $(-1,0)$ \\

Slant asymptote: $y=x+3$ \\

As $x \rightarrow -\infty$, the graph is below $y=x+3$\\

As $x \rightarrow \infty$, the graph is above $y=x+3$\\

\vfill

\columnbreak

\item $f(x) = \dfrac{x^3-3x+1}{x^2+1}$\\

Domain: $(-\infty, \infty)$\\

No vertical asymptotes \\

No holes in the graph \\

Slant asymptote: $y=x$ \\

As $x \rightarrow -\infty$, the graph is above $y=x$ \\

As $x \rightarrow \infty$, the graph is below $y=x$ \\

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\newpage

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $f(x) = \dfrac{2x^{2} + 5x - 3}{3x + 2}$\\

Domain: $\left(-\infty, -\frac{2}{3}\right) \cup \left(-\frac{2}{3}, \infty\right)$\\

Vertical asymptote: $x = -\frac{2}{3}$\\

As $x \rightarrow -\frac{2}{3}^{-}, f(x) \rightarrow \infty$\\

As $x \rightarrow -\frac{2}{3}^{+}, f(x) \rightarrow -\infty$\\

No holes in the graph \\

Slant asymptote: $y = \frac{2}{3}x + \frac{11}{9}$ \\

As $x \rightarrow -\infty$, the graph is above \small $y = \frac{2}{3}x + \frac{11}{9}$\\

\normalsize As $x \rightarrow \infty$, the graph is below \small $y = \frac{2}{3}x + \frac{11}{9}$ \normalsize \\

\vfill

\columnbreak

\item $f(x) = \dfrac{-x^{3} + 4x}{x^{2} - 9} = \dfrac{-x^{3} + 4x}{(x-3)(x+3)}$\\

Domain: $(-\infty, -3) \cup (-3, 3) \cup (3, \infty)$\\

Vertical asymptotes: $x = -3$, $x=3$\\

As $x \rightarrow -3^{-}, f(x) \rightarrow \infty$\\

As $x \rightarrow -3^{+}, f(x) \rightarrow -\infty$\\

As $x \rightarrow 3^{-}, f(x) \rightarrow \infty$\\

As $x \rightarrow 3^{+}, f(x) \rightarrow -\infty$\\

No holes in the graph \\

Slant asymptote: $y=-x$ \\

As $x \rightarrow -\infty$, the graph is above $y=-x$\\

As $x \rightarrow \infty$, the graph is below $y=-x$\\

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item \small $f(x) = \dfrac{-5x^{4} - 3x^{3} + x^{2} - 10}{x^{3} - 3x^{2} + 3x - 1} \\ \phantom{f(x)} = \dfrac{-5x^{4} - 3x^{3} + x^{2} - 10}{(x-1)^3}$ \normalsize \\

Domain: $(-\infty, 1) \cup (1, \infty)$\\

Vertical asymptotes: $x = 1$\\

As $x \rightarrow 1^{-}, f(x) \rightarrow \infty$\\

As $x \rightarrow 1^{+}, f(x) \rightarrow -\infty$\\

No holes in the graph \\

Slant asymptote: $y=-5x-18$ \\

\small As $x \rightarrow -\infty$, the graph is above $y=-5x-18$ \normalsize\\

\small As $x \rightarrow \infty$, the graph is below $y=-5x-18$ \normalsize \\

\vfill

\columnbreak

\item $f(x) = \dfrac{x^3}{1-x}$\\

Domain: $(-\infty, 1) \cup (1, \infty)$\\

Vertical asymptote: $x=1$\\

As $x \rightarrow 1^{-}, f(x) \rightarrow \infty$\\

As $x \rightarrow 1^{+}, f(x) \rightarrow -\infty$\\

No holes in the graph \\

No horizontal or slant asymptote \\

As $x \rightarrow -\infty$, $f(x) \rightarrow -\infty$\\

As $x \rightarrow \infty$, $f(x) \rightarrow -\infty$\\

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $f(x) = \dfrac{18-2x^2}{x^2-9} = -2$\\

Domain: $(-\infty, -3) \cup (-3,3) \cup (3, \infty)$\\

No vertical asymptotes \\

Holes in the graph at $(-3,-2)$ and $(3,-2)$ \\

Horizontal asymptote $y = -2$ \\

As $x \rightarrow \pm \infty$, $f(x) = -2$ \\

\vfill

\columnbreak

\item $f(x) = \dfrac{x^3-4x^2-4x-5}{x^2+x+1} = x-5$\\

Domain: $(-\infty, \infty)$\\

No vertical asymptotes \\

No holes in the graph \\

Slant asymptote: $y = x-5$ \\

$f(x) = x-5$ everywhere. \\

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item \begin{enumerate}

\item $C(25) = 590$ means it costs \$590 to remove 25\% of the fish and and$C(95)= 33630$means it would cost \$33630 to remove 95\% of the fish from the pond.

\item The vertical asymptote at $x = 100$ means that as we try to remove 100\% of the fish from the pond, the cost increases without bound; i.e., it's impossible to remove all of the fish.

\item For \$40000 you could remove about 95.76\% of the fish. \end{enumerate} \item The horizontal asymptote of the graph of$P(t) = \frac{150t}{t + 15}$is$y = 150$and it means that the model predicts the population of Sasquatch in Portage County will never exceed 150. \item \begin{enumerate} \item$\overline{C}(x) = \frac{100x+2000}{x}$,$x > 0$. \item$\overline{C}(1) = 2100$and$\overline{C}(100) = 120$. When just$1$dOpi is produced, the cost per dOpi is$\$2100$, but when $100$ dOpis are produced, the cost per dOpi is $\$120$. \item$\overline{C}(x) = 200$when$x = 20$. So to get the cost per dOpi to$\$200$, $20$ dOpis need to be produced.

\item As $x \rightarrow 0^{+}$, $\overline{C}(x) \rightarrow \infty$. This means that as fewer and fewer dOpis are produced, the cost per dOpi becomes unbounded. In this situation, there is a fixed cost of $\$2000$($C(0) = 2000$), we are trying to spread that$\$2000$ over fewer and fewer dOpis.

\item As $x \rightarrow \infty$, $\overline{C}(x) \rightarrow 100^{+}$. This means that as more and more dOpis are produced, the cost per dOpi approaches $\$100$, but is always a little more than$\$100$. Since $\$100$is the variable cost per dOpi ($C(x) = \underline{100}x+2000$), it means that no matter how many dOpis are produced, the average cost per dOpi will always be a bit higher than the variable cost to produce a dOpi. As before, we can attribute this to the$\$2000$ fixed cost, which factors into the average cost per dOpi no matter how many dOpis are produced.

\end{enumerate}

\item \begin{enumerate}

\item $~$

\centerline{\includegraphics[width=2in]{./RationalsGraphics/CIRCRAT.jpg}}

\item The maximum power is approximately $1.603 \; mW$ which corresponds to $3.9 \; k\Omega$.

\item As $x \rightarrow \infty, \; P(x) \rightarrow 0^{+}$ which means as the resistance increases without bound, the power diminishes to zero.

\end{enumerate}

\end{enumerate}

\closegraphsfile

## 4.2: Graphs of Rational Functions

\subsection{Exercises}

In Exercises \ref{sixstepfirst} - \ref{sixsteplast}, use the six-step procedure to graph the rational function. Be sure to draw any asymptotes as dashed lines.

\begin{multicols}{2}

\begin{enumerate}

\item $f(x) = \dfrac{4}{x + 2}$ \label{sixstepfirst}

\item $f(x) = \dfrac{5x}{6 - 2x}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $f(x) = \dfrac{1}{x^{2}}$

\item $f(x) = \dfrac{1}{x^{2} + x - 12}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $f(x) = \dfrac{2x - 1}{-2x^{2} - 5x + 3}$

\item $f(x) = \dfrac{x}{x^{2} + x - 12}$ \vphantom{$\dfrac{2x}{2x}$}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $f(x) = \dfrac{4x}{x^2+4}$

\item $f(x) = \dfrac{4x}{x^2-4}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $f(x) = \dfrac{x^2-x-12}{x^2+x-6}$

\item $f(x) = \dfrac{3x^2-5x-2}{x^2-9}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $f(x) = \dfrac{x^2-x-6}{x+1}$

\item $f(x) = \dfrac{x^2-x}{3-x}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $f(x) = \dfrac{x^3+2x^2+x}{x^2-x-2}$

\item $f(x) = \dfrac{-x^{3} + 4x}{x^{2} - 9}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $f(x) = \dfrac{x^3-2x^2+3x}{2x^2+2}$

\item \hspace{-0.1in}\footnote{Once you've done the six-step procedure, use your calculator to graph this function on the viewing window $[0, 12] \times [0, 0.25]$. What do you see?} $f(x) = \dfrac{x^{2} - 2x + 1}{x^{3} + x^{2} - 2x}$ \label{sixsteplast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{usetransratfirst} - \ref{usetransratlast}, graph the rational function by applying transformations to the graph of $y = \dfrac{1}{x}$.

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $f(x) = \dfrac{1}{x - 2}$ \label{usetransratfirst}

\item $g(x) = 1 - \dfrac{3}{x}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $h(x) = \dfrac{-2x + 1}{x}$ (Hint: Divide)

\item $j(x) = \dfrac{3x - 7}{x - 2}$ (Hint: Divide) \label{usetransratlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Discuss with your classmates how you would graph $f(x) = \dfrac{ax + b}{cx + d}$. What restrictions must be placed on $a, b, c$ and $d$ so that the graph is indeed a transformation of $y = \dfrac{1}{x}$?

\item In Example \ref{intropolyexample} in Section \ref{GraphsofPolynomials} we showed that $p(x) = \frac{4x+x^3}{x}$ is not a polynomial even though its formula reduced to $4 + x^{2}$ for $x \neq 0$. However, it is a rational function similar to those studied in the section. With the help of your classmates, graph $p(x)$.

\item Let $g(x) = \displaystyle \frac{x^{4} - 8x^{3} + 24x^{2} - 72x + 135}{x^{3} - 9x^{2} + 15x - 7}.\;$ With the help of your classmates, find the $x$- and $y$- intercepts of the graph of $g$. Find the intervals on which the function is increasing, the intervals on which it is decreasing and the local extrema. Find all of the asymptotes of the graph of $g$ and any holes in the graph, if they exist. Be sure to show all of your work including any polynomial or synthetic division. Sketch the graph of $g$, using more than one picture if necessary to show all of the important features of the graph.

\setcounter{HW}{\value{enumi}}

\end{enumerate}

Example \ref{calcisneededhere} showed us that the six-step procedure cannot tell us everything of importance about the graph of a rational function. Without Calculus, we need to use our graphing calculators to reveal the hidden mysteries of rational function behavior. Working with your classmates, use a graphing calculator to examine the graphs of the rational functions given in Exercises \ref{rationalneedcalcfirst} - \ref{rationalneedcalclast}. Compare and contrast their features. Which features can the six-step process reveal and which features cannot be detected by it?

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $f(x) = \dfrac{1}{x^{2} + 1}$ \vphantom{$\dfrac{2x^{3}}{2x^{2}}$} \label{rationalneedcalcfirst}

\item $f(x) = \dfrac{x}{x^{2} + 1}$ \vphantom{$\dfrac{2x^{3}}{2x^{2}}$}

\item $f(x) = \dfrac{x^{2}}{x^{2} + 1}$ \vphantom{$\dfrac{2x^{3}}{2x^{2}}$}

\item $f(x) = \dfrac{x^{3}}{x^{2} + 1}$ \vphantom{$\dfrac{2x^{3}}{2x^{2}}$} \label{rationalneedcalclast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\newpage

\begin{enumerate}

\item \begin{multicols}{2} \raggedcolumns

$f(x) = \dfrac{4}{x + 2}$\\

Domain: $(-\infty, -2) \cup (-2, \infty)$\\

No $x$-intercepts\\

$y$-intercept: $(0, 2)$\\

Vertical asymptote: $x = -2$\\

As $x \rightarrow -2^{-}, \; f(x) \rightarrow -\infty$\\

As $x \rightarrow -2^{+}, \; f(x) \rightarrow \infty$\\

Horizontal asymptote: $y = 0$\\

As $x \rightarrow -\infty, \; f(x) \rightarrow 0^{-}$\\

As $x \rightarrow \infty, \; f(x) \rightarrow 0^{+}$\\

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\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$f(x) = \dfrac{5x}{6 - 2x}$\\

Domain: $(-\infty, 3) \cup (3, \infty)$\\

$x$-intercept: $(0, 0)$\\

$y$-intercept: $(0, 0)$\\

Vertical asymptote: $x = 3$\\

As $x \rightarrow 3^{-}, \; f(x) \rightarrow \infty$\\

As $x \rightarrow 3^{+}, \; f(x) \rightarrow -\infty$\\

Horizontal asymptote: $y = -\frac{5}{2}$\\

As $x \rightarrow -\infty, \; f(x) \rightarrow -\frac{5}{2}^{+}$\\

As $x \rightarrow \infty, \; f(x) \rightarrow -\frac{5}{2}^{-}$\\

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\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$f(x) = \dfrac{1}{x^{2}}$\\

Domain: $(-\infty, 0) \cup (0, \infty)$\\

No $x$-intercepts\\

No $y$-intercepts\\

Vertical asymptote: $x = 0$\\

As $x \rightarrow 0^{-}, \; f(x) \rightarrow \infty$\\

As $x \rightarrow 0^{+}, \; f(x) \rightarrow \infty$\\

Horizontal asymptote: $y = 0$\\

As $x \rightarrow -\infty, \; f(x) \rightarrow 0^{+}$\\

As $x \rightarrow \infty, \; f(x) \rightarrow 0^{+}$\\

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\pagebreak

\item \begin{multicols}{2} \raggedcolumns

$f(x) = \dfrac{1}{x^{2} + x - 12} = \dfrac{1}{(x - 3)(x + 4)}$\\

Domain: $(-\infty, -4) \cup (-4, 3) \cup (3, \infty)$\\

No $x$-intercepts\\

$y$-intercept: $(0, -\frac{1}{12})$\\

Vertical asymptotes: $x = -4$ and $x = 3$\\

As $x \rightarrow -4^{-}, \; f(x) \rightarrow \infty$\\

As $x \rightarrow -4^{+}, \; f(x) \rightarrow -\infty$\\

As $x \rightarrow 3^{-}, \; f(x) \rightarrow -\infty$\\

As $x \rightarrow 3^{+}, \; f(x) \rightarrow \infty$\\

Horizontal asymptote: $y = 0$\\

As $x \rightarrow -\infty, \; f(x) \rightarrow 0^{+}$\\

As $x \rightarrow \infty, \; f(x) \rightarrow 0^{+}$\\

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\item \begin{multicols}{2} \raggedcolumns

$f(x) = \dfrac{2x - 1}{-2x^{2} - 5x + 3} = -\dfrac{2x - 1}{(2x - 1)(x + 3)}$\\

Domain: $(-\infty, -3) \cup (-3, \frac{1}{2}) \cup (\frac{1}{2}, \infty)$\\

No $x$-intercepts\\

$y$-intercept: $(0, -\frac{1}{3})$\\

$f(x) = \dfrac{-1}{x + 3}, \; x \neq \frac{1}{2}$\\

Hole in the graph at $(\frac{1}{2}, -\frac{2}{7})$\\

Vertical asymptote: $x = -3$\\

As $x \rightarrow -3^{-}, \; f(x) \rightarrow \infty$\\

As $x \rightarrow -3^{+}, \; f(x) \rightarrow -\infty$\\

Horizontal asymptote: $y = 0$\\

As $x \rightarrow -\infty, \; f(x) \rightarrow 0^{+}$\\

As $x \rightarrow \infty, \; f(x) \rightarrow 0^{-}$\\

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\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$f(x) = \dfrac{x}{x^{2} + x - 12} = \dfrac{x}{(x - 3)(x + 4)}$\\

Domain: $(-\infty, -4) \cup (-4, 3) \cup (3, \infty)$\\

$x$-intercept: $(0, 0)$\\

$y$-intercept: $(0, 0)$\\

Vertical asymptotes: $x = -4$ and $x = 3$\\

As $x \rightarrow -4^{-}, \; f(x) \rightarrow -\infty$\\

As $x \rightarrow -4^{+}, \; f(x) \rightarrow \infty$\\

As $x \rightarrow 3^{-}, \; f(x) \rightarrow -\infty$\\

As $x \rightarrow 3^{+}, \; f(x) \rightarrow \infty$\\

Horizontal asymptote: $y = 0$\\

As $x \rightarrow -\infty, \; f(x) \rightarrow 0^{-}$\\

As $x \rightarrow \infty, \; f(x) \rightarrow 0^{+}$\\

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\item \begin{multicols}{2} \raggedcolumns

$f(x) = \dfrac{4x}{x^{2} + 4}$\\

Domain: $(-\infty, \infty)$\\

$x$-intercept: $(0,0)$\\

$y$-intercept: $(0,0)$\\

No vertical asymptotes \\

No holes in the graph\\

Horizontal asymptote: $y = 0$ \\

As $x \rightarrow -\infty, f(x) \rightarrow 0^{-}$\\

As $x \rightarrow \infty, f(x) \rightarrow 0^{+}$\\

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\item \begin{multicols}{2} \raggedcolumns

$f(x) = \dfrac{4x}{x^{2} -4} = \dfrac{4x}{(x + 2)(x - 2)}$\\

Domain: $(-\infty, -2) \cup (-2, 2) \cup (2, \infty)$\\

$x$-intercept: $(0,0)$\\

$y$-intercept: $(0,0)$\\

Vertical asymptotes: $x = -2, x = 2$\\

As $x \rightarrow -2^{-}, f(x) \rightarrow -\infty$\\

As $x \rightarrow -2^{+}, f(x) \rightarrow \infty$\\

As $x \rightarrow 2^{-}, f(x) \rightarrow -\infty$\\

As $x \rightarrow 2^{+}, f(x) \rightarrow \infty$\\

No holes in the graph\\

Horizontal asymptote: $y = 0$ \\

As $x \rightarrow -\infty, f(x) \rightarrow 0^{-}$\\

As $x \rightarrow \infty, f(x) \rightarrow 0^{+}$\\

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\item \begin{multicols}{2} \raggedcolumns

$f(x) = \dfrac{x^2-x-12}{x^{2} +x - 6} = \dfrac{x-4}{x - 2} \, x \neq -3$\\

Domain: $(-\infty, -3) \cup (-3, 2) \cup (2, \infty)$\\

$x$-intercept: $(4,0)$\\

$y$-intercept: $(0,2)$\\

Vertical asymptote: $x = 2$\\

As $x \rightarrow 2^{-}, f(x) \rightarrow \infty$\\

As $x \rightarrow 2^{+}, f(x) \rightarrow -\infty$\\

Hole at $\left(-3, \frac{7}{5} \right)$ \\

Horizontal asymptote: $y = 1$ \\

As $x \rightarrow -\infty, f(x) \rightarrow 1^{+}$\\

As $x \rightarrow \infty, f(x) \rightarrow 1^{-}$\\

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\pagebreak

\item \begin{multicols}{2} \raggedcolumns

$f(x) = \dfrac{3x^2-5x-2}{x^{2} -9} = \dfrac{(3x+1)(x-2)}{(x + 3)(x - 3)}$\\

Domain: $(-\infty, -3) \cup (-3, 3) \cup (3, \infty)$\\

$x$-intercepts: $\left(-\frac{1}{3}, 0 \right)$, $(2,0)$\\

$y$-intercept: $\left(0, \frac{2}{9} \right)$\\

Vertical asymptotes: $x = -3, x = 3$\\

As $x \rightarrow -3^{-}, f(x) \rightarrow \infty$\\

As $x \rightarrow -3^{+}, f(x) \rightarrow -\infty$\\

As $x \rightarrow 3^{-}, f(x) \rightarrow -\infty$\\

As $x \rightarrow 3^{+}, f(x) \rightarrow \infty$\\

No holes in the graph\\

Horizontal asymptote: $y = 3$ \\

As $x \rightarrow -\infty, f(x) \rightarrow 3^{+}$\\

As $x \rightarrow \infty, f(x) \rightarrow 3^{-}$\\

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\item \begin{multicols}{2} \raggedcolumns

$f(x) = \dfrac{x^2-x-6}{x+1} = \dfrac{(x-3)(x+2)}{x+1}$\\

Domain: $(-\infty, -1) \cup (-1, \infty)$\\

$x$-intercepts: $(-2,0)$, $(3,0)$\\

$y$-intercept: $(0,-6)$\\

Vertical asymptote: $x = -1$\\

As $x \rightarrow -1^{-}, f(x) \rightarrow \infty$\\

As $x \rightarrow -1^{+}, f(x) \rightarrow -\infty$\\

Slant asymptote: $y = x-2$ \\

As $x \rightarrow -\infty$, the graph is above $y=x-2$\\

As $x \rightarrow \infty$, the graph is below $y=x-2$\\

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\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$f(x) = \dfrac{x^2-x}{3-x} = \dfrac{x(x-1)}{3-x}$\\

Domain: $(-\infty, 3) \cup (3, \infty)$\\

$x$-intercepts: $(0,0)$, $(1,0)$\\

$y$-intercept: $(0,0)$\\

Vertical asymptote: $x = 3$\\

As $x \rightarrow 3^{-}, f(x) \rightarrow \infty$\\

As $x \rightarrow 3^{+}, f(x) \rightarrow -\infty$\\

Slant asymptote: $y = -x-2$ \\

As $x \rightarrow -\infty$, the graph is above $y=-x-2$\\

As $x \rightarrow \infty$, the graph is below $y=-x-2$\\

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\normalsize

\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$f(x) = \dfrac{x^3+2x^2+x}{x^{2} -x-2} = \dfrac{x(x+1)}{x - 2} \, x \neq -1$\\

Domain: $(-\infty, -1) \cup (-1, 2) \cup (2, \infty)$\\

$x$-intercept: $(0,0)$\\

$y$-intercept: $(0,0)$\\

Vertical asymptote: $x = 2$\\

As $x \rightarrow 2^{-}, f(x) \rightarrow -\infty$\\

As $x \rightarrow 2^{+}, f(x) \rightarrow \infty$\\

Hole at $(-1,0)$ \\

Slant asymptote: $y = x+3$ \\

As $x \rightarrow -\infty$, the graph is below $y=x+3$ \\

As $x \rightarrow \infty$, the graph is above $y=x+3$\\

\begin{mfpic}[8][6]{-10}{10}{-11}{20}

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\normalsize

\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$f(x) = \dfrac{-x^{3} + 4x}{x^{2} - 9}$\\

Domain: $(-\infty, -3) \cup (-3, 3) \cup (3, \infty)$\\

$x$-intercepts: $(-2, 0), (0, 0), (2, 0)$\\

$y$-intercept: $(0, 0)$\\

Vertical asymptotes: $x = -3, x = 3$\\

As $x \rightarrow -3^{-}, \; f(x) \rightarrow \infty$\\

As $x \rightarrow -3^{+}, \; f(x) \rightarrow -\infty$\\

As $x \rightarrow 3^{-}, \; f(x) \rightarrow \infty$\\

As $x \rightarrow 3^{+}, \; f(x) \rightarrow -\infty$\\

Slant asymptote: $y = -x$\\

As $x \rightarrow -\infty$, the graph is above $y=-x$\\

As $x \rightarrow \infty$, the graph is below $y=-x$\\

\begin{mfpic}[10]{-7}{7}{-8}{8}

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\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$f(x) = \dfrac{x^3-2x^2+3x}{2x^2+2}$\\

Domain: $(-\infty,\infty)$\\

$x$-intercept: $(0,0)$\\

$y$-intercept: $(0,0)$\\

Slant asymptote: $y = \frac{1}{2}x-1$\\

As $x \rightarrow -\infty$, the graph is below $y = \frac{1}{2}x-1$\\

As $x \rightarrow \infty$, the graph is above $y = \frac{1}{2}x-1$\\

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\end{mfpic}

\end{multicols}

\pagebreak

\item \begin{multicols}{2} \raggedcolumns

$f(x) = \dfrac{x^{2} - 2x + 1}{x^{3} + x^{2} - 2x}$\\

Domain: $(-\infty, -2) \cup (-2, 0) \cup (0, 1) \cup (1, \infty)$\\

$f(x) = \dfrac{x - 1}{x(x + 2)}, \; x \neq 1$\\

No $x$-intercepts\\

No $y$-intercepts\\

Vertical asymptotes: $x = -2$ and $x = 0$\\

As $x \rightarrow -2^{-}, \; f(x) \rightarrow -\infty$\\

As $x \rightarrow -2^{+}, \; f(x) \rightarrow \infty$\\

As $x \rightarrow 0^{-}, \; f(x) \rightarrow \infty$\\

As $x \rightarrow 0^{+}, \; f(x) \rightarrow -\infty$\\

Hole in the graph at $(1, 0)$\\

Horizontal asymptote: $y = 0$\\

As $x \rightarrow -\infty, \; f(x) \rightarrow 0^{-}$\\

As $x \rightarrow \infty, \; f(x) \rightarrow 0^{+}$\\

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\gclear \circle{(1,0),0.1}

\circle{(1,0),0.1}

\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$f(x) = \dfrac{1}{x - 2}$\\

Shift the graph of $y = \dfrac{1}{x}$ \\

to the right 2 units.\\

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\normalsize

\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$g(x) = 1 - \dfrac{3}{x}$\\

Vertically stretch the graph of $y = \dfrac{1}{x}$\\

by a factor of 3.\\

Reflect the graph of $y = \dfrac{3}{x}$\\

about the $x$-axis.\\

Shift the graph of $y = -\dfrac{3}{x}$\\

up 1 unit.\\

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\xmarks{-6 step 1 until 6}

\ymarks{-5 step 1 until 7}

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\normalsize

\end{mfpic}

\end{multicols}

\pagebreak

\item \begin{multicols}{2} \raggedcolumns

$h(x) = \dfrac{-2x + 1}{x} = -2 + \dfrac{1}{x}$\\

Shift the graph of $y = \dfrac{1}{x}$\\

down 2 units.\\

\begin{mfpic}[18]{-4}{4}{-6}{2}

\arrow \reverse \arrow \function{-4,-0.25,0.1}{(1/x)-2}

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\point[3pt]{(1,-1), (-1,-3)}

\dashed \polyline{(-4,-2), (4,-2)}

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\ymarks{-5 step 1 until 1}

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\end{mfpic}

\end{multicols}

\item \begin{multicols}{2} \raggedcolumns

$j(x) = \dfrac{3x - 7}{x - 2} = 3 - \dfrac{1}{x - 2}$\\

Shift the graph of $y = \dfrac{1}{x}$\\

to the right 2 units.\\

Reflect the graph of $y = \dfrac{1}{x - 2}$\\

about the $x$-axis.\\

Shift the graph of $y = -\dfrac{1}{x - 2}$\\

up 3 units.\\

\begin{mfpic}[15]{-4}{6}{-2}{8}

\arrow \reverse \arrow \function{-4,1.8,0.1}{3 - (1/(x - 2))}

\arrow \reverse \arrow \function{2.2,6,0.1}{3 - (1/(x - 2))}

\point[3pt]{(3,2), (1,4)}

\dashed \polyline{(-4,3), (6,3)}

\dashed \polyline{(2,-2), (2,8)}

\tlabel[cc](6,-0.5){\scriptsize $x$}

\tlabel[cc](0.5,8){\scriptsize $y$}

\axes

\xmarks{-3 step 1 until 5}

\ymarks{-1 step 1 until 7}

\tiny

\tlpointsep{4pt}

\axislabels {x}{{$-3\hspace{7pt}$} -3, {$-2\hspace{7pt}$} -2, {$-1\hspace{7pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5}

\axislabels {y}{{$-1$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5, {$6$} 6, {$7$} 7}

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\end{mfpic}

\end{multicols}

\end{enumerate}

\closegraphsfile

## 4.3: Rational Inequalities and Applications

\subsection{Exercises}

In Exercises \ref{ratleqnexercisefirst} - \ref{ratleqnexerciselast}, solve the rational equation. Be sure to check for extraneous solutions.

\begin{multicols}{2}

\begin{enumerate}

\item $\dfrac{x}{5x + 4} = 3$ \label{ratleqnexercisefirst}

\item $\dfrac{3x - 1}{x^{2} + 1} = 1$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{1}{x + 3} + \dfrac{1}{x - 3} = \dfrac{x^{2} - 3}{x^{2} - 9}$

\item $\dfrac{2x + 17}{x + 1} = x + 5$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{x^{2} - 2x + 1}{x^{3} + x^{2} - 2x} = 1$

\item $\dfrac{-x^{3} + 4x}{x^{2} - 9} = 4x$ \label{ratleqnexerciselast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{ratlineqexercisefirst} - \ref{ratlineqexerciselast}, solve the rational inequality. Express your answer using interval notation.

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{1}{x + 2} \geq 0$ \label{ratlineqexercisefirst}

\item $\dfrac{x - 3}{x + 2} \leq 0$

\item $\dfrac{x}{x^{2} - 1} > 0$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{4x}{x^2+4} \geq 0$

\item $\dfrac{x^2-x-12}{x^2+x-6} > 0$

\item $\dfrac{3x^2-5x-2}{x^2-9} < 0$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{x^3+2x^2+x}{x^2-x-2} \geq 0$

\item $\dfrac{x^{2} + 5x + 6}{x^{2} - 1} > 0$

\item $\dfrac{3x - 1}{x^{2} + 1} \leq 1$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{2x + 17}{x + 1} > x + 5$

\item $\dfrac{-x^{3} + 4x}{x^{2} - 9} \geq 4x$

\item $\dfrac{1}{x^{2} + 1} < 0$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{x^4-4x^3+x^2-2x-15}{x^3-4x^2} \geq x$

\item $\dfrac{5x^3-12x^2+9x+10}{x^2-1}\geq 3x-1$ \label{ratlineqexerciselast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Carl and Mike start a 3 mile race at the same time. If Mike ran the race at 6 miles per hour and finishes the race 10 minutes before Carl, how fast does Carl run?

\item One day, Donnie observes that the wind is blowing at 6 miles per hour. A unladen swallow nesting near Donnie's house flies three quarters of a mile down the road (in the direction of the wind), turns around, and returns exactly 4 minutes later. What is the airspeed of the unladen swallow? (Here, airspeed' is the speed that the swallow can fly in still air.)

\item In order to remove water from a flooded basement, two pumps, each rated at 40 gallons per minute, are used. After half an hour, the one pump burns out, and the second pump finishes removing the water half an hour later. How many gallons of water were removed from the basement?

\item A faucet can fill a sink in 5 minutes while a drain will empty the same sink in 8 minutes. If the faucet is turned on and the drain is left open, how long will it take to fill the sink?

\item Working together, Daniel and Donnie can clean the llama pen in 45 minutes. On his own, Daniel can clean the pen in an hour. How long does it take Donnie to clean the llama pen on his own?

\item In Exercise \ref{newportaboycost}, the function $C(x) = .03x^{3} - 4.5x^{2} + 225x + 250$, for $x \geq 0$ was used to model the cost (in dollars) to produce $x$ PortaBoy game systems. Using this cost function, find the number of PortaBoys which should be produced to minimize the average cost $\overline{C}$. Round your answer to the nearest number of systems.

\item Suppose we are in the same situation as Example \ref{boxnotopfixedvolume}. If the volume of the box is to be $500$ cubic centimeters, use your calculator to find the dimensions of the box which minimize the surface area. What is the minimum surface area? Round your answers to two decimal places.

\item The box for the new Sasquatch-themed cereal, Crypt-Os', is to have a volume of $140$ cubic inches. For aesthetic reasons, the height of the box needs to be $1.62$ times the width of the base of the box.\footnote{1.62 is a crude approximation of the so-called Golden Ratio' $\phi = \frac{1 + \sqrt{5}}{2}$.} Find the dimensions of the box which will minimize the surface area of the box. What is the minimum surface area? Round your answers to two decimal places.

\item \label{fixedareaminperimetergarden} Sally is Skippy's neighbor from Exercise \ref{fixedperimetermaxareagarden} in Section \ref{QuadraticFunctions}. Sally also wants to plant a vegetable garden along the side of her home. She doesn't have any fencing, but wants to keep the size of the garden to 100 square feet. What are the dimensions of the garden which will minimize the amount of fencing she needs to buy? What is the minimum amount of fencing she needs to buy? Round your answers to the nearest foot. (Note: Since one side of the garden will border the house, Sally doesn't need fencing along that side.)

\item Another Classic Problem: A can is made in the shape of a right circular cylinder and is to hold one pint. (For dry goods, one pint is equal to $33.6$ cubic inches.)\footnote{According to \href{http://dictionary.reference.com/brow...ctionary.com}}, there are different values given for this conversion. We will stick with $33.6 \mbox{in}^{3}$ for this problem.}

\begin{enumerate}

\item Find an expression for the volume $V$ of the can in terms of the height $h$ and the base radius $r$.

\item Find an expression for the surface area $S$ of the can in terms of the height $h$ and the base radius $r$. (Hint: The top and bottom of the can are circles of radius $r$ and the side of the can is really just a rectangle that has been bent into a cylinder.)

\item Using the fact that $V = 33.6$, write $S$ as a function of $r$ and state its applied domain.

\item Use your graphing calculator to find the dimensions of the can which has minimal surface area.

\end{enumerate}

\item A right cylindrical drum is to hold 7.35 cubic feet of liquid. Find the dimensions (radius of the base and height) of the drum which would minimize the surface area. What is the minimum surface area? Round your answers to two decimal places.

\item In Exercise \ref{Sasquatchfunc1} in Section \ref{FunctionNotation}, the population of Sasquatch in Portage County was modeled by the function $P(t) = \frac{150t}{t + 15}$, where $t = 0$ represents the year 1803. When were there fewer than 100 Sasquatch in Portage County?

\setcounter{HW}{\value{enumi}}

\end{enumerate}

In Exercises \ref{varexercisefirst} - \ref{varexerciselast}, translate the following into mathematical equations.

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item At a constant pressure, the temperature $T$ of an ideal gas is directly proportional to its volume $V$. (This is \href{http://en.wikipedia.org/wiki/Charles...line{Charles's Law}}) \index{Charles's Law} \label{varexercisefirst}

\item The frequency of a wave $f$ is inversely proportional to the wavelength of the wave $\lambda$.

\item The density $d$ of a material is directly proportional to the mass of the object $m$ and inversely proportional to its volume $V$.

\item The square of the orbital period of a planet $P$ is directly proportional to the cube of the semi-major axis of its orbit $a$. (This is \href{http://en.wikipedia.org/wiki/Kepler}...rline{Kepler's Third Law of Planetary Motion }}) \index{Kepler's Third Law of Planetary Motion}

\item The drag of an object traveling through a fluid $D$ varies jointly with the density of the fluid $\rho$ and the square of the velocity of the object $\nu$.

\item Suppose two electric point charges, one with charge $q$ and one with charge $Q$, are positioned $r$ units apart. The electrostatic force $F$ exerted on the charges varies directly with the product of the two charges and inversely with the square of the distance between the charges. (This is \href{http://en.wikipedia.org/wiki/Electro...line{Coulomb's Law}}) \index{Coulomb's Law} \label{varexerciselast}

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\item According to \href{http://en.wikipedia.org/wiki/Vibrati...underline{this webpage}}, the frequency $f$ of a vibrating string is given by $f = \dfrac{1}{2L} \sqrt{\dfrac{T}{\mu}}$ where $T$ is the tension, $\mu$ is the linear mass\footnote{Also known as the linear density. It is simply a measure of mass per unit length.} of the string and $L$ is the length of the vibrating part of the string. Express this relationship using the language of variation.

\item According to the Centers for Disease Control and Prevention \href{http://www.cdc.gov}{\underline{www.cdc.gov}}, a person's Body Mass Index $B$ is directly proportional to his weight $W$ in pounds and inversely proportional to the square of his height $h$ in inches. \index{BMI, body mass index}

\begin{enumerate}

\item Express this relationship as a mathematical equation. \label{BMIfirst}

\item If a person who was $5$ feet, $10$ inches tall weighed 235 pounds had a Body Mass Index of 33.7, what is the value of the constant of proportionality? \label{BMIsecond}

\item Rewrite the mathematical equation found in part \ref{BMIfirst} to include the value of the constant found in part \ref{BMIsecond} and then find your Body Mass Index.

\end{enumerate}

\item We know that the circumference of a circle varies directly with its radius with $2\pi$ as the constant of proportionality. (That is, we know $C = 2\pi r.$) With the help of your classmates, compile a list of other basic geometric relationships which can be seen as variations.

\end{enumerate}

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\item $x = -\frac{6}{7}$

\item $x = 1, \; x = 2$

\item $x = -1$

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\begin{enumerate}

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\item $x = -6, \; x = 2$

\item No solution

\item $x = 0, \; x = \pm 2\sqrt{2}$

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\begin{enumerate}

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\item $(-2, \infty)$

\item $(-2, 3]$

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\begin{enumerate}

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\item $(-1, 0) \cup (1, \infty)$

\item $[0, \infty)$

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\begin{enumerate}

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\item $(-\infty, -3) \cup (-3,2) \cup (4, \infty)$

\item $\left(-3, -\frac{1}{3} \right) \cup (2,3)$

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\begin{enumerate}

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\item $(-1,0] \cup (2, \infty)$

\item $(-\infty, -3) \cup (-2, -1) \cup (1, \infty)$

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\end{enumerate}

\end{multicols}

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\begin{enumerate}

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\item $(-\infty, 1] \cup [2, \infty)$

\item $(-\infty, -6) \cup (-1, 2)$

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\end{enumerate}

\end{multicols}

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\begin{enumerate}

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\item $(-\infty, -3) \cup \left[-2\sqrt{2}, 0\right] \cup \left[2\sqrt{2}, 3\right)$

\item No solution

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\begin{enumerate}

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\item $[-3,0) \cup (0,4) \cup [5, \infty)$

\item $\left(-1,-\frac{1}{2}\right] \cup (1, \infty)$

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\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

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\item 4.5 miles per hour

\item 24 miles per hour

\item 3600 gallons

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\begin{multicols}{3}

\begin{enumerate}

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\item $\frac{40}{3} \approx 13.33$ minutes

\item 3 hours

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\begin{enumerate}

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\item The absolute minimum of $y=\overline{C}(x)$ occurs at $\approx (75.73, 59.57)$. Since $x$ represents the number of game systems, we check $\overline{C}(75) \approx 59.58$ and $\overline{C}(76) \approx 59.57$. Hence, to minimize the average cost, $76$ systems should be produced at an average cost of $\$59.57$per system. \item The width (and depth) should be$10.00$centimeters, the height should be$5.00$centimeters. The minimum surface area is$300.00$square centimeters. \item The width of the base of the box should be$\approx 4.12$inches, the height of the box should be$\approx 6.67$inches, and the depth of the base of the box should be$\approx 5.09$inches; minimum surface area$\approx 164.91$square inches. \item The dimensions are$\approx 7$feet by$\approx 14$feet; minimum amount of fencing required$\approx 28$feet. \item \begin{multicols}{2} \begin{enumerate} \item$V = \pi r^{2}h$\item$S = 2 \pi r^{2} + 2\pi r h$\setcounter{HWindent}{\value{enumii}} \end{enumerate} \end{multicols} \begin{multicols}{2} \begin{enumerate} \setcounter{enumii}{\value{HWindent}} \item$S(r) = 2\pi r^{2} + \frac{67.2}{r}, \;$Domain$r > 0$\item$r \approx 1.749\,$in. and$h \approx 3.498\,$in. \end{enumerate} \end{multicols} \item The radius of the drum should be$\approx 1.05$feet and the height of the drum should be$\approx 2.12$feet. The minimum surface area of the drum is$\approx 20.93$cubic feet. \item$P(t) < 100$on$(-15, 30)$, and the portion of this which lies in the applied domain is$[0,30)$. Since$t=0$corresponds to the year 1803, from 1803 through the end of 1832, there were fewer than 100 Sasquatch in Portage County. \setcounter{HW}{\value{enumi}} \end{enumerate} \begin{multicols}{3} \begin{enumerate} \setcounter{enumi}{\value{HW}} \item$T = k V$\item \hspace{-.1in} \footnote{The character$\lambda$is the lower case Greek letter lambda.'}$f = \dfrac{k}{\lambda}$\item$d = \dfrac{k m}{V}$\setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \begin{multicols}{3} \begin{enumerate} \setcounter{enumi}{\value{HW}} \item$P^2 = k a^3$\item \hspace{-.1in} \footnote{The characters$\rho$and$\nu$are the lower case Greek letters rho' and nu,' respectively.}$D = k \rho \nu^2$\item \hspace{-.1in} \footnote{Note the similarity to this formula and Newton's Law of Universal Gravitation as discussed in Example \ref{gravitylaw}.}$F = \dfrac{kqQ}{r^2}$\setcounter{HW}{\value{enumi}} \end{enumerate} \end{multicols} \begin{enumerate} \setcounter{enumi}{\value{HW}} \item Rewriting$f = \dfrac{1}{2L} \sqrt{\dfrac{T}{\mu}}$as$f = \dfrac{\frac{1}{2} \sqrt{T}}{L \sqrt{\mu}}$we see that the frequency$f$varies directly with the square root of the tension and varies inversely with the length and the square root of the linear mass. \item \begin{multicols}{3} \begin{enumerate} \item$B = \dfrac{kW}{h^{2}}$\item \hspace{-.1in} \footnote{The CDC uses 703.}$k = 702.68$\item$B = \dfrac{702.68W}{h^{2}}\$

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