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Mathematics LibreTexts

5E: Further Topics in Functions (Exercises)

  • Page ID
    4758
  • [ "article:topic", "authorname:stitzzeager" ]

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    5.1: Function Composition

    \subsection{Exercises}

    In Exercises \ref{funccompeval1first} - \ref{funccompeval1last}, use the given pair of functions to find the following values if they exist.

    \begin{multicols}{3}

    \begin{itemize}

    \item $(g\circ f)(0)$

    \item $(f\circ g)(-1)$

    \item $(f \circ f)(2)$

    \end{itemize}

    \end{multicols}

    \begin{multicols}{3}

    \begin{itemize}

    \item $(g\circ f)(-3)$

    \item $(f\circ g)\left(\frac{1}{2}\right)$

    \item $(f \circ f)(-2)$

    \end{itemize}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

    \item $f(x) = x^2$, $g(x) = 2x+1$ \label{funccompeval1first}

    \item $f(x) = 4-x$, $g(x) = 1-x^2$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $f(x) = 4-3x$, $g(x) = |x|$

    \item $f(x) = |x-1|$, $g(x) = x^2-5$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $f(x) = 4x+5$, $g(x) = \sqrt{x}$

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

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $f(x) = 6-x-x^2$, $g(x) = x\sqrt{x+10}$

    \item $f(x) = \sqrt[3]{x+1}$, $g(x) = 4x^2-x$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

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

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

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

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

    \item $f(x) =\sqrt{2x+5}$, $g(x) = \dfrac{10x}{x^2+1}$ \label{funccompeval1last}

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

    \end{enumerate}

    \end{multicols}

    In Exercises \ref{funccompexp1first} - \ref{funccompexp1last}, use the given pair of functions to find and simplify expressions for the following functions and state the domain of each using interval notation.

    \begin{multicols}{3}

    \begin{itemize}

    \item $(g \circ f)(x)$

    \item $(f \circ g)(x)$

    \item $(f \circ f)(x)$

    \end{itemize}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $f(x) = 2x+3$, $g(x) = x^2-9$ \label{funccompexp1first}

    \item $f(x) = x^2 -x+1$, $g(x) = 3x-5$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $f(x) = x^2-4$, $g(x) = |x|$

    \item $f(x) = 3x-5$, $g(x) = \sqrt{x}$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $f(x) = |x+1|$, $g(x) = \sqrt{x}$

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

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $f(x) = |x|$, $g(x) = \sqrt{4-x}$

    \item $f(x) = x^2-x-1$, $g(x) = \sqrt{x-5}$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

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

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

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

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

    \item $f(x) = \dfrac{2x}{x^2-4}$, $g(x) =\sqrt{1-x}$

    \label{funccompexp1last}

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

    \end{enumerate}

    \end{multicols}

    \pagebreak

    In Exercises \ref{threefunccompfirst} - \ref{threefunccomplast}, use $f(x) = -2x$, $g(x) = \sqrt{x}$ and $h(x) = |x|$ to find and simplify expressions for the following functions and state the domain of each using interval notation.

    \begin{multicols}{3}

    \begin{enumerate}

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

    \item $(h\circ g \circ f)(x)$ \label{threefunccompfirst}

    \item $(h\circ f \circ g)(x)$

    \item $(g\circ f \circ h)(x)$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{3}

    \begin{enumerate}

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

    \item $(g\circ h \circ f)(x)$

    \item $(f\circ h \circ g)(x)$

    \item $(f\circ g \circ h)(x)$ \label{threefunccomplast}

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

    \end{enumerate}

    \end{multicols}

    In Exercises \ref{breakdowncompexfirst} - \ref{breakdownxomexlast}, write the given function as a composition of two or more non-identity functions. (There are several correct answers, so check your answer using function composition.)

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $p(x) = (2x+3)^3$ \label{breakdowncompexfirst}

    \item $P(x) = \left(x^2-x+1\right)^5$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $h(x) = \sqrt{2x-1}$

    \item $H(x) = |7-3x|$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $r(x) = \dfrac{2}{5x+1}$

    \item $R(x) = \dfrac{7}{x^2-1}$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $q(x) = \dfrac{|x|+1}{|x|-1}$

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

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

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

    \item $w(x) = \dfrac{x^2}{x^4+1}$ \label{breakdownxomexlast}

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

    \end{enumerate}

    \end{multicols}

    \begin{enumerate}

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

    \item Write the function $F(x) = \sqrt{\dfrac{x^{3} + 6}{x^{3} - 9}}$ as a composition of three or more non-identity functions.

    \item Let $g(x) = -x, \, h(x) = x + 2, \, j(x) = 3x$ and $k(x) = x - 4$. In what order must these functions be composed with $f(x) = \sqrt{x}$ to create $F(x) = 3\sqrt{-x + 2} - 4$?

    \item What linear functions could be used to transform $f(x) = x^{3}$ into $F(x) = -\frac{1}{2}(2x - 7)^{3} + 1$? What is the proper order of composition?

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

    \end{enumerate}

    In Exercises \ref{pointcompexfirst} - \ref{pointcompexlast}, let $f$ be the function defined by \[f = \{(-3, 4), (-2, 2), (-1, 0), (0, 1), (1, 3), (2, 4), (3, -1)\}\] and let $g$ be the function defined \[g = \{(-3, -2), (-2, 0), (-1, -4), (0, 0), (1, -3), (2, 1), (3, 2)\}\]. Find the value if it exists.

    \begin{multicols}{3}

    \begin{enumerate}

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

    \item $(f \circ g)(3)$ \label{pointcompexfirst}

    \item $f(g(-1))$

    \item $(f \circ f)(0)$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{3}

    \begin{enumerate}

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

    \item $(f \circ g)(-3)$

    \item $(g \circ f)(3)$

    \item $g(f(-3))$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{3}

    \begin{enumerate}

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

    \item $(g \circ g)(-2)$

    \item $(g \circ f)(-2)$

    \item $g(f(g(0)))$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{3}

    \begin{enumerate}

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

    \item $f(f(f(-1)))$

    \item $f(f(f(f(f(1)))))$

    \item $\underbrace{(g \circ g \circ \cdots \circ g)}_{\mbox{$n$ times}}(0)$ \label{pointcompexlast}

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

    \end{enumerate}

    \end{multicols}

    %\pagebreak

    In Exercises \ref{twofuncgraphcompfirst} - \ref{twofuncgraphcomplast}, use the graphs of $y=f(x)$ and $y=g(x)$ below to find the function value.

    \begin{center}

    \begin{tabular}{cc}

    \begin{mfpic}[20]{-1}{5}{-1}{5}

    \axes

    \tlabel[cc](5,-0.5){\scriptsize $x$}

    \tlabel[cc](0.5,5){\scriptsize $y$}

    \xmarks{1,2,3,4}

    \ymarks{1,2,3,4}

    \tlpointsep{5pt}

    \scriptsize

    \axislabels {x}{{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

    \axislabels {y}{{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

    \polyline{(0,4), (1,2), (2,3), (3,3), (4,0)}

    \point[3pt]{(0,4), (1,2), (2,3), (3,3), (4,0)}

    \normalsize

    \tcaption{$y = f(x)$}

    \end{mfpic}

    &

    \hspace{1in}

    \begin{mfpic}[20]{-1}{5}{-1}{5}

    \axes

    \tlabel[cc](5,-0.5){\scriptsize $x$}

    \tlabel[cc](0.5,5){\scriptsize $y$}

    \xmarks{1,2,3,4}

    \ymarks{1,2,3,4}

    \tlpointsep{5pt}

    \scriptsize

    \axislabels {x}{{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

    \axislabels {y}{{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

    \polyline{(0,0), (1,3), (2,3), (3,0), (4,4)}

    \point[3pt]{(0,0), (1,3), (2,3), (3,0), (4,4)}

    \normalsize

    \tcaption{$y = g(x)$}

    \end{mfpic}

    \end{tabular}

    \end{center}

    \begin{multicols}{3}

    \begin{enumerate}

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

    \item $(g\circ f)(1)$ \label{twofuncgraphcompfirst}

    \item $(f \circ g)(3)$

    \item $(g\circ f)(2)$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{3}

    \begin{enumerate}

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

    \item $(f\circ g)(0)$

    \item $(f\circ f)(1)$

    \item $(g \circ g)(1)$ \label{twofuncgraphcomplast}

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

    \end{enumerate}

    \end{multicols}

    \begin{enumerate}

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

    \item The volume $V$ of a cube is a function of its side length $x$. Let's assume that $x = t + 1$ is also a function of time $t$, where $x$ is measured in inches and $t$ is measured in minutes. Find a formula for $V$ as a function of $t$.

    \item Suppose a local vendor charges $\$2$ per hot dog and that the number of hot dogs sold per hour $x$ is given by $x(t) = -4t^2+20t+92$, where $t$ is the number of hours since $10$ AM, $0 \leq t \leq 4$.

    \begin{enumerate}

    \item Find an expression for the revenue per hour $R$ as a function of $x$.

    \item Find and simplify $\left(R \circ x\right)(t)$. What does this represent?

    \item What is the revenue per hour at noon?

    \end{enumerate}

    \item Discuss with your classmates how `real-world' processes such as filling out federal income tax forms or computing your final course grade could be viewed as a use of function composition. Find a process for which composition with itself (iteration) makes sense.

    \end{enumerate}

    \newpage

    \subsection{Answers}

    \begin{enumerate}

    \item For $f(x) = x^2$ and $g(x) = 2x+1$,

    \begin{multicols}{3}

    \begin{itemize}

    \item $(g\circ f)(0) = 1$

    \item $(f\circ g)(-1) = 1$

    \item $(f \circ f)(2) = 16$

    \end{itemize}

    \end{multicols}

    \begin{multicols}{3}

    \begin{itemize}

    \item $(g\circ f)(-3) = 19$

    \item $(f\circ g)\left(\frac{1}{2}\right) = 4$

    \item $(f \circ f)(-2) = 16$

    \end{itemize}

    \end{multicols}

    \item For $f(x) = 4-x$ and $g(x) = 1-x^2$,

    \begin{multicols}{3}

    \begin{itemize}

    \item $(g\circ f)(0) = -15$

    \item $(f\circ g)(-1) = 4$

    \item $(f \circ f)(2) = 2$

    \end{itemize}

    \end{multicols}

    \begin{multicols}{3}

    \begin{itemize}

    \item $(g\circ f)(-3) = -48$

    \item $(f\circ g)\left(\frac{1}{2}\right) = \frac{13}{4}$

    \item $(f \circ f)(-2) = -2$

    \end{itemize}

    \end{multicols}

    \item For $f(x) = 4-3x$ and $g(x) = |x|$,

    \begin{multicols}{3}

    \begin{itemize}

    \item $(g\circ f)(0) = 4$

    \item $(f\circ g)(-1) = 1$

    \item $(f \circ f)(2) = 10$

    \end{itemize}

    \end{multicols}

    \begin{multicols}{3}

    \begin{itemize}

    \item $(g\circ f)(-3) = 13$

    \item $(f\circ g)\left(\frac{1}{2}\right) = \frac{5}{2}$

    \item $(f \circ f)(-2) = -26$

    \end{itemize}

    \end{multicols}

    \item For $f(x) = |x-1|$ and $g(x) = x^2-5$,

    \begin{multicols}{3}

    \begin{itemize}

    \item $(g\circ f)(0) = -4$

    \item $(f\circ g)(-1) = 5$

    \item $(f \circ f)(2) = 0$

    \end{itemize}

    \end{multicols}

    \begin{multicols}{3}

    \begin{itemize}

    \item $(g\circ f)(-3) = 11$

    \item $(f\circ g)\left(\frac{1}{2}\right) = \frac{23}{4}$

    \item $(f \circ f)(-2) = 2$

    \end{itemize}

    \end{multicols}

    \item For $f(x) = 4x+5$ and $g(x) = \sqrt{x}$,

    \begin{multicols}{3}

    \begin{itemize}

    \item $(g\circ f)(0) = \sqrt{5}$

    \item $(f\circ g)(-1)$ is not real

    \item $(f \circ f)(2) = 57$

    \end{itemize}

    \end{multicols}

    \begin{multicols}{3}

    \begin{itemize}

    \item $(g\circ f)(-3)$ is not real

    \item $(f\circ g)\left(\frac{1}{2}\right) = 5+2\sqrt{2}$

    \item $(f \circ f)(-2) = -7$

    \end{itemize}

    \end{multicols}

    \item For $f(x) = \sqrt{3-x}$ and $g(x) = x^2+1$,

    \begin{multicols}{3}

    \begin{itemize}

    \item $(g\circ f)(0) = 4$

    \item $(f\circ g)(-1) = 1$

    \item $(f \circ f)(2) = \sqrt{2}$

    \end{itemize}

    \end{multicols}

    \begin{multicols}{3}

    \begin{itemize}

    \item $(g\circ f)(-3) = 7$

    \item $(f\circ g)\left(\frac{1}{2}\right) = \frac{\sqrt{7}}{2}$

    \item $(f \circ f)(-2) = \sqrt{3 - \sqrt{5}}$

    \end{itemize}

    \end{multicols}

    \pagebreak

    \item For $f(x) = 6-x-x^2$ and $g(x) = x\sqrt{x+10}$,

    \begin{multicols}{3}

    \begin{itemize}

    \item $(g\circ f)(0) = 24$

    \item $(f\circ g)(-1) = 0$

    \item $(f \circ f)(2) = 6$

    \end{itemize}

    \end{multicols}

    \begin{multicols}{3}

    \begin{itemize}

    \item $(g\circ f)(-3) = 0$

    \item $(f\circ g)\left(\frac{1}{2}\right) = \frac{27-2\sqrt{42}}{8}$

    \item $(f \circ f)(-2) = -14$

    \end{itemize}

    \end{multicols}

    \item For $f(x) = \sqrt[3]{x+1}$ and $g(x) = 4x^2-x$,

    \begin{multicols}{3}

    \begin{itemize}

    \item $(g\circ f)(0) = 3$

    \item $(f\circ g)(-1) = \sqrt[3]{6}$

    \item $(f \circ f)(2) = \sqrt[3]{\sqrt[3]{3}+1}$

    \end{itemize}

    \end{multicols}

    \begin{multicols}{3}

    \begin{itemize}

    \item $(g\circ f)(-3) = 4\sqrt[3]{4}+\sqrt[3]{2}$

    \item $(f\circ g)\left(\frac{1}{2}\right) = \frac{\sqrt[3]{12}}{2}$

    \item $(f \circ f)(-2) = 0$

    \end{itemize}

    \end{multicols}

    \item For $f(x) = \frac{3}{1-x}$ and $g(x) = \frac{4x}{x^2+1}$,

    \begin{multicols}{3}

    \begin{itemize}

    \item $(g\circ f)(0) = \frac{6}{5}$

    \item $(f\circ g)(-1) = 1$

    \item $(f \circ f)(2) = \frac{3}{4}$

    \end{itemize}

    \end{multicols}

    \begin{multicols}{3}

    \begin{itemize}

    \item $(g\circ f)(-3) = \frac{48}{25}$

    \item $(f\circ g)\left(\frac{1}{2}\right) = -5$

    \item $(f \circ f)(-2)$ is undefined

    \end{itemize}

    \end{multicols}

    \item For $f(x) = \frac{x}{x+5}$ and $g(x) = \frac{2}{7-x^2}$,

    \begin{multicols}{3}

    \begin{itemize}

    \item $(g\circ f)(0) = \frac{2}{7}$

    \item $(f\circ g)(-1) = \frac{1}{16}$

    \item $(f \circ f)(2) = \frac{2}{37}$

    \end{itemize}

    \end{multicols}

    \begin{multicols}{3}

    \begin{itemize}

    \item $(g\circ f)(-3) = \frac{8}{19}$

    \item $(f\circ g)\left(\frac{1}{2}\right) = \frac{8}{143}$

    \item $(f \circ f)(-2) = -\frac{2}{13}$

    \end{itemize}

    \end{multicols}

    \item For $f(x) = \frac{2x}{5-x^2}$ and $g(x) = \sqrt{4x+1}$,

    \begin{multicols}{3}

    \begin{itemize}

    \item $(g\circ f)(0) = 1$

    \item $(f\circ g)(-1)$ is not real

    \item $(f \circ f)(2) = -\frac{8}{11}$

    \end{itemize}

    \end{multicols}

    \begin{multicols}{3}

    \begin{itemize}

    \item $(g\circ f)(-3) = \sqrt{7}$

    \item $(f\circ g)\left(\frac{1}{2}\right) = \sqrt{3}$

    \item $(f \circ f)(-2) = \frac{8}{11}$

    \end{itemize}

    \end{multicols}

    \item For $f(x) =\sqrt{2x+5}$ and $g(x) = \frac{10x}{x^2+1}$ ,

    \begin{multicols}{3}

    \begin{itemize}

    \item $(g\circ f)(0) = \frac{5\sqrt{5}}{3}$

    \item $(f\circ g)(-1)$ is not real

    \item $(f \circ f)(2) = \sqrt{11}$

    \end{itemize}

    \end{multicols}

    \begin{multicols}{3}

    \begin{itemize}

    \item $(g\circ f)(-3)$ is not real

    \item $(f\circ g)\left(\frac{1}{2}\right) = \sqrt{13}$

    \item $(f \circ f)(-2) = \sqrt{7}$

    \end{itemize}

    \end{multicols}

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

    \end{enumerate}

    \begin{enumerate}

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

    \item For $f(x) = 2x+3$ and $g(x) = x^2-9$

    \begin{itemize}

    \item $(g \circ f)(x) = 4x^2+12x$, domain: $(-\infty, \infty)$

    \item $(f \circ g)(x) = 2x^2-15$, domain: $(-\infty, \infty)$

    \item $(f \circ f)(x) = 4x+9$, domain: $(-\infty, \infty)$

    \end{itemize}

    \pagebreak

    \item For $f(x) = x^2 -x+1$ and $g(x) = 3x-5$

    \begin{itemize}

    \item $(g \circ f)(x) = 3x^2-3x-2$, domain: $(-\infty, \infty)$

    \item $(f \circ g)(x) =9x^2-33x+31$, domain: $(-\infty, \infty)$

    \item $(f \circ f)(x) = x^4-2x^3+2x^2-x+1$, domain: $(-\infty, \infty)$

    \end{itemize}

    \item For $f(x) = x^2-4$ and $g(x) = |x|$

    \begin{itemize}

    \item $(g \circ f)(x) = |x^2-4|$, domain: $(-\infty, \infty)$

    \item $(f \circ g)(x) =|x|^2-4 = x^2-4$, domain: $(-\infty, \infty)$

    \item $(f \circ f)(x) =x^4-8x^2+12$, domain: $(-\infty, \infty)$

    \end{itemize}

    \item For $f(x) = 3x-5$ and $g(x) = \sqrt{x}$

    \begin{itemize}

    \item $(g \circ f)(x) = \sqrt{3x-5}$, domain: $\left[ \frac{5}{3}, \infty \right)$

    \item $(f \circ g)(x) = 3\sqrt{x}-5$, domain: $[0,\infty)$

    \item $(f \circ f)(x) = 9x-20$, domain: $(-\infty, \infty)$

    \end{itemize}

    \item For $f(x) = |x+1|$ and $g(x) = \sqrt{x}$

    \begin{itemize}

    \item $(g \circ f)(x) = \sqrt{|x+1|}$, domain: $(-\infty, \infty)$

    \item $(f \circ g)(x) = |\sqrt{x}+1| = \sqrt{x}+1$, domain: $[0,\infty)$

    \item $(f \circ f)(x) = ||x+1|+1| = |x+1|+1$, domain: $(-\infty, \infty)$

    \end{itemize}

    \item For $f(x) = 3-x^2$ and $g(x) = \sqrt{x+1}$

    \begin{itemize}

    \item $(g \circ f)(x) = \sqrt{4-x^2}$, domain: $[-2,2]$

    \item $(f \circ g)(x) =2-x$, domain: $[-1, \infty)$

    \item $(f \circ f)(x) = -x^4+6x^2-6$, domain: $(-\infty, \infty)$

    \end{itemize}

    \item For $f(x) = |x|$ and $g(x) = \sqrt{4-x}$

    \begin{itemize}

    \item $(g \circ f)(x) = \sqrt{4-|x|}$, domain: $[-4,4]$

    \item $(f \circ g)(x) =|\sqrt{4-x}| = \sqrt{4-x}$, domain: $(-\infty, 4]$

    \item $(f \circ f)(x) = | |x| | = |x|$, domain: $(-\infty, \infty)$

    \end{itemize}

    \pagebreak

    \item For $f(x) = x^2-x-1$ and $g(x) = \sqrt{x-5}$

    \begin{itemize}

    \item $(g \circ f)(x) = \sqrt{x^2-x-6}$, domain: $(-\infty, -2] \cup [3,\infty)$

    \item $(f \circ g)(x) =x-6-\sqrt{x-5}$, domain: $[5,\infty)$

    \item $(f \circ f)(x) =x^4-2x^3-2x^2+3x+1$, domain: $(-\infty, \infty)$

    \end{itemize}

    \item For $f(x) = 3x-1$ and $g(x) = \frac{1}{x+3}$

    \begin{itemize}

    \item $(g \circ f)(x) = \frac{1}{3x+2}$, domain: $\left(-\infty, -\frac{2}{3}\right) \cup \left(-\frac{2}{3}, \infty\right)$

    \item $(f \circ g)(x) = -\frac{x}{x+3}$, domain: $\left(-\infty, -3\right) \cup \left(-3, \infty\right)$

    \item $(f \circ f)(x) = 9x-4$, domain: $(-\infty, \infty)$

    \end{itemize}

    \item For $f(x) = \frac{3x}{x-1}$ and $g(x) =\frac{x}{x-3}$

    \begin{itemize}

    \item $(g \circ f)(x) =x$, domain: $\left(-\infty, 1\right) \cup (1, \infty)$

    \item $(f \circ g)(x) =x$, domain: $\left(-\infty, 3\right) \cup (3,\infty)$

    \item $(f \circ f)(x) = \frac{9x}{2x+1}$, domain: $\left(-\infty, -\frac{1}{2}\right) \cup \left(-\frac{1}{2}, 1 \right) \cup \left(1,\infty \right)$

    \end{itemize}

    \item For $f(x) = \frac{x}{2x+1}$ and $g(x) = \frac{2x+1}{x}$

    \begin{itemize}

    \item $(g \circ f)(x) = \frac{4x+1}{x}$, domain: $\left(-\infty, -\frac{1}{2}\right) \cup \left(-\frac{1}{2}, 0), \cup (0, \infty\right)$

    \item $(f \circ g)(x) = \frac{2x+1}{5x+2}$, domain: $\left(-\infty, -\frac{2}{5}\right) \cup \left(-\frac{2}{5}, 0\right) \cup (0,\infty)$

    \item $(f \circ f)(x) = \frac{x}{4x+1}$, domain: $\left(-\infty, -\frac{1}{2}\right) \cup \left(-\frac{1}{2}, -\frac{1}{4} \right) \cup \left(-\frac{1}{4},\infty\right)$

    \end{itemize}

    \item For $f(x) = \frac{2x}{x^2-4}$ and $g(x) =\sqrt{1-x}$

    \begin{itemize}

    \item $(g \circ f)(x) =\sqrt{\frac{x^2-2x-4}{x^2-4}}$, domain: $\left(-\infty, -2\right) \cup \left[1-\sqrt{5}, 2\right) \cup \left[1+\sqrt{5}, \infty\right)$

    \item $(f \circ g)(x) = -\frac{2\sqrt{1-x}}{x+3}$, domain: $\left(-\infty, -3\right) \cup \left(-3, 1\right]$

    \item $(f \circ f)(x) = \frac{4x-x^3}{x^4-9x^2+16}$, domain: $\left(-\infty, -\frac{1+\sqrt{17}}{2}\right) \cup \left(-\frac{1+\sqrt{17}}{2}, -2\right) \cup \left(-2, \frac{1-\sqrt{17}}{2}\right) \cup \left(\frac{1-\sqrt{17}}{2}, \frac{-1+\sqrt{17}}{2}\right) \cup \left(\frac{-1+\sqrt{17}}{2}, 2\right) \cup \left(2, \frac{1+\sqrt{17}}{2} \right) \cup \left(\frac{1+\sqrt{17}}{2}, \infty\right)$

    \end{itemize}

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

    \end{enumerate}

    \begin{enumerate}

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

    \item $(h\circ g \circ f)(x)= |\sqrt{-2x}|= \sqrt{-2x}$, domain: $(-\infty, 0]$

    \item $(h\circ f \circ g)(x) = |-2\sqrt{x}|= 2\sqrt{x}$, domain: $[0,\infty)$

    \item $(g\circ f \circ h)(x) = \sqrt{-2|x|}$, domain: $\{0\}$

    \item $(g\circ h \circ f)(x) = \sqrt{|-2x|} = \sqrt{2|x|}$, domain: $(-\infty, \infty)$

    \item $(f\circ h \circ g)(x) = -2|\sqrt{x}| = -2\sqrt{x}$, domain: $[0,\infty)$

    \item $(f\circ g \circ h)(x) = -2\sqrt{|x|}$, , domain: $(-\infty,\infty)$

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

    \end{enumerate}

    \begin{enumerate}

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

    \item Let $f(x) = 2x+3$ and $g(x) = x^3$, then $p(x) = (g\circ f)(x)$.

    \item Let $f(x) = x^2-x+1$ and $g(x) = x^5$, $P(x) =(g\circ f)(x)$.

    \item Let $f(x) = 2x-1$ and $g(x) = \sqrt{x}$, then $h(x) = (g\circ f)(x)$.

    \item Let $f(x) = 7-3x$ and $g(x) = |x|$, then $H(x) = (g\circ f)(x)$.

    \item Let $f(x) = 5x+1$ and $g(x) = \frac{2}{x}$, then $r(x) =(g\circ f)(x)$.

    \item Let $f(x) = x^2-1$ and $g(x) = \frac{7}{x}$, then $R(x) =(g\circ f)(x)$.

    \item Let $f(x) = |x|$ and $g(x) = \frac{x+1}{x-1}$, then $q(x) =(g\circ f)(x)$.

    \item Let $f(x) = x^3$ and $g(x)= \frac{2x+1}{x-1}$, then $Q(x) =(g\circ f)(x)$.

    \item Let $f(x) =2x$ and $g(x) = \frac{x+1}{3-2x}$, then $v(x) =(g\circ f)(x)$.

    \item Let $f(x) = x^2$ and $g(x) = \frac{x}{x^2+1}$, then $w(x) =(g\circ f)(x)$.

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

    \end{enumerate}

    \begin{enumerate}

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

    \item $F(x) = \sqrt{\frac{x^{3} + 6}{x^{3} - 9}} = (h(g(f(x)))$ where $f(x) = x^{3}, \, g(x) = \frac{x + 6}{x - 9}$ and $h(x) = \sqrt{x}$.

    \item $F(x) = 3\sqrt{-x + 2} - 4 = k(j(f(h(g(x)))))$

    \item One possible solution is $F(x) = -\frac{1}{2}(2x - 7)^{3} + 1 = k(j(f(h(g(x)))))$ where $g(x) = 2x, \, h(x) = x - 7, \, j(x) = -\frac{1}{2}x$ and $k(x) = x + 1$. You could also have $F(x) = H(f(G(x)))$ where $G(x) = 2x - 7$ and $H(x) = -\frac{1}{2}x + 1$.

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

    \end{enumerate}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $(f \circ g)(3)= f(g(3)) = f(2) = 4$

    \item $f(g(-1)) = f(-4)$ which is undefined

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $(f \circ f)(0) = f(f(0)) = f(1) = 3$

    \item $(f \circ g)(-3) = f(g(-3)) = f(-2) = 2$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $(g \circ f)(3) = g(f(3)) = g(-1) = -4$

    \item $g(f(-3)) = g(4)$ which is undefined

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $(g \circ g)(-2) = g(g(-2)) = g(0) = 0$

    \item $(g \circ f)(-2) = g(f(-2)) = g(2) = 1$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $g(f(g(0))) = g(f(0)) = g(1) = -3$

    \item $f(f(f(-1))) = f(f(0)) = f(1) = 3$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $f(f(f(f(f(1))))) = f(f(f(f(3)))) =\\ f(f(f(-1))) = f(f(0)) = f(1) = 3$

    \item $\underbrace{(g \circ g \circ \cdots \circ g)}_{\mbox{$n$ times}}(0) = 0$

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

    \end{enumerate}

    \end{multicols}

    \pagebreak

    \begin{multicols}{3}

    \begin{enumerate}

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

    \item $(g\circ f)(1) = 3$

    \item $(f \circ g)(3) = 4$

    \item $(g\circ f)(2) = 0$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{3}

    \begin{enumerate}

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

    \item $(f\circ g)(0) = 4$

    \item $(f\circ f)(1) = 3$

    \item $(g \circ g)(1) = 0$

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

    \end{enumerate}

    \end{multicols}

    \begin{enumerate}

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

    \item $V(x) = x^{3}$ so $V(x(t)) = (t + 1)^{3}$

    \item \begin{enumerate}

    \item $R(x) = 2x$

    \item $\left(R \circ x \right)(t) = -8t^2+40t+184$, $0 \leq t \leq 4$. This gives the revenue per hour as a function of time.

    \item Noon corresponds to $t=2$, so $\left(R \circ x \right)(2) = 232$. The hourly revenue at noon is $\$232$ per hour.

    \end{enumerate}

    \end{enumerate}

    \closegraphsfile

    5.2: Inverse Functions

    \subsection{Exercises}

    In Exercises \ref{inversehwfirst} - \ref{inversehwlast}, show that the given function is one-to-one and find its inverse. Check your answers algebraically and graphically. Verify that the range of $f$ is the domain of $f^{-1}$ and vice-versa.

    \begin{multicols}{2}

    \begin{enumerate}

    \item $f(x) = 6x - 2$ \label{inversehwfirst}

    \item $f(x) = 42-x$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

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

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

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $f(x) = \sqrt{3x-1}+5$

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

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $f(x) = 3\sqrt{x-1}-4$

    \item $f(x) = 1 - 2\sqrt{2x+5}$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $f(x) = \sqrt[5]{3x-1}$

    \item $f(x) = 3-\sqrt[3]{x-2}$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $f(x) = x^2 - 10x$, $x \geq 5$

    \item $f(x) = 3(x + 4)^{2} - 5, \; x \leq -4$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $f(x) = x^2-6x+5, \; x \leq 3$

    \item $f(x) = 4x^2 + 4x + 1$, $x < -1$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $f(x) = \dfrac{3}{4-x}$

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

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

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

    \item $f(x) = \dfrac{4x + 2}{3x - 6}$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

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

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

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

    \end{enumerate}

    \end{multicols}

    With help from your classmates, find the inverses of the functions in Exercises \ref{genericinversefirst} - \ref{genericinverselast}.

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $f(x) = ax + b, \; a \neq 0$ \label{genericinversefirst}

    \item $f(x) = a\sqrt{x - h} + k, \; a \neq 0, x \geq h$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $f(x) = ax^{2} + bx + c$ where $a \neq 0, \, x \geq -\dfrac{b}{2a}$.

    \item $f(x) = \dfrac{ax + b}{cx + d},\;$ (See Exercise \ref{whatconditions} below.) \label{genericinverselast}

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

    \end{enumerate}

    \end{multicols}

    \begin{enumerate}

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

    \item In Example \ref{costrevenueprofitex1}, the price of a dOpi media player, in dollars per dOpi, is given as a function of the weekly sales $x$ according to the formula $p(x) = 450-15x$ for $0 \leq x \leq 30$.

    \begin{enumerate}

    \item Find $p^{-1}(x)$ and state its domain.

    \item Find and interpret $p^{-1}(105)$.

    \item In Example \ref{costrevenueprofitex1}, we determined that the profit (in dollars) made from producing and selling $x$ dOpis per week is $P(x)= -15x^2+350x-2000$, for $0 \leq x \leq 30$. Find $\left(P \circ p^{-1}\right)(x)$ and determine what price per dOpi would yield the maximum profit. What is the maximum profit? How many dOpis need to be produced and sold to achieve the maximum profit?

    \end{enumerate}

    \item Show that the Fahrenheit to Celsius conversion function found in Exercise \ref{celsiustofahr} in Section \ref{LinearFunctions} is invertible and that its inverse is the Celsius to Fahrenheit conversion function.

    \item Analytically show that the function $f(x) = x^3 + 3x + 1$ is one-to-one. Since finding a formula for its inverse is beyond the scope of this textbook, use Theorem \ref{inversefunctionprops} to help you compute $f^{-1}(1), \; f^{-1}(5), \;$ and $f^{-1}(-3)$.

    \item Let $f(x) = \frac{2x}{x^2-1}$. Using the techniques in Section \ref{RationalGraphs}, graph $y=f(x)$. Verify that $f$ is one-to-one on the interval $(-1,1)$. Use the procedure outlined on Page \pageref{inverseprocedure} and your graphing calculator to find the formula for $f^{-1}(x)$. Note that since $f(0) = 0$, it should be the case that $f^{-1}(0) = 0$. What goes wrong when you attempt to substitute $x=0$ into $f^{-1}(x)$? Discuss with your classmates how this problem arose and possible remedies.

    \item With the help of your classmates, explain why a function which is either strictly increasing or strictly decreasing on its entire domain would have to be one-to-one, hence invertible.

    \item If $f$ is odd and invertible, prove that $f^{-1}$ is also odd.

    \item \label{fcircginverse} Let $f$ and $g$ be invertible functions. With the help of your classmates show that $(f \circ g)$ is one-to-one, hence invertible, and that $(f \circ g)^{-1}(x) = (g^{-1} \circ f^{-1})(x)$.

    \item What graphical feature must a function $f$ possess for it to be its own inverse?

    \item \label{whatconditions} What conditions must you place on the values of $a, b, c$ and $d$ in Exercise \ref{genericinverselast} in order to guarantee that the function is invertible?

    \end{enumerate}

    \newpage

    \subsection{Answers}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $f^{-1}(x) = 42-x$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $f^{-1}(x) = 3x-10$

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

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

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

    \item $f^{-1}(x) = (x - 2)^{2} + 5, \; x \leq 2$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $f^{-1}(x) = \frac{1}{9}(x+4)^2+1$, $x \geq -4$

    \item $f^{-1}(x) = \frac{1}{8}(x-1)^2-\frac{5}{2}$, $x \leq 1$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

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

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

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $f^{-1}(x) = 5 + \sqrt{x+25}$

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

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

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

    \item $f^{-1}(x) =-\frac{\sqrt{x}+1}{2}$, $x > 1$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

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

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

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

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

    \item $f^{-1}(x) = \dfrac{6x + 2}{3x - 4}$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

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

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

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

    \end{enumerate}

    \end{multicols}

    \begin{enumerate}

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

    \addtocounter{enumi}{4}

    \item

    \begin{enumerate}

    \item $p^{-1}(x) = \frac{450-x}{15}$. The domain of $p^{-1}$ is the range of $p$ which is $[0,450]$

    \item $p^{-1}(105) = 23$. This means that if the price is set to $\$105$ then $23$ dOpis will be sold.

    \item $\left(P\circ p^{-1}\right)(x) = -\frac{1}{15} x^2 + \frac{110}{3} x - 5000$, $0 \leq x \leq 450$. The graph of $y = \left(P\circ p^{-1}\right)(x)$ is a parabola opening downwards with vertex $\left(275, \frac{125}{3}\right) \approx (275, 41.67)$. This means that the maximum profit is a whopping $\$41.67$ when the price per dOpi is set to $\$275$. At this price, we can produce and sell $p^{-1}(275) = 11.\overline{6}$ dOpis. Since we cannot sell part of a system, we need to adjust the price to sell either $11$ dOpis or $12$ dOpis. We find $p(11) = 285$ and $p(12) = 270$, which means we set the price per dOpi at either $\$285$ or $\$270$, respectively. The profits at these prices are $\left(P\circ p^{-1}\right)(285) = 35$ and $\left(P\circ p^{-1}\right)(270) = 40$, so it looks as if the maximum profit is $\$40$ and it is made by producing and selling $12$ dOpis a week at a price of $\$270$ per dOpi.

    \end{enumerate}

    \addtocounter{enumi}{1}

    \item Given that $f(0) = 1$, we have $f^{-1}(1) = 0$. Similarly $f^{-1}(5) = 1$ and $f^{-1}(-3) = -1$

    \end{enumerate}

    \closegraphsfile

    5.3: Other Algebraic Functions

    \subsection{Exercises}

    For each function in Exercises \ref{algfcngraphexfirst} - \ref{algfcngraphexlast} below

    \begin{itemize}

    \item Find its domain.

    \item Create a sign diagram.

    \item Use your calculator to help you sketch its graph and identify any vertical or horizontal asymptotes, `unusual steepness' or cusps.

    \end{itemize}

    \begin{multicols}{2}

    \begin{enumerate}

    \item $f(x) = \sqrt{1 - x^{2}}$ \label{algfcngraphexfirst}

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

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

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

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

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $f(x) = \sqrt[4]{\dfrac{16x}{x^{2} - 9}}$

    \item $f(x) = \dfrac{5x}{\sqrt[3]{x^{3} + 8}}$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $f(x) = x^{\frac{2}{3}}(x - 7)^{\frac{1}{3}}$

    \item $f(x) = x^{\frac{3}{2}}(x - 7)^{\frac{1}{3}}$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $f(x) = \sqrt{x(x + 5)(x - 4)}$

    \item $f(x) = \sqrt[3]{x^{3} + 3x^{2} - 6x - 8}$ \label{algfcngraphexlast}

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

    \end{enumerate}

    \end{multicols}

    In Exercises \ref{radicalgraphexfirst} - \ref{radicalgraphexlast}, sketch the graph of $y=g(x)$ by starting with the graph of $y = f(x)$ and using the transformations presented in Section \ref{Transformations}.

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $f(x) = \sqrt[3]{x}$, $g(x) = \sqrt[3]{x-1}-2$ \label{radicalgraphexfirst}

    \item $f(x) = \sqrt[3]{x}$, $g(x) = -2\sqrt[3]{x + 1} + 4$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $f(x) = \sqrt[4]{x}$, $g(x) = \sqrt[4]{x-1}-2$

    \item $f(x) = \sqrt[4]{x}$, $g(x) = 3\sqrt[4]{x - 7} - 1$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $f(x) = \sqrt[5]{x}$, $g(x) = \sqrt[5]{x + 2} + 3$

    \item $f(x) = \sqrt[8]{x}$, $g(x) = \sqrt[8]{-x} - 2$ \label{radicalgraphexlast}

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

    \end{enumerate}

    \end{multicols}

    \phantomsection

    \label{furtherequineqexercises}

    In Exercises \ref{algineqexfirst} - \ref{algineqexlast}, solve the equation or inequality.

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $x+1 = \sqrt{3x+7}$ \label{algineqexfirst}

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

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $x + \sqrt{3x+10} = -2$

    \item $3x+\sqrt{6-9x}=2$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

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

    \item $x^{\frac{3}{2}} = 8$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $x^{\frac{2}{3}} = 4$

    \item $\sqrt{x - 2} + \sqrt{x - 5} = 3$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $\sqrt{2x+1} = 3 + \sqrt{4-x}$

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

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $10-\sqrt{x-2} \leq 11$

    \item $\sqrt[3]{x} \leq x$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $2 (x-2)^{-\frac{1}{3}} -\frac{2}{3} x(x-2)^{-\frac{4}{3}} \leq 0$

    \item $-\frac{4}{3} (x-2)^{-\frac{4}{3}} + \frac{8}{9} x (x-2)^{-\frac{7}{3}} \geq 0$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $2x^{-\frac{1}{3}}(x-3)^{\frac{1}{3}} + x^{\frac{2}{3}} (x-3)^{-\frac{2}{3}} \geq 0$

    \item $\sqrt[3]{x^{3} + 3x^{2} - 6x - 8} > x + 1$

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

    \end{enumerate}

    \end{multicols}

    \begin{enumerate}

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

    \item $\frac{1}{3}x^{\frac{3}{4}}(x - 3)^{-\frac{2}{3}} + \frac{3}{4}x^{-\frac{1}{4}}(x - 3)^{\frac{1}{3}} < 0$

    \item $x^{-\frac{1}{3}} (x-3)^{-\frac{2}{3}} - x^{-\frac{4}{3}} (x-3)^{-\frac{5}{3}} (x^2-3x+2) \geq 0$

    \item $\frac{2}{3}(x + 4)^{\frac{3}{5}}(x - 2)^{-\frac{1}{3}} + \frac{3}{5}(x + 4)^{-\frac{2}{5}}(x - 2)^{\frac{2}{3}} \geq 0$ \label{algineqexlast}

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

    \end{enumerate}

    \begin{enumerate}

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

    \item Rework Example \ref{SasquatchCable} so that the outpost is 10 miles from Route 117 and the nearest junction box is 30 miles down the road for the post.

    \item The volume $V$ of a right cylindrical cone depends on the radius of its base $r$ and its height $h$ and is given by the formula $V = \frac{1}{3} \pi r^2 h$. The surface area $S$ of a right cylindrical cone also depends on $r$ and $h$ according to the formula $S = \pi r \sqrt{r^2+h^2}$. Suppose a cone is to have a volume of 100 cubic centimeters.

    \begin{enumerate}

    \item \label{heightintermsofr} Use the formula for volume to find the height $h$ as a function of $r$.

    \item Use the formula for surface area and your answer to \ref{heightintermsofr} to find the surface area $S$ as a function of $r$.

    \item Use your calculator to find the values of $r$ and $h$ which minimize the surface area. What is the minimum surface area? Round your answers to two decimal places.

    \end{enumerate}

    \item \label{WindChillTemperature} The \href{http://www.nws.noaa.gov/om/windchill...rline{National Weather Service}} uses the following formula to calculate the wind chill: \[ W = 35.74 + 0.6215 \, T_{a} - 35.75\, V^{0.16} + 0.4275 \, T_{a} \, V^{0.16} \] where $W$ is the wind chill temperature in $^{\circ}$F, $T_{a}$ is the air temperature in $^{\circ}$F, and $V$ is the wind speed in miles per hour. Note that $W$ is defined only for air temperatures at or lower than $50^{\circ}$F and wind speeds above $3$ miles per hour.

    \begin{enumerate}

    \item Suppose the air temperature is $42^{\circ}$ and the wind speed is $7$ miles per hour. Find the wind chill temperature. Round your answer to two decimal places.

    \item Suppose the air temperature is $37^{\circ}$F and the wind chill temperature is $30^{\circ}$F. Find the wind speed. Round your answer to two decimal places.

    \end{enumerate}

    \item As a follow-up to Exercise \ref{WindChillTemperature}, suppose the air temperature is $28^{\circ}$F.

    \begin{enumerate}

    \item Use the formula from Exercise \ref{WindChillTemperature} to find an expression for the wind chill temperature as a function of the wind speed, $W(V)$.

    \item \label{WindChill0} Solve $W(V) = 0$, round your answer to two decimal places, and interpret.

    \item Graph the function $W$ using your calculator and check your answer to part \ref{WindChill0}.

    \end{enumerate}

    \item \label{pendulumproblem} The period of a pendulum in seconds is given by \[T = 2\pi \sqrt{\dfrac{L}{g}}\](for small displacements) where $L$ is the length of the pendulum in meters and $g = 9.8$ meters per second per second is the acceleration due to gravity. My Seth-Thomas antique schoolhouse clock needs $T = \frac{1}{2}$ second and I can adjust the length of the pendulum via a small dial on the bottom of the bob. At what length should I set the pendulum?

    \item The Cobb-Douglas production model states that the yearly total dollar value of the production output $P$ in an economy is a function of labor $x$ (the total number of hours worked in a year) and capital $y$ (the total dollar value of all of the stuff purchased in order to make things). Specifically, $P = ax^{b}y^{1 - b}$. By fixing $P$, we create what's known as an `isoquant' and we can then solve for $y$ as a function of $x$. Let's assume that the Cobb-Douglas production model for the country of Sasquatchia is $P = 1.23x^{0.4}y^{0.6}$.

    \begin{enumerate}

    \item Let $P = 300$ and solve for $y$ in terms of $x$. If $x = 100$, what is $y$?

    \item Graph the isoquant $300 = 1.23x^{0.4}y^{0.6}$. What information does an ordered pair $(x, y)$ which makes $P = 300$ give you? With the help of your classmates, find several different combinations of labor and capital all of which yield $P = 300$. Discuss any patterns you may see.

    \end{enumerate}

    \item According to Einstein's Theory of Special Relativity, the observed mass $m$ of an object is a function of how fast the object is traveling. Specifically, \[m(x) = \dfrac{m_{r}}{\sqrt{1 - \dfrac{x^{2}}{c^{2}}}}\] where $m(0)=m_{r}$ is the mass of the object at rest, $x$ is the speed of the object and $c$ is the speed of light.

    \begin{enumerate}

    \item Find the applied domain of the function.

    \item Compute $m(.1c), \, m(.5c), \, m(.9c)$ and $m(.999c)$.

    \item As $x \rightarrow c^{-}$, what happens to $m(x)$?

    \item How slowly must the object be traveling so that the observed mass is no greater than 100 times its mass at rest?

    \end{enumerate}

    \item Find the inverse of $k(x) = \dfrac{2x}{\sqrt{x^{2} - 1}}$.

    \pagebreak

    \item \label{pursuitfurther} Suppose Fritzy the Fox, positioned at a point $(x,y)$ in the first quadrant, spots Chewbacca the Bunny at $(0,0)$. Chewbacca begins to run along a fence (the positive $y$-axis) towards his warren. Fritzy, of course, takes chase and constantly adjusts his direction so that he is always running directly at Chewbacca. If Chewbacca's speed is $v_{\mbox{\tiny$1$}}$ and Fritzy's speed is $v_{\mbox{\tiny$2$}}$, the path Fritzy will take to intercept Chewbacca, provided $v_{\mbox{\tiny$2$}}$ is directly proportional to, but not equal to, $v_{\mbox{\tiny$1$}}$ is modeled by

    \[ y = \dfrac{1}{2} \left(\dfrac{x^{1+ v_{1}/v_{2}}}{1+v_{\mbox{\tiny$1$}}/v_{\mbox{\tiny$2$}}}- \dfrac{x^{1-v_{\mbox{\tiny$1$}}/v_{\mbox{\tiny$2$}}}}{1-v_{\mbox{\tiny$1$}}/v_{\mbox{\tiny$2$}}}\right) + \dfrac{v_{\mbox{\tiny$1$}} v_{\mbox{\tiny$2$}}}{v_{\mbox{\tiny$2$}}^2-v_{\mbox{\tiny$1$}}^2} \]

    \begin{enumerate}

    \item Determine the path that Fritzy will take if he runs exactly twice as fast as Chewbacca; that is, $v_{\mbox{\tiny$2$}} = 2v_{\mbox{\tiny$1$}}$. Use your calculator to graph this path for $x \geq 0$. What is the significance of the $y$-intercept of the graph?

    \item Determine the path Fritzy will take if Chewbacca runs exactly twice as fast as he does; that is, $v_{\mbox{\tiny$1$}} = 2v_{\mbox{\tiny$2$}}$. Use your calculator to graph this path for $x > 0$. Describe the behavior of $y$ as $x \rightarrow 0^{+}$ and interpret this physically.

    \item With the help of your classmates, generalize parts (a) and (b) to two cases: $v_{\mbox{\tiny$2$}} > v_{\mbox{\tiny$1$}}$ and $v_{\mbox{\tiny$2$}} < v_{\mbox{\tiny$1$}}$. We will discuss the case of $v_{\mbox{\tiny$1$}} = v_{\mbox{\tiny$2$}}$ in Exercise \ref{pursuitlog} in Section \ref{ExpLogApplications}.

    \end{enumerate}

    \item Verify the Quotient Rule for Radicals in Theorem \ref{radicalprops}.

    \item Show that $\left(x^{\frac{3}{2}}\right)^{\frac{2}{3}} = x$ for all $x \geq 0$.

    \item Show that $\sqrt[3]{2}$ is an irrational number by first showing that it is a zero of $p(x) = x^{3} - 2$ and then showing $p$ has no rational zeros. (You'll need the Rational Zeros Theorem, Theorem \ref{RZT}, in order to show this last part.) \label{nthrootsareirrational}

    \item With the help of your classmates, generalize Exercise \ref{nthrootsareirrational} to show that $\sqrt[n]{c}$ is an irrational number for any natural numbers $c \geq 2$ and $n \geq 2$ provided that $c \neq p^{n}$ for some natural number $p$.

    \end{enumerate}

    \newpage

    \subsection{Answers}

    \begin{enumerate}

    \item \begin{multicols}{2}

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

    Domain: $[-1, 1]$\\

    \begin{mfpic}[20][10]{0}{4}{-1.5}{1.5}

    \polyline{(0,0), (4,0)}

    \xmarks{0,4}

    \tlabel[cc](0,-1){$-1 \hspace{7pt}$}

    \tlabel[cc](2,1){$(+)$}

    \tlabel[cc](0,1){$0$}

    \tlabel[cc](4,1){$0$}

    \tlabel[cc](4,-1){$1$}

    \end{mfpic}

    No asymptotes\\

    Unusual steepness at $x = -1$ and $x = 1$\\

    No cusps\\

    \vfill

    \columnbreak

    \begin{mfpic}[50]{-1.5}{1.5}{-0.15}{1.5}

    \point[3pt]{(0,1), (-1,0), (1,0)}

    \parafcn{0,3.14159,0.1}{(cos(t),sin(t))}

    \axes

    \tlabel[cc](1.5,-0.15){\scriptsize $x$}

    \tlabel[cc](0.25,1.5){\scriptsize $y$}

    \xmarks{-1,1}

    \ymarks{1}

    \tlpointsep{4pt}

    \scriptsize

    \axislabels {x}{{$-1 \hspace{6pt}$} -1, {$1$} 1}

    \axislabels {y}{{$1$} 1}

    \normalsize

    \end{mfpic}

    \end{multicols}

    \item \begin{multicols}{2}

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

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

    \begin{mfpic}[20][10]{0}{4}{-1.5}{1.5}

    \arrow \polyline{(2,0), (0,0)}

    \arrow \polyline{(3,0), (5,0)}

    \xmarks{2,3}

    \tlabel[cc](2,-1){$-1 \hspace{7pt}$}

    \tlabel[cc](1,1){$(+)$}

    \tlabel[cc](4,1){$(+)$}

    \tlabel[cc](2,1){$0$}

    \tlabel[cc](3,1){$0$}

    \tlabel[cc](3,-1){$1$}

    \end{mfpic}

    No asymptotes\\

    Unusual steepness at $x = -1$ and $x = 1$\\

    No cusps\\

    \vfill

    \columnbreak

    \begin{mfpic}[20]{-4}{4}{-1}{4}

    \point[3pt]{(-1,0), (1,0)}

    \arrow \parafcn{0,2,0.1}{(cosh(t),sinh(t))}

    \arrow \parafcn{0,2,0.1}{(-cosh(t),sinh(t))}

    \axes

    \tlabel[cc](4,-0.25){\scriptsize $x$}

    \tlabel[cc](0.25,4){\scriptsize $y$}

    \xmarks{-3,-2,-1,1,2,3}

    \ymarks{1,2,3}

    \tlpointsep{4pt}

    \scriptsize

    \axislabels {x}{{$-3 \hspace{6pt}$} -3,{$-2 \hspace{6pt}$} -2,{$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3}

    \axislabels {y}{{$1$} 1, {$2$} 2, {$3$} 3}

    \normalsize

    \end{mfpic}

    \end{multicols}

    \item \begin{multicols}{2}

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

    Domain: $[-1,1]$\\

    \begin{mfpic}[20][10]{0}{4}{-1.5}{1.5}

    \polyline{(0,0), (5,0)}

    \xmarks{0,2.5,5}

    \tlabel[cc](0,-1){$-1 \hspace{7pt}$}

    \tlabel[cc](0,1){$0$}

    \tlabel[cc](1.25,1){$(-)$}

    \tlabel[cc](2.5,-1){$0$}

    \tlabel[cc](3.75,1){$(+)$}

    \tlabel[cc](2.5,1){$0$}

    \tlabel[cc](5,-1){$1$}

    \tlabel[cc](5,1){$0$}

    \end{mfpic}

    No asymptotes\\

    Unusual steepness at $x = -1$ and $x = 1$\\

    No cusps\\

    \vfill

    \columnbreak

    \begin{mfpic}[50][40]{-1.5}{1.5}{-1}{1.5}

    \point[3pt]{(-1,0), (1,0),(0,0)}

    \parafcn{0,3.14159,0.1}{(cos(t),cos(t)*sin(t))}

    \axes

    \tlabel[cc](1.5,-0.15){\scriptsize $x$}

    \tlabel[cc](0.25,1.5){\scriptsize $y$}

    \xmarks{-1,1}

    \ymarks{-1,1}

    \tlpointsep{4pt}

    \scriptsize

    \axislabels {x}{{$-1 \hspace{6pt}$} -1, {$1$} 1}

    \axislabels {y}{{$1$} 1,{$-1$} -1}

    \normalsize

    \end{mfpic}

    \end{multicols}

    \item \begin{multicols}{2}

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

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

    \begin{mfpic}[20][10]{0}{4}{-1.5}{1.5}

    \arrow \polyline{(2,0), (0,0)}

    \arrow \polyline{(3,0), (5,0)}

    \xmarks{2,3}

    \tlabel[cc](2,-1){$-1 \hspace{7pt}$}

    \tlabel[cc](1,1){$(-)$}

    \tlabel[cc](4,1){$(+)$}

    \tlabel[cc](2,1){$0$}

    \tlabel[cc](3,1){$0$}

    \tlabel[cc](3,-1){$1$}

    \end{mfpic}

    No asymptotes\\

    Unusual steepness at $x = -1$ and $x = 1$\\

    No cusps\\

    \vfill

    \columnbreak

    \begin{mfpic}[20][15]{-4}{4}{-4}{4}

    \point[3pt]{(-1,0), (1,0)}

    \arrow \parafcn{0,2,0.1}{(cosh(t),sinh(t))}

    \arrow \parafcn{0,2,0.1}{(-cosh(t),-sinh(t))}

    \axes

    \tlabel[cc](4,-0.25){\scriptsize $x$}

    \tlabel[cc](0.25,4){\scriptsize $y$}

    \xmarks{-3,-2,-1,1,2,3}

    \ymarks{-3,-2,-1,1,2,3}

    \tlpointsep{4pt}

    \scriptsize

    \axislabels {x}{{$-3 \hspace{6pt}$} -3,{$-2 \hspace{6pt}$} -2,{$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3}

    \axislabels {y}{{$-3$} -3,{$-2$} -2,{$-1$} -1,{$1$} 1, {$2$} 2, {$3$} 3}

    \normalsize

    \end{mfpic}

    \end{multicols}

    \pagebreak

    \item \begin{multicols}{2}

    $f(x) = \sqrt[4]{\dfrac{16x}{x^2 - 9}}$\\

    Domain: $(-3, 0] \cup (3, \infty)$\\

    \begin{mfpic}[15]{-3}{6}{-1}{1}

    \polyline{(-3,0),(0,0)}

    \arrow \polyline{(3,0),(6,0)}

    \xmarks{-3,0,3}

    \tlabel[cc](-1.5,0.75){$(+)$}

    \tlabel[cc](-3,-0.75){$-3 \hspace{7pt}$}

    \tlabel[cc](-3,0.75){\textinterrobang}

    \tlabel[cc](0,-0.75){$0$}

    \tlabel[cc](0,0.75){$0$}

    \tlabel[cc](3,0.75){\textinterrobang}

    \tlabel[cc](3,-0.75){$3$}

    \tlabel[cc](4.5,0.75){$(+)$}

    \end{mfpic}

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

    Horizontal asymptote: $y = 0$\\

    Unusual steepness at $x = 0$\\

    No cusps\\

    \vfill

    \columnbreak

    \begin{mfpic}[15]{-3.5}{9}{-1}{6}

    \point[3pt]{(0,0)}

    \dashed \polyline{(-3,-1), (-3,6)}

    \dashed \polyline{(3,-1), (3,6)}

    \arrow \reverse \function{-2.93,0,0.1}{((16*x)/((x**2) - 9))**(0.25)}

    \arrow \reverse \arrow \function{3.05,9,0.1}{((16*x)/((x**2) - 9))**(0.25)}

    \axes

    \tlabel[cc](9,-0.5){\scriptsize $x$}

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

    \xmarks{-3 step 1 until 8}

    \ymarks{1,2,3,4,5}

    \tlpointsep{4pt}

    \scriptsize

    \axislabels {x}{{$-3 \hspace{6pt}$} -3, {$-2 \hspace{6pt}$} -2, {$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5, {$6$} 6, {$7$} 7, {$8$} 8}

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

    \normalsize

    \end{mfpic}

    \end{multicols}

    \item \begin{multicols}{2}

    $f(x) = x^{\frac{2}{3}}(x - 7)^{\frac{1}{3}}$\\

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

    \begin{mfpic}[10]{-3}{10}{-2}{2}

    \arrow \reverse \arrow \polyline{(-3,0),(10,0)}

    \xmarks{0,7}

    \tlabel[cc](-1.5,1){$(-)$}

    \tlabel[cc](0,-1){$0$}

    \tlabel[cc](0,1){$0$}

    \tlabel[cc](3.5,1){$(-)$}

    \tlabel[cc](7,-1){$7$}

    \tlabel[cc](7,1){$0$}

    \tlabel[cc](8.5,1){$(+)$}

    \end{mfpic}

    No vertical or horizontal asymptotes\footnote{Using Calculus it can be shown that $y = x - \frac{7}{3}$ is a slant asymptote of this graph.}\\

    Unusual steepness at $x = 7$\\

    Cusp at $x = 0$\\

    \vfill

    \columnbreak

    \begin{mfpic}[10]{-4}{10}{-5}{5.5}

    \point[3pt]{(0,0), (7,0)}

    \arrow \reverse \function{-3,0,0.1}{-((x**2)**(1/3))*((7 - x)**(1/3))}

    \function{0,7,0.1}{-((x**2)**(1/3))*((7 - x)**(1/3))}

    \arrow \function{7,9,0.1}{((x**2)**(1/3))*((x - 7)**(1/3))}

    \axes

    \tlabel[cc](10,-0.5){\scriptsize $x$}

    \tlabel[cc](0.5,5.5){\scriptsize $y$}

    \xmarks{-3 step 1 until 9}

    \ymarks{-4 step 1 until 5}

    \tlpointsep{4pt}

    \tiny

    \axislabels {x}{{$-3 \hspace{6pt}$} -3, {$-2 \hspace{6pt}$} -2, {$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5, {$6$} 6, {$7$} 7, {$8$} 8, {$9$} 9}

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

    \normalsize

    \end{mfpic}

    \end{multicols}

    \item \begin{multicols}{2}

    $f(x) = \dfrac{5x}{\sqrt[3]{x^{3} + 8}}$\\

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

    \begin{mfpic}[20]{-4}{2}{-1}{1}

    \arrow \reverse \arrow \polyline{(-4,0),(2,0)}

    \xmarks{-2,0}

    \tlabel[cc](-3, 0.5){$(+)$}

    \tlabel[cc](-2,-0.5){$-2 \hspace{7pt}$}

    \tlabel[cc](-2,0.5){\textinterrobang}

    \tlabel[cc](-1,0.5){$(-)$}

    \tlabel[cc](0,-0.5){$0$}

    \tlabel[cc](0,0.5){$0$}

    \tlabel[cc](1,0.5){$(+)$}

    \end{mfpic}

    Vertical asymptote $x = -2$\\

    Horizontal asymptote $y = 5$\\

    No unusual steepness or cusps\\

    \vfill

    \columnbreak

    \begin{mfpic}[10][8]{-5}{5}{-7}{9}

    \point[3pt]{(0,0)}

    \dashed \polyline{(-5,5), (5,5)}

    \dashed \polyline{(-2,-7), (-2,9)}

    \arrow \reverse \arrow \function{-5,-2.2,0.1}{(-5*x)/((-(x**3) - 8)**(1/3))}

    \arrow \reverse \arrow \function{-1.8,5,0.1}{(5*x)/(((x**3) + 8)**(1/3))}

    \axes

    \tlabel[cc](5,-0.5){\scriptsize $x$}

    \tlabel[cc](0.5,9){\scriptsize $y$}

    \xmarks{-4 step 1 until 4}

    \ymarks{-6 step 1 until 8}

    \tlpointsep{4pt}

    \tiny

    \axislabels {x}{{$-4 \hspace{6pt}$} -4, {$-3 \hspace{6pt}$} -3, {$-2 \hspace{6pt}$} -2, {$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4}

    \axislabels {y}{{$-6$} -6, {$-5$} -5, {$-4$} -4, {$-3$} -3, {$-2$} -2, {$-1$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5, {$6$} 6, {$7$} 7, {$8$} 8}

    \normalsize

    \end{mfpic}

    \end{multicols}

    \item \begin{multicols}{2}

    $f(x) = x^{\frac{3}{2}}(x - 7)^{\frac{1}{3}}$\\

    Domain: $[0, \infty)$\\

    \begin{mfpic}[15]{0}{10}{-1}{1}

    \reverse \arrow \polyline{(0,0),(10,0)}

    \xmarks{0, 7}

    \tlabel[cc](0,-0.5){$0$}

    \tlabel[cc](0,0.5){$0$}

    \tlabel[cc](3.5, 0.5){$(-)$}

    \tlabel[cc](7,-0.5){$7$}

    \tlabel[cc](7,0.5){$0$}

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

    \end{mfpic}

    No asymptotes\\

    Unusual steepness at $x = 7$\\

    No cusps\\

    \vfill

    \columnbreak

    \begin{mfpic}[15][3]{-1}{8.5}{-20}{30}

    \point[3pt]{(0,0), (7,0)}

    \function{0,7,0.1}{-(x**1.5)*((7 - x)**(1/3))}

    \arrow \function{7,8.5,0.1}{(x**1.5)*((x - 7)**(1/3))}

    \axes

    \tlabel[cc](8.5,-3){\scriptsize $x$}

    \tlabel[cc](0.5,30){\scriptsize $y$}

    \xmarks{1 step 1 until 8}

    \ymarks{-15 step 5 until 25}

    \tlpointsep{4pt}

    \scriptsize

    \axislabels {x}{{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5, {$6$} 6, {$7$} 7, {$8$} 8}

    \axislabels {y}{{$-15$} -15, {$-10$} -10, {$-5$} -5, {$5$} 5, {$10$} 10, {$15$} 15, {$20$} 20, {$25$} 25}

    \normalsize

    \end{mfpic}

    \end{multicols}

    \item \begin{multicols}{2}

    $f(x) = \sqrt{x(x + 5)(x - 4)}$\\

    Domain: $[-5, 0] \cup [4, \infty)$\\

    \begin{mfpic}[10]{-5}{8}{-1}{1}

    \polyline{(-5,0),(0,0)}

    \arrow \polyline{(4,0),(8,0)}

    \xmarks{-5,0,4}

    \tlabel[cc](-5,-1){$-5 \hspace{7pt}$}

    \tlabel[cc](-5,1){$0$}

    \tlabel[cc](-2.5,1){$(+)$}

    \tlabel[cc](0,-1){$0$}

    \tlabel[cc](0,1){$0$}

    \tlabel[cc](4,-1){$4$}

    \tlabel[cc](4,1){$0$}

    \tlabel[cc](6,1){$(+)$}

    \end{mfpic}

    No asymptotes\\

    Unusual steepness at $x = -5, x = 0$ and $x = 4$\\

    No cusps\\

    \vfill

    \columnbreak

    \begin{mfpic}[10]{-6}{6}{-1}{10}

    \point[3pt]{(-5,0),(0,0),(4,0)}

    \function{-5,0,0.1}{sqrt((x**3) + (x**2) - (20*x))}

    \arrow \function{4,5.5,0.1}{sqrt((x**3) + (x**2) - (20*x))}

    \axes

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

    \tlabel[cc](0.5,10){\scriptsize $y$}

    \xmarks{-5 step 1 until 5}

    \ymarks{1 step 1 until 9}

    \tlpointsep{4pt}

    \tiny

    \axislabels {x}{{$-5 \hspace{6pt}$} -5, {$-4 \hspace{6pt}$} -4, {$-3 \hspace{6pt}$} -3, {$-2 \hspace{6pt}$} -2, {$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5}

    \axislabels {y}{{$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5, {$6$} 6, {$7$} 7, {$8$} 8, {$9$} 9}

    \normalsize

    \end{mfpic}

    \end{multicols}

    \item \begin{multicols}{2}

    $f(x) = \sqrt[3]{x^{3} + 3x^{2} - 6x - 8}$\\

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

    \begin{mfpic}[10]{-8}{6}{-1}{1}

    \arrow \reverse \arrow \polyline{(-8,0),(6,0)}

    \xmarks{-4,-1,2}

    \tlabel[cc](-6,1){$(-)$}

    \tlabel[cc](-4,-1){$-4 \hspace{7pt}$}

    \tlabel[cc](-4,1){$0$}

    \tlabel[cc](-2.5,1){$(+)$}

    \tlabel[cc](-1,-1){$-1 \hspace{7pt}$}

    \tlabel[cc](-1,1){$0$}

    \tlabel[cc](0.5,1){$(-)$}

    \tlabel[cc](2,-1){$2$}

    \tlabel[cc](2,1){$0$}

    \tlabel[cc](4,1){$(+)$}

    \end{mfpic}

    No vertical or horizontal asymptotes\footnote{Using Calculus it can be shown that $y = x + 1$ is a slant asymptote of this graph.}\\

    Unusual steepness at $x = -4, x = -1$ and $x = 2$\\

    No cusps\\

    \vfill

    \columnbreak

    \begin{mfpic}[10]{-6}{6}{-5}{7}

    \point[3pt]{(-4,0),(-1,0),(2,0)}

    \arrow \reverse \function{-6,-4,0.1}{-((-((x**3) + (3*(x**2)) - (6*x) - 8))**(1/3))}

    \function{-4,-1,0.1}{((x**3) + (3*(x**2)) - (6*x) - 8)**(1/3)}

    \function{-1,2,0.1}{-((-((x**3) + (3*(x**2)) - (6*x) - 8))**(1/3))}

    \arrow \function{2,6,0.1}{((x**3) + (3*(x**2)) - (6*x) - 8)**(1/3)}

    \axes

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

    \tlabel[cc](0.5,7){\scriptsize $y$}

    \xmarks{-5 step 1 until 5}

    \ymarks{-4 step 1 until 6}

    \tlpointsep{4pt}

    \tiny

    \axislabels {x}{{$-5 \hspace{6pt}$} -5, {$-4 \hspace{6pt}$} -4, {$-3 \hspace{6pt}$} -3, {$-2 \hspace{6pt}$} -2, {$-1 \hspace{6pt}$} -1, {$1$} 1, {$2$} 2, {$3$} 3, {$4$} 4, {$5$} 5}

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

    \normalsize

    \end{mfpic}

    \end{multicols}

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

    \end{enumerate}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $g(x) = \sqrt[3]{x-1}-2$ \\

    \begin{mfpic}[8][13]{-10}{12}{-5}{1}

    \arrow \reverse \arrow \parafcn{-4.2,0.2,0.1}{(((t + 2)**3) + 1,t)}

    \axes

    \tlabel[cc](12,-0.5){\scriptsize $x$}

    \tlabel[cc](0.75,1){\scriptsize $y$}

    \point[3pt]{(-7, -4), (0,-3), (1,-2), (2,-1), (9,0)}

    \ymarks{-4,-3,-2,-1}

    \xmarks{-9 step 1 until 11}

    \tiny

    \tlpointsep{4pt}

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

    \axislabels {x}{{$-9 \hspace{6pt}$} -9, {$-7 \hspace{6pt}$} -7, {$-5 \hspace{6pt}$} -5, {$-3 \hspace{6pt}$} -3, {$-1 \hspace{6pt}$} -1, {$1$} 1, {$3$} 3, {$5$} 5, {$7$} 7, {$9$} 9, {$11$} 11}

    \normalsize

    \end{mfpic}

    \vfill

    \columnbreak

    \item $g(x) = -2\sqrt[3]{x + 1} + 4$\\

    \begin{mfpic}[10][9]{-7}{9}{-1}{8}

    \point[3pt]{(-2,6),(-1,4),(0,2),(7,0)}

    \arrow \reverse \function{-7,-1,0.1}{2*((-x - 1)**(1/3)) + 4}

    \arrow \function{-1,8.5,0.1}{-2*((x + 1)**(1/3)) + 4}

    \axes

    \tlabel[cc](9,-0.5){\scriptsize $x$}

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

    \xmarks{-6 step 1 until 8}

    \ymarks{1 step 1 until 7}

    \tlpointsep{4pt}

    \tiny

    \axislabels {x}{ {$-5 \hspace{6pt}$} -5, {$-3 \hspace{6pt}$} -3, {$-1 \hspace{6pt}$} -1, {$1$} 1, {$3$} 3, {$5$} 5, {$7$} 7 }

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

    \normalsize

    \end{mfpic}

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $g(x) = \sqrt[4]{x-1}-2$\\

    \begin{mfpic}[8][25]{-1}{22}{-3}{1}

    \arrow \parafcn{-2,0.12,0.1}{(((t + 2)**4) + 1,t)}

    \axes

    \tlabel[cc](22,-0.75){\scriptsize $x$}

    \tlabel[cc](0.5,1){\scriptsize $y$}

    \point[3pt]{(1,-2),(2,-1),(17,0)}

    \ymarks{-2,-1}

    \xmarks{1 step 1 until 21}

    \tiny

    \tlpointsep{4pt}

    \axislabels {y}{{$-2$} -2, {$-1$} -1}

    \axislabels {x}{{$1$} 1, {$3$} 3, {$5$} 5, {$7$} 7, {$9$} 9, {$11$} 11, {$13$} 13, {$15$} 15, {$17$} 17, {$19$} 19, {$21$} 21}

    \normalsize

    \end{mfpic}

    \vfill

    \columnbreak

    \item $g(x) = 3\sqrt[4]{x - 7} - 1$\\

    \begin{mfpic}[5][13]{-1}{25}{-2}{6}

    \point[3pt]{(7,-1),(8,2),(23,5)}

    \arrow \function{7,25,0.1}{3*((x - 7)**(0.25)) - 1}

    \axes

    \tlabel[cc](25,-0.5){\scriptsize $x$}

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

    \xmarks{1 step 1 until 23}

    \ymarks{-1 step 1 until 5}

    \tlpointsep{4pt}

    \tiny

    \axislabels {x}{{$7$} 7, {$8$} 8, {$23$} 23}

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

    \normalsize

    \end{mfpic}

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{2}

    \begin{enumerate}

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

    \item $g(x) = \sqrt[5]{x + 2} + 3$\\

    \begin{mfpic}[2][10]{-37}{33}{-1}{6}

    \point[2pt]{(-34,1),(-3,2),(-2,3),(-1,4),(30,5)}

    \arrow \function{-2,33,0.1}{((x + 2)**(0.20)) + 3}

    \arrow \reverse \function{-37,-2,0.1}{(-((-x - 2)**(0.20))) + 3}

    \axes

    \tlabel[cc](33,-0.5){\scriptsize $x$}

    \tlabel[cc](2,6){\scriptsize $y$}

    \xmarks{-34,-2,30}

    \ymarks{1 step 1 until 5}

    \tlpointsep{4pt}

    \tiny

    \axislabels {x}{{$-34 \hspace{5pt}$} -34, {$-2 \hspace{5pt}$} -2, {$30$} 30}

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

    \normalsize

    \end{mfpic}

    \item $g(x) = \sqrt[8]{-x} - 2$\\

    \begin{mfpic}[3][15]{-45}{5}{-3}{1}

    \point[2pt]{(0,-2),(-1,-1)}

    \arrow \reverse \function{-45,0,0.1}{((-x)**0.125) - 2}

    \axes

    \tlabel[cc](5,-0.5){\scriptsize $x$}

    \tlabel[cc](1.5,1){\scriptsize $y$}

    \xmarks{-40,-30,-20,-10}

    \ymarks{-2,-1}

    \tlpointsep{4pt}

    \tiny

    \axislabels {x}{{$-40 \hspace{5pt}$} -40, {$-30 \hspace{5pt}$} -30, {$-20 \hspace{5pt}$} -20, {$-10 \hspace{5pt}$} -10}

    \axislabels {y}{{$-2$} -2, {$-1$} -1}

    \normalsize

    \end{mfpic}

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{3}

    \begin{enumerate}

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

    \item $x=3$

    \item $x = \frac{1}{4}$

    \item $x=-3$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{3}

    \begin{enumerate}

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

    \item $x = -\frac{1}{3}, \; \frac{2}{3}$

    \item $x = \frac{5 + \sqrt{57}}{8}$

    \item $x = 4$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{3}

    \begin{enumerate}

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

    \item $x = \pm 8$

    \item $x = 6$

    \item $x = 4$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{3}

    \begin{enumerate}

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

    \item $x=-2, 6$

    \item $[2, \infty)$

    \item $[-1, 0] \cup [1, \infty)$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{3}

    \begin{enumerate}

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

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

    \item $(2,6]$

    \item $(-\infty, 0) \cup [2,3) \cup (3, \infty)$

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

    \end{enumerate}

    \end{multicols}

    \begin{multicols}{3}

    \begin{enumerate}

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

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

    \item $\left(0, \frac{27}{13} \right)$

    \item $(-\infty, 0) \cup (0,3)$

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

    \end{enumerate}

    \end{multicols}

    \begin{enumerate}

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

    \item $(-\infty, -4) \cup \left(-4, -\frac{22}{19}\right] \cup (2, \infty)$

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

    \end{enumerate}

    \begin{enumerate}

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

    \item $C(x) = 15x+20\sqrt{100+(30-x)^2}$, $0 \leq x \leq 30$. The calculator gives the absolute minimum at $\approx (18.66, 582.29)$. This means to minimize the cost, approximately 18.66 miles of cable should be run along Route 117 before turning off the road and heading towards the outpost. The minimum cost to run the cable is approximately $\$582.29$.

    \item

    \begin{enumerate}

    \item $h(r) = \frac{300}{\pi r^2}$, $r > 0$.

    \item $S(r) = \pi r \sqrt{r^2+\left(\frac{300}{\pi r^2}\right)^2} = \frac{\sqrt{\pi^2 r^6+90000}}{r}$, $r>0$

    \item The calculator gives the absolute minimum at the point $\approx (4.07, 90.23)$. This means the radius should be (approximately) 4.07 centimeters and the height should be 5.76 centimeters to give a minimum surface area of 90.23 square centimeters.

    \end{enumerate}

    \item

    \begin{enumerate}

    \item $W \approx 37.55^{\circ}$F.

    \item $V \approx 9.84$ miles per hour.

    \end{enumerate}

    \item

    \begin{enumerate}

    \item $W(V) = 53.142 - 23.78 V^{0.16}$. Since we are told in Exercise \ref{WindChillTemperature} that wind chill is only effect for wind speeds of more than 3 miles per hour, we restrict the domain to $V > 3$.

    \item $W(V)=0$ when $V \approx 152.29$. This means, according to the model, for the wind chill temperature to be $0^{\circ}$F, the wind speed needs to be $152.29$ miles per hour.

    \item The graph is below. \\

    \centerline{\includegraphics[width=1.75in]{./FurtherGraphics/WINDCHILL.jpg}}

    \end{enumerate}

    \item $9.8 \left(\dfrac{1}{4\pi}\right)^{2} \approx 0.062$ meters or $6.2$ centimeters

    \item \begin{enumerate}

    \item First rewrite the model as $P = 1.23x^{\frac{2}{5}}y^{\frac{3}{5}}$. Then $300 = 1.23x^{\frac{2}{5}}y^{\frac{3}{5}}$ yields $y = \left( \dfrac{300}{1.23x^{\frac{2}{5}}} \right)^{\frac{5}{3}}$. If $x = 100$ then $y \approx 441.93687$.

    \end{enumerate}

    \item \begin{enumerate}

    \item $[0, c)$

    \item $~$

    \begin{tabular}{ll}

    $m(.1c) = \dfrac{m_{r}}{\sqrt{.99}} \approx 1.005m_{r}$ & $m(.5c) = \dfrac{m_{r}}{\sqrt{.75}} \approx 1.155m_{r}$\\ \smallskip

    $m(.9c) = \dfrac{m_{r}}{\sqrt{.19}} \approx 2.294m_{r}$ & $m(.999c) = \dfrac{m_{r}}{\sqrt{.0.001999}} \approx 22.366m_{r}$ \\ \end{tabular}

    \item As $x \rightarrow c^{-}, \, m(x) \rightarrow \infty$

    \item If the object is traveling no faster than approximately $0.99995$ times the speed of light, then its observed mass will be no greater than $100m_{r}$.

    \end{enumerate}

    \item $k^{-1}(x) = \dfrac{x}{\sqrt{x^{2} - 4}}$

    \item \begin{enumerate}

    \item $y = \frac{1}{3}x^{3/2} - \sqrt{x} + \frac{2}{3}$. The point $\left(0,\frac{2}{3}\right)$ is when Fritzy's path crosses Chewbacca's path - in other words, where Fritzy catches Chewbacca.

    \item $y = \frac{1}{6}x^3+\frac{1}{2x} - \frac{2}{3}$. Using the techniques from Chapter \ref{Rationals}, we find as $x \rightarrow 0^{+}$, $y \rightarrow \infty$ which means, in this case, Fritzy's pursuit never ends; he never catches Chewbacca. This makes sense since Chewbacca has a head start and is running faster than Fritzy.

    \begin{center}

    \begin{tabular}{cc}

    \includegraphics[width=2in]{./FurtherGraphics/PURSUIT01.jpg} & \hspace{1in} \includegraphics[width=2in]{./FurtherGraphics/PURSUIT02.jpg} \\

    $y = \frac{1}{3}x^{3/2} - \sqrt{x} + \frac{2}{3}$ & \hspace{1in} $y = \frac{1}{6}x^3+\frac{1}{2x} - \frac{2}{3}$ \\

    \end{tabular}

    \end{center}

    \end{enumerate}

    \end{enumerate}

    \closegraphsfile