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diff --git a/Master/texmf-dist/doc/support/latexindent/success/sampleBEFORE.tex b/Master/texmf-dist/doc/support/latexindent/success/sampleBEFORE.tex index bf62bb9ee1f..ae704bd4bc1 100644 --- a/Master/texmf-dist/doc/support/latexindent/success/sampleBEFORE.tex +++ b/Master/texmf-dist/doc/support/latexindent/success/sampleBEFORE.tex @@ -25,7 +25,7 @@ a \emph{decreasing} function. Similarly, if $a>0$ then $g$ is \emph{concave up} and if $a<0$ then $g$ is \emph{concave down}. Graphical representations of these statements are given in \cref{poly:fig:linquad}. - + \begin{figure}[!htb] \setlength{\figurewidth}{.2\textwidth} \begin{subfigure}{\figurewidth} @@ -94,18 +94,18 @@ \caption{Typical graphs of linear and quadratic functions.} \label{poly:fig:linquad} \end{figure} - + Let's look a little more closely at the formulas for $f$ and $g$ in \cref{poly:eq:linquad}. Note that the \emph{degree} of $f$ is $1$ since the highest power of $x$ that is present in the formula for $f(x)$ is $1$. Similarly, the degree of $g$ is $2$ since the highest power of $x$ that is present in the formula for $g(x)$ is $2$. - + In this section we will build upon our knowledge of these elementary functions. In particular, we will generalize the functions $f$ and $g$ to a function $p$ that has any degree that we wish. - + %=================================== % Author: Hughes % Date: March 2012 @@ -161,7 +161,7 @@ \end{subproblem} \end{problem} \end{essentialskills} - + \subsection*{Power functions with positive exponents} The study of polynomials will rely upon a good knowledge of power functions| you may reasonably ask, what is a power function? @@ -171,17 +171,17 @@ f(x) = a_n x^n \] where $n$ can be any real number. - + Note that for this section we will only be concerned with the case when $n$ is a positive integer. \end{pccdefinition} - + You may find assurance in the fact that you are already very comfortable with power functions that have $n=1$ (linear) and $n=2$ (quadratic). Let's explore some power functions that you might not be so familiar with. As you read \cref{poly:ex:oddpow,poly:ex:evenpow}, try and spot as many patterns and similarities as you can. - + %=================================== % Author: Hughes % Date: March 2012 @@ -199,12 +199,12 @@ the long-run behavior of each of the functions is the same, and in particular \begin{align*} f(x)\rightarrow\infty & \text{ as } x\rightarrow\infty \\ - \mathllap{\text{and }} f(x)\rightarrow-\infty & \text{ as } x\rightarrow-\infty + \mathllap{\text{and }} f(x)\rightarrow-\infty & \text{ as } x\rightarrow-\infty \end{align*} The same results hold for $g$ and $h$. \end{pccsolution} \end{pccexample} - + \begin{figure}[!htb] \begin{minipage}{.45\textwidth} \begin{tikzpicture} @@ -254,7 +254,7 @@ \label{poly:fig:evenpow} \end{minipage}% \end{figure} - + %=================================== % Author: Hughes % Date: March 2012 @@ -271,12 +271,12 @@ of each of the functions is the same, and in particular \begin{align*} F(x)\rightarrow\infty & \text{ as } x\rightarrow\infty \\ - \mathllap{\text{and }} F(x)\rightarrow\infty & \text{ as } x\rightarrow-\infty + \mathllap{\text{and }} F(x)\rightarrow\infty & \text{ as } x\rightarrow-\infty \end{align*} The same result holds for $G$ and $H$. \end{pccsolution} \end{pccexample} - + \begin{doyouunderstand} \begin{problem} Repeat \cref{poly:ex:oddpow,poly:ex:evenpow} using (respectively) @@ -285,7 +285,7 @@ \begin{shortsolution} The functions $f$, $g$, and $h$ have domain $(-\infty,\infty)$ and are graphed below. - + \begin{tikzpicture} \begin{axis}[ framed, @@ -303,11 +303,11 @@ \legend{$f$,$g$,$h$} \end{axis} \end{tikzpicture} - + Note that \begin{align*} f(x)\rightarrow-\infty & \text{ as } x\rightarrow\infty \\ - \mathllap{\text{and }} f(x)\rightarrow\infty & \text{ as } x\rightarrow-\infty + \mathllap{\text{and }} f(x)\rightarrow\infty & \text{ as } x\rightarrow-\infty \end{align*} The same is true for $g$ and $h$. \end{shortsolution} @@ -317,7 +317,7 @@ \begin{shortsolution} The functions $F$, $G$, and $H$ have domain $(-\infty,\infty)$ and are graphed below. - + \begin{tikzpicture} \begin{axis}[ framed, @@ -335,18 +335,18 @@ \legend{$F$,$G$,$H$} \end{axis} \end{tikzpicture} - + Note that \begin{align*} F(x)\rightarrow-\infty & \text{ as } x\rightarrow\infty \\ - \mathllap{\text{and }} F(x)\rightarrow-\infty & \text{ as } x\rightarrow-\infty + \mathllap{\text{and }} F(x)\rightarrow-\infty & \text{ as } x\rightarrow-\infty \end{align*} The same is true for $G$ and $H$. \end{shortsolution} \end{subproblem} \end{problem} \end{doyouunderstand} - + \subsection*{Polynomial functions} Now that we have a little more familiarity with power functions, we can define polynomial functions. Provided that you were comfortable @@ -357,7 +357,7 @@ and quadratic functions. Once you've studied the examples and problems in this section, you'll hopefully agree that polynomial functions are remarkably predictable. - + %=================================== % Author: Hughes % Date: May 2011 @@ -376,11 +376,11 @@ \end{itemize} In particular, we call $a_n$ the \emph{leading} coefficient, and $a_nx^n$ the \emph{leading term}. - + Note that if a polynomial is given in factored form, then the degree can be found by counting the number of linear factors. \end{pccdefinition} - + %=================================== % Author: Hughes % Date: March 2012 @@ -416,7 +416,7 @@ \end{enumerate} \end{pccsolution} \end{pccexample} - + %=================================== % Author: Hughes % Date: March 2012 @@ -440,7 +440,7 @@ shape and long-run behavior to the functions described in \cref{poly:ex:oddpow}. \end{itemize} \end{pccexample} - + %=================================== % Author: Hughes % Date: May 2011 @@ -536,7 +536,7 @@ \caption{Graphs to illustrate typical curves of polynomial functions.} \label{poly:fig:typical} \end{figure} - + %=================================== % Author: Hughes % Date: March 2012 @@ -550,7 +550,7 @@ to guide you. \begin{shortsolution} $a_1<0$: - + \begin{tikzpicture} \begin{axis}[ framed, @@ -563,9 +563,9 @@ \addplot expression[domain=-10:8]{-(x+2)}; \end{axis} \end{tikzpicture} - + $a_2<0$ - + \begin{tikzpicture} \begin{axis}[ framed, @@ -578,9 +578,9 @@ \addplot expression[domain=-4:4]{-(x^2-6)}; \end{axis} \end{tikzpicture} - + $a_3<0$ - + \begin{tikzpicture} \begin{axis}[ framed, @@ -593,9 +593,9 @@ \addplot expression[domain=-7.5:7.5]{-0.05*(x+6)*x*(x-6)}; \end{axis} \end{tikzpicture} - + $a_4<0$ - + \begin{tikzpicture} \begin{axis}[ framed, @@ -608,9 +608,9 @@ \addplot expression[domain=-2.35:5.35,samples=100]{-0.2*(x-5)*x*(x-3)*(x+2)}; \end{axis} \end{tikzpicture} - + $a_5<0$ - + \begin{tikzpicture} \begin{axis}[ framed, @@ -626,11 +626,11 @@ \end{shortsolution} \end{problem} \end{doyouunderstand} - + \fixthis{poly: Need a more basic example here- it can have a similar format to the multiple zeros example, but just keep it simple; it should be halfway between the 2 examples surrounding it} - + %=================================== % Author: Hughes % Date: May 2011 @@ -642,24 +642,24 @@ \begin{align*} p(x) & =(x-3)^2(x+4)^2 \\ q(x) & =x(x+2)^2(x-1)^2(x-3) \\ - r(x) & =x(x-3)^3(x+1)^2 + r(x) & =x(x-3)^3(x+1)^2 \end{align*} Find the degree of $p$, $q$, and $r$, and decide if the functions bounce off or cut through the horizontal axis at each of their zeros. \begin{pccsolution} The degree of $p$ is 4. Referring to \cref{poly:fig:bouncep}, the curve bounces off the horizontal axis at both zeros, $3$ and $4$. - + The degree of $q$ is 6. Referring to \cref{poly:fig:bounceq}, the curve bounces off the horizontal axis at $-2$ and $1$, and cuts through the horizontal axis at $0$ and $3$. - + The degree of $r$ is 6. Referring to \cref{poly:fig:bouncer}, the curve bounces off the horizontal axis at $-1$, and cuts through the horizontal axis at $0$ and at $3$, although is flattened immediately to the left and right of $3$. \end{pccsolution} \end{pccexample} - + \setlength{\figurewidth}{0.25\textwidth} \begin{figure}[!htb] \begin{subfigure}{\figurewidth} @@ -712,7 +712,7 @@ \caption{} \label{poly:fig:moremultiple} \end{figure} - + \begin{pccdefinition}[Multiple zeros]\label{poly:def:multzero} Let $p$ be a polynomial that has a repeated linear factor $(x-a)^n$. Then we say that $p$ has a multiple zero at $a$ of multiplicity $n$ and @@ -724,7 +724,7 @@ \end{itemize} If $n=1$, then we say that $p$ has a \emph{simple} zero at $a$. \end{pccdefinition} - + %=================================== % Author: Hughes % Date: May 2011 @@ -786,7 +786,7 @@ Let's check if the formula we have written satisfies this requirement \begin{align*} p(1) & = (1)(4)(2)(-1) \\ - & = -8 + & = -8 \end{align*} which is clearly not correct| it is close though. We can correct this by multiplying $p$ by a constant $k$; so let's assume that @@ -807,7 +807,7 @@ evaluate $p(2)$ \begin{align*} p(2) & =k(4)^2(-1) \\ - & =-16k + & =-16k \end{align*} We solve the equation $4=-8k$ and obtain $k=-\frac{1}{4}$ and conclude that the formula for $q(x)$ is @@ -817,8 +817,8 @@ \end{enumerate} \end{pccsolution} \end{pccexample} - - + + \fixthis{Chris: need sketching polynomial problems} \begin{pccspecialcomment}[Steps to follow when sketching polynomial functions] \begin{steps} @@ -865,12 +865,12 @@ \item We draw the details we have obtained so far on \cref{poly:fig:simplecubicp1}. Given that the curve of $p$ looks like the curve of $x^3$ in the long-run, we are able to complete a sketch of the graph of $p$ in \cref{poly:fig:simplecubicp2}. - + Note that we can not find the coordinates of the local minimums, local maximums, and inflection points| for the moment we make reasonable guesses as to where these points are (you'll find how to do this in calculus). \end{steps} - + \begin{figure}[!htbp] \begin{subfigure}{.45\textwidth} \begin{tikzpicture} @@ -909,7 +909,7 @@ \end{figure} \end{pccsolution} \end{pccexample} - + %=================================== % Author: Hughes % Date: May 2012 @@ -934,7 +934,7 @@ \item We mark the details we have found so far on \cref{poly:fig:degree5p1}. Given that the curve of $q$ looks like the curve of $-x^5$ in the long-run, we can complete \cref{poly:fig:degree5p2}. \end{steps} - + \begin{figure}[!htbp] \begin{subfigure}{.45\textwidth} \begin{tikzpicture} @@ -973,7 +973,7 @@ \end{figure} \end{pccsolution} \end{pccexample} - + %=================================== % Author: Hughes % Date: May 2012 @@ -1000,7 +1000,7 @@ the curve of $r$ looks like the curve of $x^6$ in the long-run, we complete the graph of $r$ in \cref{poly:fig:degree6p2}. \end{steps} - + \begin{figure}[!htbp] \begin{subfigure}{.45\textwidth} \begin{tikzpicture} @@ -1038,7 +1038,7 @@ \end{figure} \end{pccsolution} \end{pccexample} - + %=================================== % Author: Hughes % Date: March 2012 @@ -1058,7 +1058,7 @@ $x$ represents the length of a side, and $V(x)$ represents the volume of the box, we necessarily require both values to be positive; we illustrate the part of the curve that applies to this problem using a solid line. - + \begin{figure}[!htb] \centering \begin{tikzpicture} @@ -1080,21 +1080,21 @@ \caption{$y=V(x)$} \label{poly:fig:opentoppedbox} \end{figure} - + According to \cref{poly:fig:opentoppedbox}, the maximum volume of such a box is approximately $\unit[4000]{cm^2}$, and we achieve it using a base of length approximately $\unit[20]{cm}$. Since the base is square and each sheet of cardboard is $\unit[1200]{cm^2}$, we conclude that the dimensions of each box are $\unit[20]{cm}\times\unit[20]{cm}\times\unit[30]{cm}$. \end{pccsolution} \end{pccexample} - + \subsection*{Complex zeros} There has been a pattern to all of the examples that we have seen so far| the degree of the polynomial has dictated the number of \emph{real} zeros that the polynomial has. For example, the function $p$ in \cref{poly:ex:simplecubic} has degree $3$, and $p$ has $3$ real zeros; the function $q$ in \cref{poly:ex:degree5} has degree $5$ and $q$ has $5$ real zeros. - + You may wonder if this result can be generalized| does every polynomial that has degree $n$ have $n$ real zeros? Before we tackle the general result, let's consider an example that may help motivate it. @@ -1113,7 +1113,7 @@ x^2+1=0 \end{equation} The solutions to \cref{poly:eq:complx} are $\pm i$. - + We conclude that $c$ has $3$ zeros: $0$ and $\pm i$; we note that \emph{not all of them are real}. \end{pccexample} @@ -1143,7 +1143,7 @@ We begin by factoring $p$ \begin{align*} p(x) & =x^4-2x^3+5x^2 \\ - & =x^2(x^2-2x+5) + & =x^2(x^2-2x+5) \end{align*} We note that $0$ is a zero of $p$ with multiplicity $2$. The other zeros of $p$ can be found by solving the equation @@ -1154,7 +1154,7 @@ \begin{align*} x & =\frac{2\pm\sqrt{(-2)^2}-20}{2(1)} \\ & =\frac{2\pm\sqrt{-16}}{2} \\ - & =1\pm 2i + & =1\pm 2i \end{align*} We conclude that $p$ has $4$ zeros: $0$ (multiplicity $2$), and $1\pm 2i$ (simple). \end{pccsolution} @@ -1169,13 +1169,13 @@ We know that the zeros of a polynomial can be found by analyzing the linear factors. We are given the zeros, and have to work backwards to find the linear factors. - + We begin by assuming that $p$ has the form \begin{align*} p(x) & =(x-(2-i\sqrt{2}))(x-(2+i\sqrt{2})) \\ & =x^2-x(2+i\sqrt{2})-x(2-i\sqrt{2})+(2-i\sqrt{2})(2+i\sqrt{2}) \\ & =x^2-4x+(4-2i^2) \\ - & =x^2-4x+6 + & =x^2-4x+6 \end{align*} We conclude that a possible formula for a polynomial function, $p$, that has zeros at $2\pm i\sqrt{2}$ is @@ -1235,8 +1235,8 @@ \end{enumerate} \end{shortsolution} \end{problem} - - + + \begin{figure}[!htb] \setlength{\figurewidth}{0.3\textwidth} \begin{subfigure}{\figurewidth} @@ -1287,10 +1287,10 @@ \caption{} \label{poly:fig:findformula} \end{figure} - - - - + + + + \begin{exercises} %=================================== % Author: Hughes @@ -1459,14 +1459,14 @@ \begin{align*} p(x) & = (x-1)(x+2)(x-3) \\ m(x) & = -(x-1)(x+2)(x-3) \\ - n(x) & = (x-1)(x+2)(x-3)(x+1)(x+4) + n(x) & = (x-1)(x+2)(x-3)(x+1)(x+4) \end{align*} Note that for our present purposes we are not concerned with the vertical scale of the graphs. \begin{subproblem} Identify both on the graph {\em and} algebraically, the zeros of each polynomial. \begin{shortsolution} $y=p(x)$ is shown below. - + \begin{tikzpicture} \begin{axis}[ xmin=-5,xmax=5, @@ -1477,9 +1477,9 @@ \addplot[soldot] coordinates{(-2,0)(1,0)(3,0)}; \end{axis} \end{tikzpicture} - + $y=m(x)$ is shown below. - + \begin{tikzpicture} \begin{axis}[ xmin=-5,xmax=5, @@ -1490,9 +1490,9 @@ \addplot[soldot] coordinates{(-2,0)(1,0)(3,0)}; \end{axis} \end{tikzpicture} - + $y=n(x)$ is shown below. - + \begin{tikzpicture} \begin{axis}[ xmin=-5,xmax=5, @@ -1503,7 +1503,7 @@ \addplot[soldot] coordinates{(-4,0)(-2,0)(-1,0)(1,0)(3,0)}; \end{axis} \end{tikzpicture} - + The zeros of $p$ are $-2$, $1$, and $3$; the zeros of $m$ are $-2$, $1$, and $3$; the zeros of $n$ are $-4$, $-2$, $-1$, and $3$. \end{shortsolution} @@ -1520,7 +1520,7 @@ \end{shortsolution} \end{subproblem} \end{problem} - + \begin{figure}[!htb] \begin{widepage} \setlength{\figurewidth}{0.3\textwidth} @@ -1773,7 +1773,7 @@ \end{shortsolution} \end{subproblem} \end{problem} - + %=================================== % Author: Hughes % Date: May 2011 @@ -1792,7 +1792,7 @@ $\dd\lim_{x\rightarrow\infty}s(x)=\infty$, \end{shortsolution} \end{problem} - + %=================================== % Author: Hughes % Date: May 2011 @@ -1917,7 +1917,7 @@ is positive. \begin{shortsolution} Assuming that $a_3>0$: - + \begin{tikzpicture} \begin{axis}[ xmin=-10,xmax=10, @@ -1937,7 +1937,7 @@ is negative. \begin{shortsolution} Assuming that $a_3<0$: - + \begin{tikzpicture} \begin{axis}[ xmin=-10,xmax=10, @@ -1961,7 +1961,7 @@ coefficient of $q$ is positive. Hint: only one of the zeros is simple. \begin{shortsolution} Assuming that $a_4>0$ there are $2$ different options: - + \begin{tikzpicture} \begin{axis}[ xmin=-10,xmax=10, @@ -1982,7 +1982,7 @@ coefficient of $q$ is negative. \begin{shortsolution} Assuming that $a_4<0$ there are $2$ different options: - + \begin{tikzpicture} \begin{axis}[ xmin=-10,xmax=10, @@ -2194,7 +2194,7 @@ \end{subproblem} \end{multicols} \end{problem} - + %=================================== % Author: Hughes % Date: July 2012 @@ -2238,7 +2238,7 @@ \end{shortsolution} \end{subproblem} \end{problem} - + %=================================== % Author: Hughes % Date: July 2012 @@ -2290,7 +2290,7 @@ \end{subproblem} \end{multicols} \end{problem} - + %=================================== % Author: Hughes % Date: May 2011 @@ -2298,7 +2298,7 @@ \begin{problem}[Find a formula from a table]\label{poly:prob:findformula} \Crefrange{poly:tab:findformulap}{poly:tab:findformulas} show values of polynomial functions, $p$, $q$, $r$, and $s$. - + \begin{table}[!htb] \centering \begin{widepage} @@ -2382,7 +2382,7 @@ \end{subtable} \end{widepage} \end{table} - + \begin{subproblem} Assuming that all of the zeros of $p$ are shown (in \cref{poly:tab:findformulap}), how many zeros does $p$ have? \begin{shortsolution} @@ -2433,7 +2433,7 @@ \end{subproblem} \end{problem} \end{exercises} - + \section{Rational functions} \subsection*{Power functions with negative exponents} The study of rational functions will rely upon a good knowledge @@ -2455,21 +2455,21 @@ the long-run behavior of each of the functions is the same, and in particular \begin{align*} f(x)\rightarrow 0 & \text{ as } x\rightarrow\infty \\ - \mathllap{\text{and }} f(x)\rightarrow 0 & \text{ as } x\rightarrow-\infty + \mathllap{\text{and }} f(x)\rightarrow 0 & \text{ as } x\rightarrow-\infty \end{align*} The same results hold for $g$ and $h$. Note also that each of the functions has a \emph{vertical asymptote} at $0$. We see that \begin{align*} f(x)\rightarrow -\infty & \text{ as } x\rightarrow 0^- \\ - \mathllap{\text{and }} f(x)\rightarrow \infty & \text{ as } x\rightarrow 0^+ + \mathllap{\text{and }} f(x)\rightarrow \infty & \text{ as } x\rightarrow 0^+ \end{align*} The same results hold for $g$ and $h$. - + The curve of a function that has a vertical asymptote is necessarily separated into \emph{branches}| each of the functions $f$, $g$, and $h$ have $2$ branches. \end{pccsolution} \end{pccexample} - + \begin{figure}[!htb] \begin{minipage}{.45\textwidth} \begin{tikzpicture} @@ -2525,8 +2525,8 @@ \label{rat:fig:evenpow} \end{minipage}% \end{figure} - - + + %=================================== % Author: Hughes % Date: May 2011 @@ -2543,7 +2543,7 @@ the long-run behavior of each of the functions is the same, and in particular \begin{align*} F(x)\rightarrow 0 & \text{ as } x\rightarrow\infty \\ - \mathllap{\text{and }} f(x)\rightarrow 0 & \text{ as } x\rightarrow-\infty + \mathllap{\text{and }} f(x)\rightarrow 0 & \text{ as } x\rightarrow-\infty \end{align*} As in \cref{rat:ex:oddpow}, $F$ has a horizontal asymptote that has equation $y=0$. @@ -2551,7 +2551,7 @@ has a \emph{vertical asymptote} at $0$. We see that \begin{align*} F(x)\rightarrow \infty & \text{ as } x\rightarrow 0^- \\ - \mathllap{\text{and }} F(x)\rightarrow \infty & \text{ as } x\rightarrow 0^+ + \mathllap{\text{and }} F(x)\rightarrow \infty & \text{ as } x\rightarrow 0^+ \end{align*} The same results hold for $G$ and $H$. Each of the functions $F$, $G$, and $H$ have $2$ branches. @@ -2569,7 +2569,7 @@ \begin{shortsolution} The functions $k$, $m$, and $n$ have domain $(-\infty,0)\cup (0,\infty)$, and are graphed below. - + \begin{tikzpicture} \begin{axis}[ framed, @@ -2590,14 +2590,14 @@ \legend{$k$,$m$,$n$} \end{axis} \end{tikzpicture} - + Note that \begin{align*} k(x)\rightarrow 0 & \text{ as } x\rightarrow\infty \\ \mathllap{\text{and }} k(x)\rightarrow 0 & \text{ as } x\rightarrow-\infty \\ \intertext{and also} k(x)\rightarrow \infty & \text{ as } x\rightarrow 0^- \\ - \mathllap{\text{and }} k(x)\rightarrow -\infty & \text{ as } x\rightarrow 0^+ + \mathllap{\text{and }} k(x)\rightarrow -\infty & \text{ as } x\rightarrow 0^+ \end{align*} The same are true for $m$ and $n$. \end{shortsolution} @@ -2607,7 +2607,7 @@ \begin{shortsolution} The functions $K$, $M$, and $N$ have domain $(-\infty,0)\cup (0,\infty)$, and are graphed below. - + \begin{tikzpicture} \begin{axis}[ framed, @@ -2628,21 +2628,21 @@ \legend{$K$,$M$,$N$} \end{axis} \end{tikzpicture} - + Note that \begin{align*} K(x)\rightarrow 0 & \text{ as } x\rightarrow\infty \\ \mathllap{\text{and }} K(x)\rightarrow 0 & \text{ as } x\rightarrow-\infty \\ \intertext{and also} K(x)\rightarrow -\infty & \text{ as } x\rightarrow 0^- \\ - \mathllap{\text{and }} K(x)\rightarrow -\infty & \text{ as } x\rightarrow 0^+ + \mathllap{\text{and }} K(x)\rightarrow -\infty & \text{ as } x\rightarrow 0^+ \end{align*} The same are true for $M$ and $N$. \end{shortsolution} \end{subproblem} \end{problem} \end{doyouunderstand} - + \subsection*{Rational functions} \begin{pccdefinition}[Rational functions]\label{rat:def:function} Rational functions have the form @@ -2650,7 +2650,7 @@ r(x) = \frac{p(x)}{q(x)} \] where both $p$ and $q$ are polynomials. - + Note that \begin{itemize} \item the domain or $r$ will be all real numbers, except those that @@ -2658,13 +2658,13 @@ \item the zeros of $r$ are the zeros of $p$, i.e the real numbers that make the \emph{numerator}, $p(x)$, equal to $0$. \end{itemize} - + \Cref{rat:ex:oddpow,rat:ex:evenpow} are particularly important because $r$ will behave like $\frac{1}{x}$, or $\frac{1}{x^2}$ around its vertical asymptotes, depending on the power that the relevant term is raised to| we will demonstrate this in what follows. \end{pccdefinition} - + %=================================== % Author: Hughes % Date: May 2011 @@ -2699,7 +2699,7 @@ \end{enumerate} \end{pccsolution} \end{pccexample} - + %=================================== % Author: Hughes % Date: May 2011 @@ -2779,7 +2779,7 @@ \caption{} \label{rat:fig:whichiswhich} \end{figure} - + \begin{pccsolution} Let's start with the function $r$. Note that domain of $r$ is $(-\infty,3)\cup(0,3)$, so we search for a function that has a vertical asymptote at $3$. There @@ -2787,18 +2787,18 @@ but note that the function in \cref{rat:fig:which3} also has a vertical asymptote at $-2$ which is not consistent with the formula for $r(x)$. Therefore, $y=r(x)$ is graphed in \cref{rat:fig:which2}. - + The function $q$ has domain $(-\infty,-5)\cup(-5,\infty)$, so we search for a function that has a vertical asymptote at $-5$. The only candidate is the curve shown in \cref{rat:fig:which1}; note that the curve also goes through $(2,0)$, which is consistent with the formula for $q(x)$, since $q(2)=0$, i.e $q$ has a zero at $2$. - + The function $k$ has domain $(-\infty,-2)\cup(-2,3)\cup(3,\infty)$, and has vertical asymptotes at $-2$ and $3$. This is consistent with the graph in \cref{rat:fig:which3} (and is the only curve that has $3$ branches). - + We note that each function behaves like $\frac{1}{x}$ around its vertical asymptotes, because each linear factor in each denominator is raised to the power $1$; if (for example) the definition of $r$ was instead @@ -2809,7 +2809,7 @@ the graph of $r$ would be very different. We will deal with these cases in the examples that follow. \end{pccsolution} \end{pccexample} - + %=================================== % Author: Hughes % Date: May 2011 @@ -2827,7 +2827,7 @@ so we are not surprised to see that each curve has $3$ branches. We also note that the numerator of each function is the same, which tells us that each function has only $1$ zero at $2$. - + The functions $g$ and $h$ are different from those that we have considered previously, because they have a repeated factor in the denominator. Notice in particular the way that the functions behave around their asymptotes: @@ -2910,7 +2910,7 @@ \caption{} \label{rat:fig:repfactd} \end{figure} - + \Cref{rat:def:function} says that the zeros of the rational function $r$ that has formula $r(x)=\frac{p(x)}{q(x)}$ are the zeros of $p$. Let's explore this a little more. @@ -2929,9 +2929,9 @@ x+5=0 \] The zero of $\alpha$ is $-5$. - + Similarly, we may solve $9-x=0$ to find the zero of $\beta$, which is clearly $9$. - + The zeros of $\gamma$ satisfy the equation \[ 17x^2-10=0 @@ -2943,7 +2943,7 @@ The zeros of $\gamma$ are $\pm\frac{10}{17}$. \end{pccsolution} \end{pccexample} - + \subsection*{Long-run behavior} Our focus so far has been on the behavior of rational functions around their \emph{vertical} asymptotes. In fact, rational functions also @@ -2966,7 +2966,7 @@ \end{pccdefinition} We will concentrate on functions that have horizontal asymptotes until we reach \cref{rat:sec:oblique}. - + %=================================== % Author: Hughes % Date: May 2012 @@ -2979,7 +2979,7 @@ and obtained the curves shown in \cref{rat:fig:horizasymp}. Kebede decides to test his knowledgeable friend \pccname{Oscar}, and asks him to match the formulas to the graphs. - + \begin{figure}[!htb] \setlength{\figurewidth}{0.3\textwidth} \begin{subfigure}{\figurewidth} @@ -3050,7 +3050,7 @@ \caption{Horizontal asymptotes} \label{rat:fig:horizasymp} \end{figure} - + Oscar notices that each function has a vertical asymptote at $3$ and a zero at $-1$. The main thing that catches Oscar's eye is that each function has a different coefficient in the numerator, and that each curve has a different horizontal asymptote. @@ -3064,14 +3064,14 @@ that since the degree of the numerator and the degree of the denominator is the same for each of the functions $r$, $s$, and $t$, the horizontal asymptote will be determined by evaluating the ratio of their leading coefficients. - + Oscar therefore says that $r$ should have a horizontal asymptote $y=\frac{1}{1}=1$, $s$ should have a horizontal asymptote $y=\frac{2}{1}=2$, and $t$ should have a horizontal asymptote $y=\frac{3}{1}=3$. Kebede is able to finish the problem from here, and says that $r$ is shown in \cref{rat:fig:horizasymp2}, $s$ is shown in \cref{rat:fig:horizasymp1}, and $t$ is shown in \cref{rat:fig:horizasymp3}. \end{pccexample} - + %=================================== % Author: Hughes % Date: May 2012 @@ -3080,7 +3080,7 @@ \pccname{Xiao} and \pccname{Dwayne} saw \cref{rat:ex:horizasymp} but are a little confused about horizontal asymptotes. What does it mean to say that a function $r$ has a horizontal asymptote? - + They decide to explore the concept by constructing a table of values for the rational functions $R$ and $S$ that have formulas \[ @@ -3119,13 +3119,13 @@ \end{tabular} \end{minipage} \end{table} - + Xiao and Dwayne study \cref{rat:tab:plusinfty,rat:tab:minusinfty} and decide that the functions $R$ and $S$ never actually touch their horizontal asymptotes, but they do get infinitely close. They also feel as if they have a better understanding of what it means to study the behavior of a function as $x\rightarrow\pm\infty$. \end{pccexample} - + %=================================== % Author: Hughes % Date: May 2011 @@ -3142,7 +3142,7 @@ \] We also notice that the numerators of each function are quite similar| indeed, each function has a zero at $2$, but how does each function behave around their zero? - + Using \cref{rat:fig:repfactn} to guide us, we note that \begin{itemize} \item $f$ has a horizontal intercept $(2,0)$, but the curve of @@ -3152,7 +3152,7 @@ \item $h$ has a horizontal intercept $(2,0)$, and the curve of $h$ also cuts the axis, but appears flattened as it does so. \end{itemize} - + We can further enrich our study by discussing the long-run behavior of each function. Using the tools of \cref{rat:def:longrun}, we can deduce that \begin{itemize} @@ -3162,7 +3162,7 @@ study this more in \cref{rat:sec:oblique}). \end{itemize} \end{pccexample} - + \begin{figure}[!htb] \setlength{\figurewidth}{0.3\textwidth} \begin{subfigure}{\figurewidth} @@ -3235,7 +3235,7 @@ \caption{} \label{rat:fig:repfactn} \end{figure} - + \subsection*{Holes} Rational functions have a vertical asymptote at $a$ if the denominator is $0$ at $a$. What happens if the numerator is $0$ at the same place? In this case, we say that the rational @@ -3250,7 +3250,7 @@ $(a,r(a))$ on the curve $y=r(x)$ by using a hollow circle, $\circ$. \end{pccdefinition} - + %=================================== % Author: Hughes % Date: March 2012 @@ -3262,12 +3262,12 @@ \] in their calculators, and can not decide if the correct graph is \cref{rat:fig:hole} or \cref{rat:fig:hole1}. - + Luckily for them, Oscar is nearby, and can help them settle the debate. Oscar demonstrates that \begin{align*} r(x) & =\frac{(x+3)(x-2)}{(x-2)} \\ - & = x+3 + & = x+3 \end{align*} but only when $x\ne 2$, because the function is undefined at $2$. Oscar says that this necessarily means that the domain or $r$ is @@ -3275,7 +3275,7 @@ (-\infty,2)\cup(2,\infty) \] and that $r$ must have a hole at $2$. - + Mohammed and Sue are very grateful for the clarification, and conclude that the graph of $r$ is shown in \cref{rat:fig:hole1}. \begin{figure}[!htb] @@ -3319,7 +3319,7 @@ \end{minipage}% \end{figure} \end{pccexample} - + %=================================== % Author: Hughes % Date: May 2011 @@ -3333,12 +3333,12 @@ make the denominator equal to $0$. Notice that \begin{align*} f(x) & = \frac{x(x+3)}{x(x-4)} \\ - & = \frac{x+3}{x-4} + & = \frac{x+3}{x-4} \end{align*} provided that $x\ne 0$. Since $0$ makes the numerator and the denominator 0 at the same time, we say that $f$ has a hole at $(0,-\nicefrac{3}{4})$. Note that this necessarily means that $f$ does not have a vertical intercept. - + We also note $f$ has a vertical asymptote at $4$; the function is graphed in \cref{rat:fig:holeex}. \begin{figure}[!htb] \centering @@ -3362,9 +3362,9 @@ \label{rat:fig:holeex} \end{figure} \end{pccexample} - - - + + + %=================================== % Author: Hughes % Date: March 2012 @@ -3374,7 +3374,7 @@ if a rational function has a vertical asymptote, then it can not possibly have local minimums and maximums, nor can it have global minimums and maximums. - + Trang says this statement is not always true. She plots the functions $f$ and $g$ that have formulas \[ @@ -3383,7 +3383,7 @@ in \cref{rat:fig:minmax1,rat:fig:minmax2} and shows them to Seamus. On seeing the graphs, Seamus quickly corrects himself, and says that $f$ has a local (and global) maximum of $2$ at $0$, and that $g$ has a local (and global) minimum of $-2$ at $0$. - + \begin{figure}[!htb] \begin{minipage}{.45\textwidth} \begin{tikzpicture}[/pgf/declare function={f=-32*(x-1)*(x+1)/(( x-2)^2*(x+2)^2);}] @@ -3427,19 +3427,19 @@ \label{rat:fig:minmax2} \end{minipage}% \end{figure} - + Seamus also notes that (in its domain) the function $f$ is always concave down, and that (in its domain) the function $g$ is always concave up. Furthermore, Trang observes that each function behaves like $\frac{1}{x^2}$ around each of its vertical asymptotes, because each linear factor in the denominator is raised to the power $2$. - + \pccname{Oscar} stops by and reminds both students about the long-run behavior; according to \cref{rat:def:longrun} since the degree of the denominator is greater than the degree of the numerator (in both functions), each function has a horizontal asymptote at $y=0$. \end{pccexample} - - + + \investigation*{} %=================================== % Author: Pettit/Hughes @@ -3448,12 +3448,12 @@ \begin{problem}[The spaghetti incident] The same Queen from \vref{exp:prob:queenschessboard} has recovered from the rice experiments, and has called her loyal jester for another challenge. - + The jester has an $11-$inch piece of uncooked spaghetti that he puts on a table; he uses a book to cover $\unit[1]{inch}$ of it so that $\unit[10]{inches}$ hang over the edge. The jester then produces a box of $\unit{mg}$ weights that can be hung from the spaghetti. - + The jester says it will take $\unit[y]{mg}$ to break the spaghetti when hung $\unit[x]{inches}$ from the edge, according to the rule $y=\frac{100}{x}$. \begin{margintable} @@ -3537,7 +3537,7 @@ note that this necessarily means that you will not be able to plot all of the points. \begin{shortsolution} The graph of $y=\frac{100}{x}$ is shown below. - + \begin{tikzpicture} \begin{axis}[ framed, @@ -3567,9 +3567,9 @@ \end{subproblem} The Queen looks forward to more food-related investigations from her jester. \end{problem} - - - + + + %=================================== % Author: Adams (Hughes) % Date: March 2012 @@ -3593,17 +3593,17 @@ Paying off the debt in $2$ years, we use \begin{align*} M & = \frac{2000\cdot 0.015}{1-(1+0.015)^{-24}} \\ - & \approx 99.85 + & \approx 99.85 \end{align*} The monthly payments are \$99.85. - + Paying off the debt in $1$ year, we use \begin{align*} M & = \frac{2000\cdot 0.015}{1-(1+0.015)^{-12}} \\ - & \approx 183.36 + & \approx 183.36 \end{align*} The monthly payments are \$183.36 - + In the $2$-year model we would pay a total of $\$99.85\cdot 12=\$2396.40$. In the $1$-year model we would pay a total of $\$183.36\cdot 12=\$2200.32$. We would therefore save $\$196.08$ if we went with the $1$-year model instead of the $2$-year model. @@ -3619,20 +3619,20 @@ For the $20$-year loan we use \begin{align*} M & = \frac{300000\cdot \frac{0.052}{12}}{1-\left( 1+\frac{0.052}{12} \right)^{-12\cdot 20}} \\ - & \approx 2013.16 + & \approx 2013.16 \end{align*} The monthly payments are \$2013.16. - + For the $30$-year loan we use \begin{align*} M & = \frac{300000\cdot \frac{0.052}{12}}{1-\left( 1+\frac{0.052}{12} \right)^{-12\cdot 30}} \\ - & \approx 1647.33 + & \approx 1647.33 \end{align*} The monthly payments are \$1647.33. - + The total amount paid during the $20$-year loan is $\$2013.16\cdot 12\cdot 20=\$483,158.40$. The total amount paid during the $30$-year loan is $\$1647.33\cdot 12\cdot 30=\$593,038.80$. - + Recommendation: if you can afford the payments, choose the $20$-year loan. \end{shortsolution} \end{subproblem} @@ -3662,7 +3662,7 @@ This means that the monthly payments will be calculated using \begin{align*} M & = \frac{14000\cdot \frac{0.04}{12}}{1-\left( 1+\frac{0.04}{12} \right)^{-12\cdot 5}} \\ - & \approx 257.83 + & \approx 257.83 \end{align*} The monthly payments will be $\$257.83$. The total amount paid will be $\$257.83\cdot 5\cdot 12=\$15,469.80$, of which $\$1469.80$ is interest. @@ -3670,7 +3670,7 @@ This means that the monthly payments will be calculated using \begin{align*} M & = \frac{12000\cdot \frac{0.08}{12}}{1-\left( 1+\frac{0.08}{12} \right)^{-12\cdot 5}} \\ - & \approx 243.32 + & \approx 243.32 \end{align*} The monthly payments will be $\$243.32$. The total amount paid will be $\$243.32\cdot 5\cdot 12 =\$14,599.20$, of which $\$2599.2$ is @@ -3681,7 +3681,7 @@ \end{shortsolution} \end{subproblem} \end{problem} - + \begin{exercises} %=================================== % Author: Hughes @@ -3783,7 +3783,7 @@ $\begin{aligned}[t] r(0) & =\frac{(0-2)(0+3)}{(0+5)(0-7)} \\ & =\frac{-6}{-35} \\ - & =\frac{6}{35} + & =\frac{6}{35} \end{aligned}$ \end{shortsolution} \end{subproblem} @@ -3793,7 +3793,7 @@ $\begin{aligned}[t] r(1) & =\frac{(1-2)(1+3)}{(1+5)(1-7)} \\ & =\frac{-4}{-36} \\ - & =\frac{1}{9} + & =\frac{1}{9} \end{aligned}$ \end{shortsolution} \end{subproblem} @@ -3803,7 +3803,7 @@ $\begin{aligned}[t] r(2) & =\frac{(2-2)(2+3)}{(2+5)(2-7)} \\ & = \frac{0}{-50} \\ - & =0 + & =0 \end{aligned}$ \end{shortsolution} \end{subproblem} @@ -3813,7 +3813,7 @@ $\begin{aligned}[t] r(4) & =\frac{(4-2)(4+3)}{(4+5)(4-7)} \\ & =\frac{14}{-27} \\ - & =-\frac{14}{27} + & =-\frac{14}{27} \end{aligned}$ \end{shortsolution} \end{subproblem} @@ -3822,9 +3822,9 @@ \begin{shortsolution} $\begin{aligned}[t] r(7) & =\frac{(7-2)(7+3)}{(7+5)(7-7)} \\ - & =\frac{50}{0} + & =\frac{50}{0} \end{aligned}$ - + $r(7)$ is undefined. \end{shortsolution} \end{subproblem} @@ -3834,7 +3834,7 @@ $\begin{aligned}[t] r(-3) & =\frac{(-3-2)(-3+3)}{(-3+5)(-3-7)} \\ & =\frac{0}{-20} \\ - & =0 + & =0 \end{aligned}$ \end{shortsolution} \end{subproblem} @@ -3843,9 +3843,9 @@ \begin{shortsolution} $\begin{aligned}[t] r(-5) & =\frac{(-5-2)(-5+3)}{(-5+5)(-5-7)} \\ - & =\frac{14}{0} + & =\frac{14}{0} \end{aligned}$ - + $r(-5)$ is undefined. \end{shortsolution} \end{subproblem} @@ -3856,7 +3856,7 @@ r\left( \frac{1}{2} \right) & = \frac{\left( \frac{1}{2}-2 \right)\left( \frac{1}{2}+3 \right)}{\left( \frac{1}{2}+5 \right)\left( \frac{1}{2}-7 \right)} \\ & =\frac{-\frac{3}{2}\cdot\frac{7}{2}}{\frac{11}{2}\left( -\frac{13}{2} \right)} \\ & =\frac{-\frac{21}{4}}{-\frac{143}{4}} \\ - & =\frac{37}{143} + & =\frac{37}{143} \end{aligned}$ \end{shortsolution} \end{subproblem} @@ -3908,7 +3908,7 @@ \end{subproblem} \end{multicols} \end{problem} - + %=================================== % Author: Hughes % Date: May 2011 @@ -3925,7 +3925,7 @@ \end{itemize} \end{shortsolution} \end{problem} - + \begin{figure}[!htb] \begin{widepage} \setlength{\figurewidth}{0.3\textwidth} @@ -3997,7 +3997,7 @@ \label{rat:fig:findformula} \end{widepage} \end{figure} - + %=================================== % Author: Hughes % Date: May 2011 @@ -4034,7 +4034,7 @@ \end{shortsolution} \end{subproblem} \end{problem} - + %=================================== % Author: Hughes % Date: May 2011 @@ -4099,7 +4099,7 @@ \end{subproblem} \end{multicols} \end{problem} - + %=================================== % Author: Hughes % Date: May 2012 @@ -4181,7 +4181,7 @@ \end{subproblem} \end{multicols} \end{problem} - + %=================================== % Author: Hughes % Date: May 2011 @@ -4212,7 +4212,7 @@ \end{shortsolution} \end{subproblem} \end{problem} - + %=================================== % Author: Hughes % Date: Feb 2011 @@ -4265,8 +4265,8 @@ \end{shortsolution} \end{subproblem} \end{problem} - - + + %=================================== % Author: Hughes % Date: May 2011 @@ -4309,7 +4309,7 @@ Sketch a graph of $r$. \begin{shortsolution} A graph of $r$ is shown below. - + \begin{tikzpicture} \begin{axis}[ framed, @@ -4329,7 +4329,7 @@ \end{shortsolution} \end{subproblem} \end{problem} - + %=================================== % Author: Hughes % Date: July 2012 @@ -4367,8 +4367,8 @@ \end{subproblem} \end{multicols} \end{problem} - - + + %=================================== % Author: Hughes % Date: July 2012 @@ -4419,8 +4419,8 @@ \end{subproblem} \end{multicols} \end{problem} - - + + %=================================== % Author: Hughes % Date: May 2011 @@ -4428,7 +4428,7 @@ \begin{problem}[Find a formula from a table]\label{rat:prob:findformula} \Crefrange{rat:tab:findformular}{rat:tab:findformulau} show values of rational functions $r$, $q$, $s$, and $t$. Assume that any values marked with an X are undefined. - + \begin{table}[!htb] \begin{widepage} \centering @@ -4525,7 +4525,7 @@ r(-4) & = \frac{-4-3}{-4+2} \\ & = \frac{7}{2} \\ \end{aligned}$ - + $r(-3)=\ldots$ etc \end{shortsolution} \end{subproblem} @@ -4541,9 +4541,9 @@ \begin{shortsolution} $\begin{aligned}[t] s(-4) & =\frac{-4+2}{(-4-3)(-4+1)} \\ - & =-\frac{2}{21} + & =-\frac{2}{21} \end{aligned}$ - + $s(-3)=\ldots$ etc \end{shortsolution} \end{subproblem} @@ -4563,13 +4563,13 @@ \end{subproblem} \end{problem} \end{exercises} - + \section{Graphing rational functions (horizontal asymptotes)} \reformatstepslist{R} % the steps list should be R1, R2, \ldots We studied rational functions in the previous section, but were not asked to graph them; in this section we will demonstrate the steps to be followed in order to sketch graphs of the functions. - + Remember from \vref{rat:def:function} that rational functions have the form \[ @@ -4583,7 +4583,7 @@ in this section \emph{will have a horizontal asymptote} (see \vref{rat:def:longrun}). The cases in which the degree of $p$ is greater than the degree of $q$ is covered in the next section. - + Before we begin, it is important to remember the following: \begin{itemize} \item Our sketches will give a good representation of the overall @@ -4612,10 +4612,10 @@ find the exact coordinates of local minimums, local maximums, and points of inflection. \end{pccspecialcomment} - + The examples that follow show how \crefrange{rat:step:first}{rat:step:last} can be applied to a variety of different rational functions. - + %=================================== % Author: Hughes % Date: May 2012 @@ -4641,7 +4641,7 @@ \end{steps} \end{pccsolution} \end{pccexample} - + \begin{figure}[!htbp] \begin{subfigure}{.45\textwidth} \begin{tikzpicture} @@ -4678,14 +4678,14 @@ \end{subfigure}% \caption{$y=\dfrac{1}{x-2}$} \end{figure} - + The function $r$ in \cref{rat:ex:1overxminus2p2} has a horizontal asymptote which has equation $y=0$. This asymptote lies on the horizontal axis, and you might (understandably) find it hard to distinguish between the two lines (\cref{rat:fig:1overxminus2p2}). When faced with such a situation, it is perfectly acceptable to draw the horizontal axis as a dashed line| just make sure to label it correctly. We will demonstrate this in the next example. - + %=================================== % Author: Hughes % Date: May 2012 @@ -4711,12 +4711,12 @@ is, because we know what the overall shape will be. Let's compute $v(2)$ \begin{align*} v(2) & =\dfrac{10}{2} \\ - & = 5 + & = 5 \end{align*} We therefore mark the point $(2,5)$ on \cref{rat:fig:1overxp2}, and then complete the sketch using the details we found in the previous steps. \end{steps} - + \begin{figure}[!htbp] \begin{subfigure}{.45\textwidth} \begin{tikzpicture} @@ -4760,7 +4760,7 @@ \end{figure} \end{pccsolution} \end{pccexample} - + %=================================== % Author: Hughes % Date: May 2012 @@ -4778,7 +4778,7 @@ \begin{align*} u(x) & =\frac{-4(x^2-9)}{x^2-8x+15} \\ & =\frac{-4(x+3)(x-3)}{(x-5)(x-3)} \\ - & =\frac{-4(x+3)}{x-5} + & =\frac{-4(x+3)}{x-5} \end{align*} provided that $x\ne 3$. Therefore $u$ has a vertical asymptote at $5$ and a hole at $3$. The curve of $u$ has $2$ branches. @@ -4788,7 +4788,7 @@ \item We put the details we have obtained so far on \cref{rat:fig:1overxminus2p1}. Notice that there is only one way to complete the graph, which we have done in \cref{rat:fig:1overxminus2p2}. \end{steps} - + \begin{figure}[!htbp] \begin{subfigure}{.45\textwidth} \begin{tikzpicture} @@ -4833,12 +4833,12 @@ \end{figure} \end{pccsolution} \end{pccexample} - + \Cref{rat:ex:1overxminus2p2,rat:ex:1overxp1,rat:ex:asympandholep1} have focused on functions that only have one vertical asymptote; the remaining examples in this section concern functions that have more than one vertical asymptote. We will demonstrate that \crefrange{rat:step:first}{rat:step:last} still apply. - + %=================================== % Author: Hughes % Date: May 2012 @@ -4862,20 +4862,20 @@ of the numerator and denominator, we say that $w$ has a horizontal asymptote with equation $y=\frac{2}{1}=2$. \item We put the details we have obtained so far on \cref{rat:fig:sketchtwoasymptp1}. - + The function $w$ is a little more complicated than the functions that we have considered in the previous examples because the curve has $3$ branches. When graphing such functions, it is generally a good idea to start with the branch for which you have the most information| in this case, that is the \emph{middle} branch on the interval $(-5,4)$. - + Once we have drawn the middle branch, there is only one way to complete the graph (because of our observations about the behavior of $w$ around its vertical asymptotes), which we have done in \cref{rat:fig:sketchtwoasymptp2}. \end{steps} \end{pccsolution} \end{pccexample} - + \begin{figure}[!htbp] \begin{subfigure}{.45\textwidth} \begin{tikzpicture} @@ -4921,12 +4921,12 @@ \end{subfigure}% \caption{$y=\dfrac{2(x+3)(x-5)}{(x+5)(x-4)}$} \end{figure} - + The rational functions that we have considered so far have had simple factors in the denominator; each function has behaved like $\frac{1}{x}$ around each of its vertical asymptotes. \Cref{rat:ex:2asympnozeros,rat:ex:2squaredasymp} consider functions that have a repeated factor in the denominator. - + %=================================== % Author: Hughes % Date: May 2012 @@ -4949,17 +4949,17 @@ denominator of $f$ is $2$. $f$ has a horizontal asymptote with equation $y=0$. \item We put the details we have obtained so far on \cref{rat:fig:2asympnozerosp1}. - + The function $f$ is similar to the function $w$ that we considered in \cref{rat:ex:sketchtwoasymp}| it has two vertical asymptotes and $3$ branches, but in contrast to $w$ it does not have any zeros. - + We sketch $f$ in \cref{rat:fig:2asympnozerosp2}, using the middle branch as our guide because we have the most information about the function on the interval $(-5,4)$. - + Once we have drawn the middle branch, there is only one way to complete the graph because of our observations about the behavior of $f$ around its vertical asymptotes (it behaves like $\frac{1}{x}$), which we have done in \cref{rat:fig:2asympnozerosp2}. - + Note that we are not yet able to find the local minimum of $f$ algebraically on the interval $(-5,4)$, so we make a reasonable guess as to where it is| we can be confident that it is above the horizontal axis since $f$ has no zeros. You may think that this is unsatisfactory, but once we have the tools of calculus, we will @@ -4967,7 +4967,7 @@ \end{steps} \end{pccsolution} \end{pccexample} - + \begin{figure}[!htbp] \begin{subfigure}{.45\textwidth} \begin{tikzpicture} @@ -5011,7 +5011,7 @@ \end{subfigure}% \caption{$y=\dfrac{100}{(x+5)(x-4)^2}$} \end{figure} - + %=================================== % Author: Hughes % Date: May 2012 @@ -5037,19 +5037,19 @@ \cref{rat:fig:2squaredasymp1}. The function $g$ is similar to the functions we considered in \cref{rat:ex:sketchtwoasymp,rat:ex:2asympnozeros} because it has $2$ vertical asymptotes and $3$ branches. - + We sketch $g$ using the middle branch as our guide because we have the most information about $g$ on the interval $(-3,5)$. Note that there is no other way to draw this branch without introducing other zeros which $g$ does not have. - + Once we have drawn the middle branch, there is only one way to complete the graph because of our observations about the behavior of $g$ around its vertical asymptotes| it behaves like $\frac{1}{x^2}$. - + \end{steps} \end{pccsolution} \end{pccexample} - + \begin{figure}[!htbp] \begin{subfigure}{.45\textwidth} \begin{tikzpicture} @@ -5093,14 +5093,14 @@ \end{subfigure}% \caption{$y=\dfrac{50(2-x)}{(x+3)^2(x-5)^2}$} \end{figure} - + Each of the rational functions that we have considered so far has had either a \emph{simple} zero, or no zeros at all. Remember from our work on polynomial functions, and particularly \vref{poly:def:multzero}, that a \emph{repeated} zero corresponds to the curve of the function behaving differently at the zero when compared to how the curve behaves at a simple zero. \Cref{rat:ex:doublezero} details a function that has a non-simple zero. - + %=================================== % Author: Hughes % Date: June 2012 @@ -5127,20 +5127,20 @@ \cref{rat:fig:doublezerop1}. The function $h$ is different from the functions that we have considered in previous examples because of the multiplicity of the zero at $3$. - + We sketch $h$ using the middle branch as our guide because we have the most information about $h$ on the interval $(-4,6)$. Note that there is no other way to draw this branch without introducing other zeros which $h$ does not have| also note how the curve bounces off the horizontal axis at $3$. - + Once we have drawn the middle branch, there is only one way to complete the graph because of our observations about the behavior of $h$ around its vertical asymptotes| it behaves like $\frac{1}{x}$. - + \end{steps} \end{pccsolution} \end{pccexample} - + \begin{figure}[!htbp] \begin{subfigure}{.45\textwidth} \begin{tikzpicture} @@ -5199,7 +5199,7 @@ at $b$, and a vertical asymptote at $c$. Furthermore, these functions behave like $\frac{1}{x}$ around their vertical asymptote, and the curve of each function will have $2$ branches. - + Katie has been working with $3$ functions that have the form given in \cref{rat:eq:deducecurve}, and has followed \crefrange{rat:step:first}{rat:step:penultimate}; her results are shown in \cref{rat:fig:deducecurve}. There is just one @@ -5207,7 +5207,7 @@ Help Katie finish each graph by deducing the curve of each function. \begin{shortsolution} \Vref{rat:fig:deducecurve1} - + \begin{tikzpicture}[/pgf/declare function={f=3*(x+4)/(x+5);}] \begin{axis}[ xmin=-10,xmax=10, @@ -5222,9 +5222,9 @@ \addplot[pccplot] expression[domain=-4.76923:10,samples=50]{f}; \end{axis} \end{tikzpicture} - + \Vref{rat:fig:deducecurve2} - + \begin{tikzpicture}[/pgf/declare function={f=-3*(x-2)/(x-4);}] \begin{axis}[ xmin=-10,xmax=10, @@ -5239,9 +5239,9 @@ \addplot[pccplot] expression[domain=4.85714:10]{f}; \end{axis} \end{tikzpicture} - + \Vref{rat:fig:deducecurve4} - + \begin{tikzpicture}[/pgf/declare function={f=2*(x-6)/(x-4);}] \begin{axis}[ xmin=-10,xmax=10, @@ -5258,7 +5258,7 @@ \end{tikzpicture} \end{shortsolution} \end{problem} - + \begin{figure}[!htb] \begin{widepage} \setlength{\figurewidth}{0.3\textwidth} @@ -5316,7 +5316,7 @@ \label{rat:fig:deducecurve} \end{widepage} \end{figure} - + %=================================== % Author: Hughes % Date: June 2012 @@ -5331,13 +5331,13 @@ at $b$ and $c$, and vertical asymptotes at $d$ and $e$. Furthermore, these functions behave like $\frac{1}{x}$ around both vertical asymptotes, and the curve of the function will have $3$ branches. - + David has followed \crefrange{rat:step:first}{rat:step:penultimate} for $3$ separate functions, and drawn the results in \cref{rat:fig:deducehard}. Help David finish each graph by deducing the curve of each function. \begin{shortsolution} \Vref{rat:fig:deducehard1} - + \begin{tikzpicture}[/pgf/declare function={f=(x-6)*(x+3)/( (x-4)*(x+1));}] \begin{axis}[ xmin=-10,xmax=10, @@ -5354,9 +5354,9 @@ \addplot[pccplot] expression[domain=4.24276:10]{f}; \end{axis} \end{tikzpicture} - + \Vref{rat:fig:deducehard2} - + \begin{tikzpicture}[/pgf/declare function={f=3*(x-2)*(x+3)/( (x-6)*(x+5));}] \begin{axis}[ xmin=-10,xmax=10, @@ -5373,9 +5373,9 @@ \addplot[pccplot] expression[domain=7.34324:10]{f}; \end{axis} \end{tikzpicture} - + \Vref{rat:fig:deducehard3} - + \begin{tikzpicture}[/pgf/declare function={f=2*(x-7)*(x+3)/( (x+6)*(x-5));}] \begin{axis}[ xmin=-10,xmax=10, @@ -5392,10 +5392,10 @@ \addplot[pccplot] expression[domain=5.25586:10]{f}; \end{axis} \end{tikzpicture} - + \end{shortsolution} \end{problem} - + \begin{figure}[!htb] \begin{widepage} \setlength{\figurewidth}{0.3\textwidth} @@ -5470,7 +5470,7 @@ $y=\dfrac{4}{x+2}$ \begin{shortsolution} Vertical intercept: $(0,2)$; vertical asymptote: $x=-2$, horizontal asymptote: $y=0$. - + \begin{tikzpicture} \begin{axis}[ framed, @@ -5494,7 +5494,7 @@ Vertical intercept:$\left( 0,\frac{1}{9} \right)$; horizontal intercept: $\left( \frac{1}{2},0 \right)$; vertical asymptotes: $x=-3$, $x=3$, horizontal asymptote: $y=0$. - + \begin{tikzpicture} \begin{axis}[ framed, @@ -5519,7 +5519,7 @@ \begin{shortsolution} Vertical intercept $\left( 0,-\frac{3}{5} \right)$; horizontal intercept: $(-3,0)$; vertical asymptote: $x=5$; horizontal asymptote: $y=1$. - + \begin{tikzpicture} \begin{axis}[ framed, @@ -5544,7 +5544,7 @@ \begin{shortsolution} Vertical intercept: $(0,-3)$; horizontal intercept: $\left( -\frac{3}{2},0 \right)$; vertical asymptote: $x=\frac{1}{3}$, horizontal asymptote: $y=\frac{2}{3}$. - + \begin{tikzpicture}[/pgf/declare function={f=(2*x+3)/(3*x-1);}] \begin{axis}[ framed, @@ -5568,7 +5568,7 @@ Vertical intercept: $\left( 0,-\frac{4}{9} \right)$; horizontal intercepts: $(2,0)$, $(-2,0)$; vertical asymptotes: $x=-3$, $x=3$; horizontal asymptote: $y=-1$. - + \begin{tikzpicture}[/pgf/declare function={f=(4-x^2)/(x^2-9);}] \begin{axis}[ framed, @@ -5594,7 +5594,7 @@ Vertical intercept: $\left( 0,\frac{4}{5} \right)$; horizontal intercepts: $\left( -\frac{5}{4},0 \right)$, $\left( \frac{4}{3},0 \right)$; vertical asymptotes: $x=-\frac{5}{2}$, $x=5$; horizontal asymptote: $y=6$. - + \begin{tikzpicture}[/pgf/declare function={f=(4*x+5)*(3*x-4)/((2*x+5)*(x-5));}] \begin{axis}[ framed, @@ -5724,7 +5724,7 @@ R(x)= \begin{dcases} \frac{2}{x+3}, & x<-5 \\ - \frac{x-4}{x-10}, & x\geq -5 + \frac{x-4}{x-10}, & x\geq -5 \end{dcases} \] Evaluate each of the following. @@ -5762,7 +5762,7 @@ \end{subproblem} \end{problem} \end{exercises} - + \section{Graphing rational functions (oblique asymptotes)}\label{rat:sec:oblique} \begin{subproblem} $y=\dfrac{x^2+1}{x-4}$ @@ -5771,7 +5771,7 @@ \item $\left( 0,-\frac{1}{4} \right)$ \item Vertical asymptote: $x=4$. \item A graph of the function is shown below - + \begin{tikzpicture}[/pgf/declare function={f=(x^2+1)/(x-4);}] \begin{axis}[ framed, @@ -5798,7 +5798,7 @@ \item $(0,0)$, $(-3,0)$ \item Vertical asymptote: $x=5$, horizontal asymptote: none. \item A graph of the function is shown below - + \begin{tikzpicture}[/pgf/declare function={f=x^3*(x+3)/(x-5);}] \begin{axis}[ framed, |