\hfill \thepage} %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \begin{document} \subsection{MATC 1200, Calculus for Business and Social Sciences} \subsection{Lab 3: Week of April 13, 2009} \subsection{Secant and Tangent Lines\protect\bigskip} This lab reinforces the idea of the tangent line to the graph of a function at a point. \ In particular, you'll explore how the slope of the tangent line is \emph{defined} in terms of slopes of secant lines.\vspace{0.1in} \begin{enumerate} \item \textsf{Using the slope of secants to approximate the slope of the tangent at a point. }\ Recall from lecture that the slope of the tangent line at $x=a$ \ (or $f^{\text{ }\prime }(a)$) is defined to be\medskip \[ f^{\text{ }\prime }(a)=\lim_{h\rightarrow 0}\frac{f\left( a+h\right) -f\left( a\right) }{h}\text{, \ \ if the limit exists.} \]% \medskip That is, the derivative is the limit of the difference quotient where the difference quotient gives us the slope of the secant line between $% (a,$ $f(a))$ and $(a+h,$ $f(a+h))$.\medskip \[ \text{Difference Quotient }=\frac{f\left( a+h\right) -f\left( a\right) }{h}% \text{, where }h\neq 0\text{.} \]% \medskip \begin{enumerate} \item Here is the graph of $f\left( x\right) =x^{2}-9x+24$.\medskip \FRAME{% dtbpFX}{2.7233in}{2.7233in}{0pt}{}{}{Plot}{\special{language "Scientific Word";type "MAPLEPLOT";width 2.7233in;height 2.7233in;depth 0pt;display "USEDEF";plot_snapshots TRUE;mustRecompute FALSE;lastEngine "MuPAD";xmin "0";xmax "9";xviewmin "0";xviewmax "9";yviewmin "0";yviewmax "24";viewset"XY";rangeset"X";plottype 4;plotticks 1;num-x-ticks 10;num-y-ticks 20;axesFont "Times New Roman,12,0000000000,useDefault,normal";numpoints 100;plotstyle "patch";axesstyle "normal";axestips FALSE;gridLines TRUE;xis \TEXUX{x};var1name \TEXUX{$x$};function \TEXUX{$x^{2}-9x+24$};linecolor "black";linestyle 1;pointstyle "point";linethickness 1;lineAttributes "Solid";var1range "0,9";num-x-gridlines 100;curveColor "[flat::RGB:0000000000]";curveStyle "Line";VCamFile 'KHWKN403.xvz';valid_file "T";tempfilename 'KHRLZ00D.wmf';tempfile-properties "XPR";}}\medskip We want to explore the idea of the tangent line to the graph of $f(x)$ at $% x=3$. \ Draw the point $(3,f(3))$ on the graph and sketch the tangent line to the graph at $x=3$.\medskip \item Use \textsf{SN} to \textbf{evaluate} $f(x)$ at the following values.\medskip \begin{tabular}{|c|c|c|c|c|c|} \hline $x$ & $3$ & $4$ & $5$ & $6$ & $7$ \\ \hline $f(x)$ & \hspace{1in} & \hspace{1in} & \hspace{1in} & \hspace{1in} & \hspace{% 1in} \\ \hline \end{tabular}% \medskip \item Draw the following four secant lines on the graph of $f(x)$:\vspace{% 0.1in} \begin{enumerate} \item The secant line between $(3,f(3))$ and $(7,f(7))$\medskip \item The secant line between $(3,f(3))$ and $(6,f(6))$\medskip \item The secant line between $(3,f(3))$ and $(5,f(5))$\medskip \item The secant line between $(3,f(3))$ and $(4,f(4))$\medskip \end{enumerate} Notice that the closer the points get together, the secant lines make a better approximation to the tangent line and we cannot draw a secant line between $(3,f(3))$ and $(3,f(3))$ because we need two distinct points.% \vspace{0.1in} \item Now compute the slopes of the 4 given secant lines. \ Show your work and use \textsf{SN} to assist in the calculations if needed.\vspace{0.1in} \begin{tabular}{|c|c|} \hline Secant line between & Slope \\ \hline $(3,f(3))$ and $(7,f(7))$ & \hspace{1in} \\ \hline $(3,f(3))$ and $(6,f(6))$ & \\ \hline $(3,f(3))$ and $(5,f(5))$ & \\ \hline $(3,f(3))$ and $(4,f(4))$ & \\ \hline \end{tabular}% \vspace{0.1in}\vspace{0.1in}\vspace{0.1in} \item We know that the difference quotient also gives us the slope of a secant line. \ Notice that% \[ \text{Difference Quotient }=\frac{f\left( 3+h\right) -f\left( 3\right) }{h}% \text{, where }h\neq 0\text{.} \] gives us the slope of the secant line between $(3,f(3))$ and $(3+h,f(3+h))$.% \vspace{0.1in} Now express the difference quotient in terms of the function $f\left( x\right) =x^{2}-9x+24$ for $x=3$. \textbf{\ Do not simplify - yet.}\vspace{% 0.1in} \item By hand, simplify the difference quotient for $f(x)=x^{2}-9x+24$ at $% x=3$. \ Show all of your steps.\vspace{0.1in} \item Now use \textsf{SN} to simplify your difference quotient from part (e) and write down the result. \ Note: \ Your answers from part (f) and part (g) should agree!\vspace{0.1in} \item The difference quotient you found corresponds to the slope of the secant line between $(3,f(3))$ and $(3+h,f(3+h))$. To compare to each of the secant line slopes you found in part (d), you need to find the corresponding $"h"$ which represents the (directed) distance between the two $x-$% coordinates. \ Complete the table below. \ Notice the first calculation has been done for you. \[ \begin{tabular}{|c|c|} \hline $(3,f(3)$ and $(3+h,f(3+h))$ & $h$ \\ \hline $(3,f(3))$ and $(7,f(7))$ & $7-3=4$ \\ \hline $(3,f(3))$ and $(6,f(6))$ & \\ \hline $(3,f(3))$ and $(5,f(5))$ & \\ \hline $(3,f(3))$ and $(4,f(4))$ & \\ \hline \end{tabular}% \]% \vspace{0.1in} \item Use \textsf{SN} (if needed) to \textbf{evaluate} the difference quotient for the values of $h$ that you found and the other decreasing values of $h$ given in the table in the solution file.\vspace{0.1in}% \[ \begin{tabular}{|l|l|l|} \hline $(3,f(3)$ and $(3+h,f(3+h))$ & $h$ & $\dfrac{f(3+h)-f(3)}{h}$ \\ \hline\hline $(3,f(3))$ and $(7,f(7)$ & & \hspace{1in} \\ \hline $(3,f(3))$ and $(6,f(6))$ & & \\ \hline $(3,f(3))$ and $(5,f(5))$ & & \\ \hline $(3,f(3))$ and $(4,f(4))$ & & \\ \hline $(3,f(3))$ and $(3.5,f(3.5))$ & $0.5$ & \\ \hline $(3,f(3))$ and $(3.1,f(3.1))$ & $0.1$ & \\ \hline $(3,f(3))$ and $(3.01,f(3.01))$ & $0.01$ & \\ \hline $(3,f(3))$ and $(3.001,f(3.001))$ & $0.001$ & \\ \hline $(3,f(3))$ and $(3.0001,f(3.001))$ & $0.00001$ & \\ \hline \end{tabular}% \]% $\medskip $ \item Check to see that the first 4 difference quotients agree with the slopes you found in part (d). \ From the table, what is a reasonable guess for the limit of the difference quotients as $h\rightarrow 0$?\vspace{0.1in} \item From the table, what is a reasonable guess for $f^{\,\prime }(3)$? \ Or, equivalently, what is a reasonable guess for the slope of the tangent line to $f(x)$ at $x=3$?\vspace{0.1in} \item Label the slope of the tangent line on your graph of $f(x)$.\vspace{% 0.1in} \end{enumerate} \item \textsf{The Limit Definition of the Derivative. \ }Let's look at how \textsf{SN} can help us to find the derivative using the formal limit definition. \ Recall that the slope of the tangent line or $g^{\prime }(x)$ is defined to be\vspace{0.1in}% \[ g^{\prime }(x)=\lim_{h\rightarrow 0}\frac{g\left( x+h\right) -g\left( x\right) }{h}\text{, \ \ if the limit exists.} \]% \vspace{0.1in}That is, the derivative is the limit of the difference quotient where the difference quotient gives us the slope of the secant line between $(x,$ $g(x))$ and $(x+h,$ $g(x+h))$.\vspace{0.1in}% \[ \text{Difference Quotient }=\frac{g\left( x+h\right) -g\left( x\right) }{h}% \text{, where }h\neq 0\text{.} \]% \vspace{0.1in} \begin{enumerate} \item Here is the graph of a function $g(x)=x^{3}-2x+1$.\FRAME{dtbpFX}{% 2.8963in}{1.9303in}{0pt}{}{}{Plot}{\special{language "Scientific Word";type "MAPLEPLOT";width 2.8963in;height 1.9303in;depth 0pt;display "USEDEF";plot_snapshots TRUE;mustRecompute FALSE;lastEngine "MuPAD";xmin "-3";xmax "3";xviewmin "-3";xviewmax "3";yviewmin "-10";yviewmax "10";viewset"XY";rangeset"X";plottype 4;axesFont "Times New Roman,12,0000000000,useDefault,normal";numpoints 100;plotstyle "patch";axesstyle "normal";axestips FALSE;xis \TEXUX{x};var1name \TEXUX{$x$};function \TEXUX{$x^{3}-2x+1$};linecolor "black";linestyle 1;pointstyle "point";linethickness 1;lineAttributes "Solid";var1range "-3,3";num-x-gridlines 100;curveColor "[flat::RGB:0000000000]";curveStyle "Line";VCamFile 'KHRL360E.xvz';valid_file "T";tempfilename 'KHRL360B.wmf';tempfile-properties "XPR";}} \item Sketch the tangent line to $g(x)$ at $x=-2$. \ Then sketch the tangent line to $g(x)$ at $x=1.5$.$\medskip $ \item To find the slope of each tangent line, it would be tedious to repeat the process in question 1 for each value of $x=a$. \ For that reason, we will do things more generally. \ Write down the difference quotient specific to $g(x)=x^{3}-2x+1$.\vspace{0.1in} \item Simplify your different quotient expression by hand. \ Use \textsf{SN} to expand the $(x+h)^{3}$ term in your difference quotient if needed.\vspace{% 0.1in} \item Now \textbf{simplify }the difference quotient expression in part (c) using \textsf{SN}. \ This should agree with your answer in part (d)!\vspace{% 0.1in} \item \textbf{By hand}, find the limit as $h\rightarrow 0$ of the simplified difference quotient.\vspace{0.1in} \item Using parts (c)-(f), what is $g^{\prime }(x)$?\vspace{0.1in} \item \textbf{By hand,} evaluate $g^{\prime }(-2)$ and $g^{\prime }(1.5)$.% \vspace{0.1in} \item Label the two tangent lines with the appropriate slopes and check that they make sense.\medskip \item Now just consider the tangent line to the graph at $x=-2$. \ What is the equation of the tangent line to the graph of $g(x)$ when $x=-2$? \ Give your final answer in slope-intercept form.\vspace{0.1in} \item Graph $g(x)$ and your tangent line together in \textsf{SN}. \ Use the $% xy$ window $[-3,3]\times \lbrack -10,10]$.\medskip \item We say that a function is differentiable at $x=a$ if $g^{\prime }(a)$ exists. \ Informally, this means the graph of the function is locally linear at $x=a$ or if you zoom into the graph at $x=a$ the graph looks like a non-vertical line. \ Now double click on your graph of $g(x)$ with the tangent line and select the large mountain icon. \ Note that by clicking and dragging you can zoom in on any part of your graph. \ (You can use the small mountain icon to zoom out.). \ Zoom in on your graph around the point given by $x=-2$. \ Stop when you cannot distinguish between your tangent line and the graph of $g(x)$. \ Record your final $xy$ viewing window below. \end{enumerate} \end{enumerate} \end{document} %%%%%%%%%%%%%%%%%%%%%% End /document/lab3Post.tex %%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%% Start /document/KHRLZ00D.wmf %%%%%%%%%%%%%%%%%%%% WwlqZB@@@@@@@xhBFj@zC@@@@@@[VH@@I@@@Cl_B@@`A@pA@@@@@@P@@@@p@A`@@E@@@@l`@@@@ @@T@@@@@CB\rCs|@E@@@@[@@@X@@vK@@@PB@A@`@@@@`@@x}_B@@@@~_@@xI@@D@@@@t R@@@@A@@@@pG@@@`@@@@`~B@@@@@@@@@@@@@@A@@@@mD@@@\@@@@@B@@@@@@@@@@@D@@@@tR@A 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