\hfill \hfill } %} \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]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \geometry{left=0.5in,right=0.4in,top=0.5in,bottom=0.3in} \pagestyle{fancy} \lhead{\color{blue} \Large Math 207} \lfoot{\small \copyright $\;$ Hidegkuti, 2013} \cfoot{} \chead{\LARGE Extreme Value Theorems} \rfoot{\small Last revised: October 22, 2015} \rhead{\large page \thepage } \textwidth 7.5in \textheight 9.6in \setlength{\headheight}{29pt} \setlength{\parindent}{0pt} \begin{document} Definition: \ Let $f$ be a function with domain $D$. \ Then $f$ has a \textbf{relative maximum value} at a point $c$ if $f\left( x\right) \leq f\left( c\right) $ for all $x$ in $D$ lying in some open interval containing $c$. \ A function $f$ has a \textbf{relative minimum value} at a point $c$ if $f\left( x\right) \geq f\left( c\right) $ for all $x$ in $D$ lying in some open interval containing $c$.\bigskip \bigskip Theorem: (First Derivative Theorem for Local Extreme Values) If $f$ is has a relative maximum or minimum value at an interior point $c$ of its domain, and if $f^{\prime }$ is defined at $c$, then $f^{\prime }\left( c\right) =0$% .\bigskip Proof: \ Suppose that $f$ is differentiable at $c$ and $f$ has a local maximum at $c$. \ Then $f^{\prime }\left( c\right) =\lim\limits_{h\rightarrow 0}\dfrac{f\left( c+h\right) -f\left( c\right) }{h} $ exists and is a two-sided limit. \ For a sufficiently small positive value of $h$, $f\left( c+h\right) $ exists and $f\left( c+h\right) \leq f\left( c\right) $ since $f$ has a local maximum value at $c$. \ Then $f\left( c+h\right) -f\left( c\right) \leq 0$. \ Divide that by positive $h$ and get that $\dfrac{f\left( c+h\right) -f\left( c\right) }{h}\leq 0$ and so \begin{equation*} \lim\limits_{h\rightarrow 0^{+}}\dfrac{f\left( c+h\right) -f\left( c\right) }{h}\leq 0 \end{equation*}% Now let $h$ be a very small negative number. \ Then by the same argument, $% f\left( c+h\right) \leq f\left( c\right) $. \ Divide that by a negative $h$ and get that $\dfrac{f\left( c+h\right) -f\left( c\right) }{h}\geq 0$ and so \begin{equation*} \lim\limits_{h\rightarrow 0^{-}}\dfrac{f\left( c+h\right) -f\left( c\right) }{h}\geq 0 \end{equation*}% For the two-sided limit $f^{\prime }\left( c\right) $ to exists, we must have \begin{equation*} \lim\limits_{h\rightarrow 0^{-}}\dfrac{f\left( c+h\right) -f\left( c\right) }{h}=\lim\limits_{h\rightarrow 0^{+}}\dfrac{f\left( c+h\right) -f\left( c\right) }{h} \end{equation*}% Since one side is less than or equal to zero and the other is greater than or equal to zero, they both must be zero.\bigskip Definition: A \textbf{critical number} of a function $f$ is a number $c$ in its domain such that either $f^{\prime }\left( c\right) =0$ or $f^{\prime }\left( c\right) $ does not exist.\bigskip Theorem: (Fermat) \ If $f$ has a local maximum or minimum at $c$, then $c$ is a critical number of $f$.\bigskip Definition: \ Let $f$ be a function with domain $D$. \ Then $f$ has an \textbf{absolute maximum value} on $D$ at a point $c$ if $f\left( x\right) \leq f\left( c\right) $ for all $x$ in $D$ and an \textbf{absolute minimum value} on $D$ at a point $c$ if $f\left( x\right) \geq f\left( c\right) $ for all $x$ in $D$.\bigskip Theorem: (\textbf{Extreme Value Theorem}) If $f$ is continuous on a closed interval $\left[ a,b\right] $, then $f$ attains an absolute maximum value $% f\left( c\right) $ and an absolute minimum value $f\left( d\right) $ at some numbers $c$ and $d$ in $\left[ a,b\right] $.\bigskip While this theorem is very important and fundamental, its proof is difficult and so it will not be covered in this class.\bigskip \bigskip \textbf{Closed interval method: \ }To find absolute extrema of a continuous function $f$ on a closed interval $\left[ a,b\right] $. \qquad 1) \ Find the values of $f$ at the critical numbers of $f$ in $\left[ a,b\right] .$ \qquad 2) \ Find the values of $f$ at the endpoints of the interval. \qquad 3) \ The largest of the values from Steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value.\bigskip \bigskip \pagebreak Example: \ Find all absolute and relative extrema for the function \ $% f\left( x\right) =x^{3}+3x^{2}-24x+24$ \ defined\ on the closed interval $% \left[ -6,5\right] $.\bigskip Since this is a cubic polynomial with a positive leading coefficient, we have our initial expectations on end-behavior and one relative maximum, followed by a relative minimum. \ We first find these relative extrema. \ Since the function is dfferentiable everywhere, all critical numbers will occur where the derivative is zero. \begin{eqnarray*} f^{\prime }\left( x\right) &=&3x^{2}+6x-24=3\left( x+4\right) \left( x-2\right) \\ f^{\prime }\left( x\right) &=&0~~\implies ~~x_{1}=-4\text{ and }x_{2}=2 \end{eqnarray*}% Based on the graph of $f^{\prime }$, we conclude that $f^{\prime }$ changes sign from positive to negative at $x=-4$ and from negative to positive at $% x=2$. \ Thus $f$ has a relative maximum at $x=-4$ and a relative minimum at $% x=2$.\FRAME{dtbpFX}{2.0081in}{1.0032in}{0pt}{}{}{Plot}{\special{language "Scientific Word";type "MAPLEPLOT";width 2.0081in;height 1.0032in;depth 0pt;display "USEDEF";plot_snapshots TRUE;mustRecompute FALSE;lastEngine "MuPAD";xmin "-7.01";xmax "5.01";xviewmin "-7.01";xviewmax "5.01";yviewmin "-27.10310";yviewmax "81.10800";viewset"XY";rangeset"X";plottype 4;plotticks 1;num-x-ticks 12;numpoints 100;plotstyle "patch";axesstyle "normal";xis \TEXUX{x};yis \TEXUX{y};var1name \TEXUX{$x$};var2name \TEXUX{$y$};function \TEXUX{$3\left( x+4\right) \left( x-2\right) $};linecolor "black";linestyle 1;pointstyle "point";linethickness 2;lineAttributes "Solid";var1range "-7.01,5.01";num-x-gridlines 100;curveColor "[flat::RGB:0000000000]";curveStyle "Line";valid_file "T";tempfilename 'NWMKT706.wmf';tempfile-properties "XPR";}}To find the absolute extremas, we just need to compare the function values at the critical numbers, $-4$ and $% 2 $ and at the endpoints of the domain, $-6$ and $5$. We evaluate the function at these numbers and find that $\ f\left( -6\right) =60$,$\ \ f\left( -4\right) =104$,\ \ $\ f\left( 2\right) =-4$, \ and \ $f\left( 5\right) =104. $ \bigskip Based on this, $f$ has an absolute minimum at $x=-4$ and an absolute maximum at $x=-4$ and $x=5$. \FRAME{dtbpFX}{4.0145in}{2.0081in}{0pt}{}{}{Plot}{\special{language "Scientific Word";type "MAPLEPLOT";width 4.0145in;height 2.0081in;depth 0pt;display "USEDEF";plot_snapshots TRUE;mustRecompute FALSE;lastEngine "MuPAD";xmin "-7.01";xmax "7.01";xviewmin "-7.01";xviewmax "7.01";yviewmin "-30";yviewmax "120";viewset"XY";rangeset"X";plottype 4;plotticks 1;num-x-ticks 15;numpoints 100;plotstyle "patch";axesstyle "normal";xis \TEXUX{x};yis \TEXUX{y};var1name \TEXUX{$x$};var2name \TEXUX{$y$};function \TEXUX{$x^{3}+3x^{2}-24x+24$};linecolor "gray";linestyle 1;pointstyle "point";linethickness 1;lineAttributes "Solid";var1range "-7.01,7.01";num-x-gridlines 100;curveColor "[flat::RGB:0x00c0c0c0]";curveStyle "Line";function \TEXUX{$x^{3}+3x^{2}-24x+24$};linecolor "black";linestyle 1;pointstyle "point";linethickness 3;lineAttributes "Solid";var1range "-6,5";num-x-gridlines 100;curveColor "[flat::RGB:0000000000]";curveStyle "Line";rangeset"X";function \TEXUX{$3\left( x+4\right) \left( x-2\right) $};linecolor "gray";linestyle 2;pointstyle "point";linethickness 1;lineAttributes "Dash";var1range "-7.01,7.01";num-x-gridlines 100;curveColor "[flat::RGB:0x00c0c0c0]";curveStyle "Line";valid_file "T";tempfilename 'NWMKTG07.wmf';tempfile-properties "XPR";}}In summary: \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{tabular}{lllll} relative minimum: & $\left( 2,-4\right) $ & ~~~~~~~~~~ & absolute minimum: & $\left( 2,-4\right) $ \\ relative maximum: & $\left( -4,104\right) $ & & absolute maximum: & $\left( -4,104\right) $ and $\left( 5,104\right) $% \end{tabular} \bigskip \vspace{1.7in} {\small %TCIMACRO{\TeXButton{\small}{\small}}% %BeginExpansion \small% %EndExpansion } \href{https://teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html}{% For more documents like this, visit our page at\ https://teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail questions or comments to mhidegkuti@ccc.edu.} \end{document}