\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 Definitions, Theorems} \rfoot{\small Last revised: October 20, 2015} \rhead{\large page \thepage } \textwidth 7.5in \textheight 9.6in \setlength{\headheight}{29pt} \setlength{\parindent}{0pt} \begin{document} Theorem: \ (\textbf{The Intermediate Value Theorem}) \ If $f$ is continuous on a closed interval $\left[ a,b\right] $ and if $f\left( a\right) <0$ and $% f\left( b\right) >0$, then there exists $c$ in $\left( a,b\right) $ so that $% f\left( c\right) =0$.\bigskip Theorem: \ (\textbf{The Intermediate value Theorem for Continuous Functions}% ) \ If $f$ is continuous on a closed interval $\left[ a,b\right] $ and if $% y_{0}$ is any value between $f\left( a\right) $ and $f\left( b\right) $, then $y_{0}=f\left( c\right) $ for some $c$ in $\left[ a,b\right] $.\bigskip \bigskip \bigskip \bigskip \begin{center} {\Large Applications of the Intermediate Value Theorem\bigskip }\bigskip \bigskip \bigskip \end{center} \begin{enumerate} \item Prove that the function $f\left( x\right) =x^{5}-3x^{4}+8x^{3}-x-2$ \ has at least one zero in the interval $\left[ 0,1\right] $. \item Prove that the function $f\left( x\right) =6x^{4}+x^{3}-25x^{2}-4x+4$ \ has at least two zeroes in the interval $\left[ -1,1\right] $. \item Prove that all polynomials with an odd degree have at least one zero. \bigskip \end{enumerate} \bigskip \bigskip \bigskip \bigskip \begin{center} {\Large Solutions} {\Large \bigskip } \end{center} \begin{enumerate} \item $f$ is continuous on $\left[ 0,1\right] $, $f\left( 0\right) =-2$,\ and $f\left( 1\right) =3$. \ By the Intermediate Value Theorem, $f$ has a zero in $\left[ 0,1\right] $. \item $f$ is continuous on $\left[ -1,1\right] .$ \ \ $f\left( -1\right) =-12 $ \ and\ \ $f\left( 0\right) =4$. By the Intermediate Value Theorem, $f$ has a zero in $\left[ -1,0\right] .$ \ Also,\ $f\left( 0\right) =4$ and \ $% f\left( 1\right) =-18$. \ By the Intermediate Value Theorem, $f$ has a zero in $\left[ 0,1\right] .$ \item Suppose that $f$ is an odd degree polynomial with a positive leading coefficient. \ Then $\lim\limits_{x\rightarrow -\infty }f\left( x\right) =-\infty $ and $\lim\limits_{x\rightarrow -\infty }f\left( x\right) =-\infty $. \ Then there exist sufficiently a large negative value of $x$ (denoted by $a$) so that $f\left( a\right) $ is negative. \ Also, there exist sufficiently a large positive value of $x$ (denoted by $b$) so that $f\left( b\right) $ is positive. \ Since $f$ is a polynomial, it is continuous on $% %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion $ and thus on $\left[ a,b\right] $. \ By the Intermediate Value Theorem, $f$ must have a zero in the interval $\left[ a,b\right] $. \ The proof goes similarly for functions with negative leading coefficients.\bigskip \bigskip \end{enumerate} \pagebreak \begin{center} {\Large Maximum-Minimum Theorems\bigskip } \end{center} 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 \bigskip Closed interval method: \ To find absolute extrema of a continuous function $% f$ on a closed interval $\left[ a,b\right] $.\newline 1) \ Find the values of $f$ at the critical numbers of $f$ in $\left[ a,b% \right] .$\newline 2) \ Find the values of $f$ at the endpoints of the interval.\newline 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 \begin{center} {\Large The Mean Value Theorem\bigskip } \end{center} Theorem: (Rolle) Suppose that $f$ is continuous on $\left[ a,b\right] $ and differentiable on $\left( a,b\right) $. \ If $f\left( a\right) =f\left( b\right) ,$ then there exists $c$ in $\left( a,b\right) $ with $f^{\prime }\left( c\right) =0$.\bigskip Proof: \ Suppos that that $f$ is continuous on $\left[ a,b\right] $, differentiable on $\left( a,b\right) $, and $f\left( a\right) =f\left( b\right) $. \ By the extreme value theorem, $f$ has an absolute maximum \ and minimum on $\left[ a,b\right] $. \ If that maximum or minimum is at an interior point $c$, then $f^{\prime }\left( c\right) =0$. \ If both the maximum and minimum are at an endpoint, then $f\left( a\right) =f\left( b\right) $ means that the function is constant and so at all interior point $% c$ we have $f^{\prime }\left( c\right) =0$.\bigskip Theorem: (\textbf{Mean Value Theorem}) \ Suppose that $f$ is continuous on $% \left[ a,b\right] $ and differentiable on $\left( a,b\right) $. \ Then there exists $c$ in $\left( a,b\right) $ with $f^{\prime }\left( c\right) =\dfrac{% f\left( b\right) -f\left( a\right) }{b-a}$.\bigskip Proof: \ Suppose that $f$ is continuous on $\left[ a,b\right] $ and differentiable on $\left( a,b\right) $. \ Define $g$ a linear function to be the line connecting $f\left( a\right) $ and $f\left( b\right) $. \ The equation of this line is% \begin{eqnarray*} g\left( x\right) &=&m\left( x-a\right) +f\left( a\right) \text{ \ \ \ where }% m=\dfrac{f\left( b\right) -f\left( a\right) }{b-a}\text{ \ and so} \\ g\left( x\right) &=&\dfrac{f\left( b\right) -f\left( a\right) }{b-a}\left( x-a\right) +f\left( a\right) \end{eqnarray*}% Clearly $g$ is differentiable on $% %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion $. \ Now define the difference function $h$ between $f$ and $g$.% \begin{eqnarray*} h\left( x\right) &=&f\left( x\right) -g\left( x\right) =f\left( x\right) - \left[ \dfrac{f\left( b\right) -f\left( a\right) }{b-a}\left( x-a\right) +f\left( a\right) \right] \\ h\left( x\right) &=&f\left( x\right) -\dfrac{f\left( b\right) -f\left( a\right) }{b-a}\left( x-a\right) -f\left( a\right) \end{eqnarray*}% We will be able to apply Rolle's Theorem to this function. \ Since both $f$ and $g$ are continuous on $\left[ a,b\right] $ and differentiable on $\left( a,b\right) $, so is $h$. \ Furthermore, $h\left( a\right) =h\left( b\right) =0$.% \begin{eqnarray*} h\left( a\right) &=&f\left( a\right) -g\left( a\right) =f\left( a\right) -\left( \dfrac{f\left( b\right) -f\left( a\right) }{b-a}\left( a-a\right) +f\left( a\right) \right) =f\left( a\right) -f\left( a\right) =0\text{ and} \\ h\left( b\right) &=&f\left( b\right) -g\left( b\right) =f\left( b\right) -\left( \dfrac{f\left( b\right) -f\left( a\right) }{b-a}\left( b-a\right) +f\left( a\right) \right) =f\left( b\right) -f\left( b\right) =0 \end{eqnarray*}% and so the conditions hold for Rolle's Theorem. \ By this theorem, there exists $c$ in $\left( a,b\right) $ so that $h^{\prime }\left( c\right) =0$. \ This means that there exists $c$ for which $h^{\prime }\left( c\right) =0$% , where $h\left( x\right) =f\left( x\right) -\dfrac{f\left( b\right) -f\left( a\right) }{b-a}\left( x-a\right) -f\left( a\right) $% \begin{eqnarray*} h^{\prime }\left( x\right) &=&f^{\prime }\left( x\right) -\dfrac{f\left( b\right) -f\left( a\right) }{b-a}\text{ \ and \ for some }c\text{ in }\left( a,b\right) \\ 0 &=&h^{\prime }\left( c\right) =f^{\prime }\left( c\right) -\dfrac{f\left( b\right) -f\left( a\right) }{b-a} \\ 0 &=&f^{\prime }\left( c\right) -\dfrac{f\left( b\right) -f\left( a\right) }{% b-a} \\ f^{\prime }\left( c\right) &=&\dfrac{f\left( b\right) -f\left( a\right) }{b-a% } \end{eqnarray*}% which completes our proof.\bigskip \pagebreak Corollary 1. \ If $f^{\prime }\left( x\right) =0$ at each point of an open interval $\left( a,b\right) $, then $f\left( x\right) =C$ for all $x$ in $% \left( a,b\right) $ for some constant $C$. \bigskip Proof: \ Suppose that $f^{\prime }\left( x\right) =0$ at each point of an open interval $\left( a,b\right) $. \ \ Let $x_{1}$ and $x_{2}$ be any two different points in $\left( a,b\right) $. \ Since $f$ is differentiable, it is also continuous on $\left[ x_{1},x_{2}\right] $. \ We will apply the Mean Value Theorem. \ The slope $\dfrac{f\left( x_{2}\right) -f\left( x_{1}\right) }{x_{2}-x_{1}}$ \ is achieved as a derivative somewhere in the interval $\left[ x_{1},x_{2}\right] $. \ There exists $c$ between $x_{1}$ and $x_{2}$ so that \begin{equation*} f^{\prime }\left( c\right) =\dfrac{f\left( x_{2}\right) -f\left( x_{1}\right) }{x_{2}-x_{1}} \end{equation*}% By the hypotheses, $f^{\prime }\left( c\right) =0$. \ This gives us% \begin{eqnarray*} \dfrac{f\left( x_{2}\right) -f\left( x_{1}\right) }{x_{2}-x_{1}} &=&0 \\ f\left( x_{2}\right) -f\left( x_{1}\right) &=&0 \\ f\left( x_{2}\right) &=&f\left( x_{1}\right) \end{eqnarray*}% and so $f$ is constant on $\left( a,b\right) $.\bigskip Corollary 2. \ If $f^{\prime }\left( x\right) =g^{\prime }\left( x\right) $ at each point $x$ in an open interval $\left( a,b\right) $, then there exists a constant $C$ such that $f\left( x\right) =g\left( x\right) +C$ for all $x$ in $\left( a,b\right) $. \ That is, $f-g$ is constant on $\left( a,b\right) $.\bigskip Proof: \ Suppose that $f^{\prime }\left( x\right) =g^{\prime }\left( x\right) $ on $\left( a,b\right) $. \ Define $h\left( x\right) =f\left( x\right) -g\left( x\right) $ on $\left( a,b\right) $. \ Then $h$ is differentiable and \begin{equation*} h^{\prime }\left( x\right) =\left( f\left( x\right) -g\left( x\right) \right) ^{\prime }=f^{\prime }\left( x\right) -g^{\prime }\left( x\right) =0 \end{equation*}% By the previous corollary, $h^{\prime }\left( x\right) =0$ on $\left( a,b\right) $ inplies that $h\left( x\right) =C$ on $\left( a,b\right) $ for some constant $C$. \ Thus \begin{eqnarray*} h\left( x\right) &=&C\text{ \ \ on }\left( a,b\right) \\ f\left( x\right) -g\left( x\right) &=&C \\ f\left( x\right) &=&g\left( x\right) +C\text{ \ on }\left( a,b\right) \end{eqnarray*}% which completes our proof. \bigskip \bigskip \bigskip \bigskip \bigskip \vspace{0.5in} \bigskip \bigskip \vspace{1.2in} \bigskip \bigskip \href{http://www.teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html% }{For more documents like this, visit our page at\ http://www.teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail questions or comments to mhidegkuti@ccc.edu.} \end{document}