\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=0.4in,right=0.5in,top=0.5in,bottom=0.4in} \pagestyle{fancy} \lhead{\color{blue} \Large Math 207} \lfoot{\small \copyright $\;$ Hidegkuti 2015} \cfoot{} \chead{\LARGE Definition of Continuity} \rfoot{\small Last revised: October 1, 2015} \rhead{ page \thepage } \textwidth 7.5in \textheight 9.6in \setlength{\headheight}{29pt} \setlength{\parindent}{0pt} \begin{document} Definitions: \ (\textit{Interior point}) \ A function $y=f\left( x\right) $ is continuous at an interior point $c$ of its domain if $\lim\limits_{x\rightarrow c}f\left( x\right) =f\left( c\right) $.\newline (\textit{Endpoint}) \ A function $y=f\left( x\right) $ is continuous at a left endpoint $a$ or is continuous at a right endpoint $b$ of its domain if \begin{equation*} \lim\limits_{x\rightarrow a^{+}}f\left( x\right) =f\left( a\right) \text{ \ \ or \ \ }\lim\limits_{x\rightarrow b^{-}}f\left( x\right) =f\left( b\right) \text{, \ respectively.} \end{equation*} A function $f$ is said to be continuous on an open interval $\left( a,b\right) $ if it is continuous at every point inside the interval. \ A function $f$ is said to be continuous on a closed interval $\left[ a,b\right] $ if it is continuous at every point in the interval as described above. \ (End-points of the interval require only one-sided limits.) \bigskip We can use operations on functions to create new functions. \ For example, given the functions $f$ and $g$, we can define new functions as follows. \ $% f+g$, $f-g$, $fg$, and $\dfrac{f}{g}$ are defined as. \begin{eqnarray*} \left( f+g\right) \left( x\right) &=&f\left( x\right) +g\left( x\right) \text{ \ \ \ \ \ \ \ \ \ for all }x\text{ in the domains of both }f\text{ and }g \\ \left( f-g\right) \left( x\right) &=&f\left( x\right) -g\left( x\right) \text{ \ \ \ \ \ \ \ \ \ for all }x\text{ in the domains of both }f\text{ and }g \\ \left( fg\right) \left( x\right) &=&f\left( x\right) g\left( x\right) \text{ \ \ \ \ \ \ \ \ \ \ \ \ \ for all }x\text{ in the domains of both }f\text{ and }g \\ \left( \dfrac{f}{g}\right) \left( x\right) &=&\dfrac{f\left( x\right) }{% g\left( x\right) }\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ for all }x\text{ in the domains of both }f\text{ and }g\text{ and where }g\left( x\right) \not=0 \end{eqnarray*}% Another operation on functions is to compose them, that is to apply the rule for first one function and then the other. \ The new function that is created is denoted by $f\circ g$ and is defined by \begin{equation*} \left( f\circ g\right) \left( x\right) =f\left( g\left( x\right) \right) \text{ \ where }x\text{ is in the domain of }g\text{ and }g\left( x\right) \text{ is in the domain of }f \end{equation*}% Notice that composing functions is not commutative, i.e. very often $f\circ g\not=g\circ f$.\bigskip Theorem: \ If $f$ is continuous at $c$ and $g$ is continuous at $f\left( c\right) $, then $g\circ f$ is continuous at $c$. \bigskip Theorem: \ If $g$ is continuous at the point $b$ and $\lim\limits_{x% \rightarrow c}f\left( x\right) =b$, then \begin{equation*} \lim\limits_{x\rightarrow c}g\left( f\left( x\right) \right) =g\left( b\right) =g\left( \lim\limits_{x\rightarrow c}f\left( x\right) \right) \end{equation*}% \bigskip Continuous functions have very nice properties and we easily visualize continuous functions. \ However, many, many functions are not continuous and their study might be more difficult. \ Functions that are continuous on a closed interval have especially nice properties.\bigskip \bigskip Theorem: \ If $f$ is continuous on a closed interval $\left[ a,b\right] ,$ where $f\left( a\right) $ is negative and $f\left( b\right) $ is positive, then \ there exists $c$ in $\left( a,b\right) $ such that $f\left( c\right) =0$.\bigskip \bigskip Theorem: \ (\textbf{The Intemediate 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 \pagebreak Example 1: \ Suppose that $f$ is a function defined as $\ f\left( x\right) =\left\{ \begin{array}{ccc} mx-10 & \text{if} & x<-2 \\ x^{2}+9x-8 & \text{if} & x\geq -2% \end{array}% \right. .$ \ Find the value of $m$ if we know that $f$ is continuous everywhere. \bigskip Solution: \ If $x<-2$, then the function is continuous for all $x.$ \ Similarly, $f$ is also continuous on all $x$ with $x\geq -2$. \ The only questionable point is at $x=-2.$ \ For a continuous function, we need the left limit and the right limit to exist and have the same value. \ \begin{eqnarray*} \lim_{x\rightarrow -2^{-}}f\left( x\right) &=&\lim_{x\rightarrow -2^{+}}f\left( x\right) \\ \lim_{x\rightarrow -2^{-}}\left( mx-10\right) &=&\lim_{x\rightarrow -2^{+}}\left( x^{2}+9x-8\right) \end{eqnarray*}% By the various properties of limits, this equation can be simplified as follows:% \begin{eqnarray*} m\left( -2\right) -10 &=&\left( -2\right) ^{2}+9\left( -2\right) -8 \\ -2m-10 &=&-22 \\ -2m &=&-12 \\ m &=&6 \end{eqnarray*}% And so $m=6$ is the value for which $f$ is continuous on the entire number line. \bigskip \bigskip \begin{center} {\Large Practice Problems}\bigskip \end{center} \begin{enumerate} \item Suppose that $f$ is a function defined as $\ f\left( x\right) =\left\{ \begin{array}{ccc} mx-13 & \text{if} & x<-10 \\ x^{2}+5x-3 & \text{if} & x\geq -10% \end{array}% \right. .$ \ Find the value of $m$ if we know that $f$ is continuous everywhere. \item Suppose that $f$ is a function defined as $\ f\left( x\right) =\left\{ \begin{array}{ccc} 8x-4 & \text{if} & x\leq 4 \\ -2x+b & \text{if} & x>4% \end{array}% \right. .$ \ Find the value of $b$ if we know that $f$ is continuous everywhere. \item Suppose that $f$ is a function defined as $\ f\left( x\right) =\left\{ \begin{array}{ccc} mx-11 & \text{if} & x<-6 \\ x^{2}+4x-5 & \text{if} & x\geq -6% \end{array}% \right. .$ \ Find the value of $m$ if we know that $f$ is continuous everywhere. \item Suppose that $f$ is a function defined as $\ f\left( x\right) =\left\{ \begin{array}{ccc} 2x+b & \text{if} & x<7 \\ \sqrt{x+2} & \text{if} & x\geq 7% \end{array}% \right. .$ \ Find the value of $b$ if we know that $f$ is continuous everywhere. \end{enumerate} \bigskip \bigskip \begin{center} {\Large Answers - Practice Problems\bigskip } \end{center} \bigskip 1.) \ $-6$ \ \ \ \ \ \ 2.) \ $36$\ \ \ \ \ \ \ 3.) \ $-3$ \ \ \ \ \ 4) \ $-11 $ \end{document}