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%TCIDATA{<META NAME="Title" CONTENT="Problem Set 1 - long - Math 207 - Spring 2011">}
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\lhead{\color{blue} \Large Math 207}
\lfoot{\small   \copyright $\;$ copyright  Hidegkuti,  Powell,  2012}
\cfoot{}
\chead{\LARGE Definitions, Theorems}
\rfoot{\small Last revised: July 22, 2012}
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\begin{document}


\begin{center}
{\Large Continuity\bigskip }\bigskip 
\end{center}

Definition: \qquad (\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) $.

\qquad \qquad \qquad \qquad (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*}

\bigskip 

Theorem: \ Properties of continuous functions?

\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 

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 

\begin{center}
\bigskip 

{\Large Maximum-minimum theorems\bigskip }
\end{center}

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 

Definition: A 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

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

\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

Theorem: (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

Corollary: If $f^{\prime }\left( x\right) =0$ on an interval $\left(
a,b\right) ,$ then $f\left( x\right) =c$ on $\left( a,b\right) $ for a
constant $c$.\bigskip

Theorem: \ If $f^{\prime }\left( x\right) =g^{\prime }\left( x\right) $ on $%
\left( a,b\right) $, then there exists $c$ with $f\left( x\right) =g\left(
x\right) +c$ on $\left( a,b\right) $.\bigskip \bigskip \bigskip \pagebreak

\begin{center}
{\Large The Fundamental Theorem\bigskip }
\end{center}

Theorem: (\textbf{Mean value Theorem for Definite Integrals}) \ If $f$ is
continuous on $\left[ a,b\right] $, then there exists $c$ in $\left[ a,b%
\right] $ such that%
\begin{equation*}
f\left( c\right) =\dfrac{1}{b-a}\dint\limits_{a}^{b}f\left( x\right) dx
\end{equation*}%
\bigskip 

Theorem: (\textbf{Fundamental Theorem of Calculus, Part 1}) \ If $f$ is
continuous on $\left[ a,b\right] $, then $F\left( x\right)
=\dint\limits_{a}^{x}f\left( t\right) dt$ is continuous on $\left[ a,b\right]
$, \ differentiable on $\left( a,b\right) $, and its derivative is $f\left(
x\right) $:%
\begin{equation*}
F^{\prime }\left( x\right) =\dfrac{d}{dx}\dint\limits_{a}^{x}f\left(
t\right) dt=f\left( x\right) 
\end{equation*}%
\bigskip Theorem: (\textbf{Fundamental Theorem of Calculus, Part 2}) \ If $f$
is continuous on $\left[ a,b\right] $, and $F$ is any antiderivative of $f$
on $\left[ a,b\right] $, then%
\begin{equation*}
\dint\limits_{a}^{b}f\left( x\right) dx=F\left( b\right) -F\left( a\right) 
\end{equation*}%
\bigskip Theorem: (Net Change Theorem) \ The net change in a function $%
F\left( x\right) $ over an interval $\left[ a,b\right] $ is the integral of
its rate of change:%
\begin{equation*}
F\left( b\right) -F\left( a\right) =\dint\limits_{a}^{b}F^{\prime }\left(
x\right) dx
\end{equation*}%
\bigskip 

\pagebreak

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