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\lhead{\color{blue} \Large Math 208}
\lfoot{\small   \copyright $\;$ copyright  Hidegkuti,  Powell,  2012}
\cfoot{}
\chead{\LARGE Definitions, Theorems}
\rfoot{\small Last revised: August 28, 2012}
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\begin{document}


\begin{center}
{\Large The Fundamental Theorem\bigskip }
\end{center}

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 \bigskip 

For all positive $x$, define $f\left( x\right) =\dint\limits_{1}^{x}\dfrac{1%
}{t}dt$

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\begin{enumerate}
\item $f$ is continuous, differentiable, with $f^{\prime }\left( x\right) =%
\dfrac{1}{x}$

\item $f^{\prime }>0$ \ $\Longrightarrow $ \ $f$ is strictly increasing \ $%
\Longrightarrow $ \ one-to-one

\item $f\left( 1\right) =0$ \ If $x>1,$ then $f\left( x\right) $ is positive
and 

if $0<x<1$, then $f\left( x\right) $ is negative. \ 

This is because \ all area is above $x-$axis

\item $f\left( xy\right) =f\left( x\right) +f\left( y\right) $

Fix $y>0$ and define $g\left( x\right) =f\left( xy\right) -f\left( x\right) $

$g^{\prime }\left( x\right) =\dfrac{1}{xy}\cdot y-\dfrac{1}{x}=0$ 

thus $g\left( x\right) $ is a constant function

$g\left( 1\right) =f\left( y\right) -f\left( 1\right) =f\left( y\right) $

$f\left( y\right) =f\left( xy\right) -f\left( x\right) $ for all $x$

\item $f\left( \dfrac{1}{x}\right) =-f\left( x\right) $

$0=f\left( 1\right) =f\left( x\dfrac{1}{x}\right) =f\left( x\right) +f\left( 
\dfrac{1}{x}\right) $

\item $f\left( \dfrac{x}{y}\right) =f\left( x\right) -f\left( y\right) $

$f\left( \dfrac{x}{y}\right) =f\left( x\cdot \dfrac{1}{y}\right) =f\left(
x\right) +f\left( \dfrac{1}{y}\right) =f\left( x\right) +\left( -f\left(
y\right) \right) =f\left( x\right) -f\left( y\right) $

\item $\lim\limits_{x\rightarrow \infty }f\left( x\right) =\infty $ \ and \ $%
\lim\limits_{x\rightarrow 0^{+}}f\left( x\right) =-\infty $ \ \ \ $%
\Longrightarrow $ \ \ range of $f$ is $%
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\mathbb{R}
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$

by the intermedite value theorem

pf: \ $\ln 2>\dfrac{1}{2}$ \ picture \ \ Then $\ln 2^{2N}=\ln 2+\ln
2+....+\ln 2>N$ \ and $\ln \dfrac{1}{2^{2N}}<-N$
\end{enumerate}

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\bigskip 
\begin{equation*}
\text{For all positive }x\text{, define }\ln \left( x\right)
=\dint\limits_{1}^{x}\dfrac{1}{t}dt
\end{equation*}

\pagebreak 

Define $\exp x$ to be the inverse function of $\ln x$\bigskip 

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\begin{enumerate}
\item domain is $%
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\mathbb{R}
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$, range is $\left( 0,\infty \right) $

\item $\exp \left( \ln x\right) =x$ and $\ln \left( \exp x\right) =x$

\item $\exp \left( x+y\right) =\left( \exp x\right) \left( \exp y\right) $

pf. \ $\left( \exp x\right) \left( \exp y\right) =\exp \left( \ln \left(
\left( \exp x\right) \left( \exp y\right) \right) \right) =\exp \left( \ln
\left( \exp x\right) +\ln \left( \exp y\right) \right) =\exp \left(
x+y\right) $

\item $\exp \left( x-y\right) =\dfrac{\exp x}{\exp y}$

pf. \ $\dfrac{\exp x}{\exp y}=\exp \left( \ln \left( \dfrac{\exp x}{\exp y}%
\right) \right) =\exp \left( \ln \exp x-\ln \exp y\right) =\exp \left(
x-y\right) $

\item $\dfrac{d}{dx}\exp x=\exp x$ \ $\Longrightarrow $ $\exp x$ is strictly
increasing, one-to-one

\item $\lim\limits_{x\rightarrow -\infty }\exp x=0$ \ and \ $%
\lim\limits_{x\rightarrow \infty }\exp x=\infty $ follows from inverse
functions
\end{enumerate}

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\bigskip 

define $e=\exp 1$%
\begin{equation*}
\text{denote }\exp x\text{ by }e^{x}
\end{equation*}

\bigskip 

7) $\ln e=1$

\bigskip 

Define $\exp _{b}x=\exp \left( x\ln b\right) $

\bigskip 

7) \ $\exp _{e}x=\exp x$

\end{document}