\documentclass{article} \usepackage{amssymb} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %TCIDATA{OutputFilter=LATEX.DLL} %TCIDATA{Created=Monday, June 25, 2001 17:45:57} %TCIDATA{LastRevised=Saturday, June 01, 2002 19:44:15} %TCIDATA{} %TCIDATA{} %TCIDATA{CSTFile=On line bluem.cst} %TCIDATA{PageSetup=72,72,72,72,0} %TCIDATA{AllPages= %H=36 %F=36,\PARA{038
\QTR{small}{Matematick\U{e1} anal\U{fd}za I online - D\U{f4}kazy\dotfill \thepage }} %} \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]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \begin{document} \author{A. U. Thor} \title{Lab Report} \date{The Date } \maketitle \begin{abstract} A Laboratory report created with Scientific Notebook \end{abstract} \section{Aplik\'{a}cie deriv\'{a}ci\'{\i}} \subsection{D\^{o}kaz Cauchyho vety} \textbf{D\^{o}kaz: }Nech $h:\left\langle a,b\right\rangle \longrightarrow \mathbf{R},\,h\left( x\right) =\left[ f\left( b\right) -f\left( a\right) % \right] g\left( x\right) -\left[ g\left( b\right) -g\left( a\right) \right] f\left( x\right) $, potom $h$ je spojit\'{a} na $\left\langle a,b\right\rangle $ a diferencovate\v{l}n\'{a} na $(a,b)$. Plat\'{\i} $% h(a)=f(b)g(a)-g(b)f(a)=h(b)$. Potom pod\v{l}a \hyperref{Rolleho vety}{}{}{% M62.tex#3} existuje $c\in (a,b)$ tak\'{e}, \v{z}e $h^{\prime }(c)=0$ t.j. $% [f(b)-f(a)]g^{\prime }(c)-[g(b)-g(a)]f\,$\/$\,^{\prime }(c)=0.$ Pod\v{l}a predpokladu $g^{\prime }(x)\neq 0$ pre $x\in (a,b)$, teda aj $g^{\prime }(c)\neq 0$. Potom z \hyperref{Rolleho vety}{}{}{M62.tex#3} m\'{a}me, \v{z}e aj $g(b)-g(a)\neq 0$, teda \[ \frac{f\left( b\right) -f\left( a\right) }{g\left( b\right) -g\left( a\right) }=\frac{f\,\/\,^{\prime }\left( c\right) }{g^{\prime }\left( c\right) }.\blacksquare \] \begin{center} \begin{tabular}{|c|} \hline \hyperref{{\small Sp\"{a}\v{t}}}{}{}{M62.tex#2} \\ \hline \end{tabular} \end{center} \end{document}