%% %% Ein Beispiel der DANTE-Edition %% Mathematiksatz mit LaTeX %% 3. Auflage %% Beispiel 09-21-1 auf Seite 203. %% Copyright (C) 2018 Herbert Voss %% %% It may be distributed and/or modified under the conditions %% of the LaTeX Project Public License, either version 1.3 %% of this license or (at your option) any later version. %% See http://www.latex-project.org/lppl.txt for details. %% %% ==== % Show page(s) 1,2 %% %% \documentclass[12pt]{screxa} \pagestyle{empty} \setlength\textwidth{170.40707pt} \AtBeginDocument{\setlength\parindent{0pt}} \usepackage[ngerman]{babel} \usepackage[a5paper,margin=1cm]{geometry} %StartShownPreambleCommands \usepackage{xcolor,mdframed,amsmath} \newcommand*\diff{\mathop{}\!\mathrm{d}} \allowdisplaybreaks %StopShownPreambleCommands \begin{document} \mdfdefinestyle{theoremstyle}{linecolor=red,linewidth=2pt,frametitlerule=true,% frametitlebackgroundcolor=gray!20,innertopmargin=\topskip} \mdtheorem[style=theoremstyle]{definition}{Definition} \begin{definition}[Quadratische Gleichung] \addtolength\jot{5pt} \begin{align} f(x) &= g(x)\\ -x^{2}+6x-5 &= -\frac{1}{3}x^{2}+\frac{4}{3}x+\frac{5}{3}\\ 0 &= \frac{2}{3}x^{2}-\frac{14}{3}x+ \frac{20}{3}\label{eq:fg}\\ 0 &= x^{2}-7x+10\label{eq:4}\\ 0 &= (x-2)(x-5)\\ \mathbb{L} &= \{2;\ 5\}\\[10pt] F&=\int\limits_2^5\left(g(x)-f(x)\right)\diff x \\[10pt] F &= \int\limits_2^5\left(\frac{2}{3}x^{2}\frac{14}{3}x+\frac{20}{3}\right)\diff x\\ &= \left[\frac{2}{9}x^{3}-\frac{14}{6}x^{2}+\frac{20}{3}x\right]_{2}^{5}\\ &= \underbrace{{\vphantom{\left(\frac{2}{9}\right)} \frac{2}{9}\cdot125-\frac{14}{6}\cdot25+\frac{20}{3} \cdot5}}_{\textrm{obere Grenze}}\underbrace{\vphantom{% \left(\frac{2}{9}\right)}-}_{-}\underbrace{\left(\frac{2}{9} \cdot8-\frac{14}{6}\cdot4+\frac{20}{3}\cdot 2\right)}_{\textrm{untere Grenze}}\\[10pt] &= \frac{500-1050+600}{18}-\frac{32-168+240}{18}\\ &= \frac{50}{18}-\frac{104}{18}\\ &= -\frac{54}{18} = -3\\[10pt] f(x) &= -x^{2}+6x-5=-\left(x^{2}-6x+5\right)\\ &= -\left(x-1\right)\left(x-5\right)\\[10pt] f(x) &= -x^{2}+6x-5=-\left(x^{2}-6x+5\right)\\ &= -\left(x^2-2\cdot3\cdot x+3^2-3^2+5\right)\\[-\normalbaselineskip] & \phantom{\mathrel{{}=-{}}(}\underbrace{\phantom{x^{2}-2\cdot3\cdot x+3^2}}_{\left(x-3\right)^2}\underbrace{\phantom{-3^2+5}}_{-4}\nonumber\\[10pt] g(x) &= -\frac{1}{3}x^{2}+\frac{4}{3}x+\frac{5}{3}\\ &= -\frac{1}{3}\left(x^{2}-4x-5\right)\\ &= -\frac{1}{3}\left(x+1\right)\left(x-5\right)\qquad\boxed{\mathbb{L}=\{-1;5\}}\\ g(x) &= -\frac{1}{3}x^{2}+\frac{4}{3}x+\frac{5}{3}\\ &= -\frac{1}{3}\left(x^{2}-4x-5\right)\\ &= -\frac{1}{3}\left(x^{2}-2\cdot2\cdot x+2^{2}-2^{2}-5\right)\\ &= -\frac{1}{3}\left(\left(x-2\right)^{2}-9\right)\\ &= -\frac{1}{3}\left(x-2\right)^{2}+3\qquad\boxed{SP(2;3)} \end{align} \end{definition} \end{document}