%
%	gbb60166@gmail.com	https://unilab.gbb60166.jp/prekou/prekou.htm
%
% aspectratio= は 1610, 169, 149, 54, 32 の中から選べる（省略時は 43）
% C:\texlive\2024\texmf-dist\tex\latex\beamer\beamer.cls
%\documentclass[17pt,dvipdfmx,usepdftitle=false,hyperref={pdfpagemode=FullScreen},aspectratio=169]{beamer}
%\documentclass[17pt,dvipdfmx,usepdftitle=false,hyperref={pdfstartview={XYZ null null 8.00}},aspectratio=169]{beamer}
\documentclass[17pt,dvipdfmx,usepdftitle=false,aspectratio=169]{beamer}
% pdfの栞の字化けを防ぐ
%\AtBeginDvi{\special{pdf:tounicode EUC-UCS2}}
% テーマ
\usetheme{Copenhagen}
% navi. symbolsは目立たないが，dvipdfmxを使うと機能しないので非表示に
\setbeamertemplate{navigation symbols}{} 
\usepackage{bxdpx-beamer,pxjahyper,minijs}
%\usepackage{graphicx}
%\usepackage{tkokugo,furikana,tsayusen,shiika,sfkanbun,jdkintou,plext}
\usepackage{type1cm}
\usepackage[deluxe]{otf}
%\usepackage{tikzsymbols}
%\usepackage[dvipdfmx]{graphicx}%
\def\pgfsysdriver{pgfsys-dvipdfmx.def}%（graphicxパッケージを使用しない場合はこの行を有効に）
%\def\pgfsysdriver{pgfsys-dvips.def}%デフォルト
%\usepackage[e]{esvect}
\usepackage{tikz}%（これで、pgfとpgfforが読み込まれます。）
%\usepackage{tikz-3dplot}
\usetikzlibrary{calc}
%\usetikzlibrary{arrows.meta,patterns,angles,intersections,backgrounds}
%\usepackage{animate}
%\usepackage{pgfplots}
%\pgfplotsset{compat=newest}
%\pgfplotsset{compat=newest, every axis/.append style={line width=1pt}}
%\PassOptionsToPackage{dvipdfmx}{graphicx}
\nofiles

\hypersetup{% Jul 18 2023
	pdftitle={プレ高数学科（LaTeX Beamer で汎用PDFプレゼンを作る）},
	pdfauthor={gbb60166@gmail.com},
	pdfsubject={https://unilab.gbb60166.jp/prekou/prekou.htm}
	}

% フォントはお好みで
%\usepackage{txfonts}

\mathversion{bold}
\renewcommand{\familydefault}{\sfdefault}
\renewcommand{\seriesdefault}{bx}
\renewcommand{\kanjifamilydefault}{\gtdefault}
\renewcommand{\kanjiseriesdefault}{bx}
\setbeamerfont{title}{size=\normalsize,series=\bfseries}
\setbeamerfont{frametitle}{size=\normalsize,series=\bfseries}
\setbeamertemplate{frametitle}[default][center]
\usefonttheme{professionalfonts}

% 参考にしたURL
% http://windom.phys.hirosaki-u.ac.jp/fswiki/wiki.cgi?page=LaTeX+Beamer%A4%C7%A5%D7%A5%EC%A5%BC%A5%F3%A5%C6%A1%BC%A5%B7%A5%E7%A5%F3

\newcommand{\Slash}[1]{\ooalign{\hfil\kern-3pt／\hfil\crcr$#1$}}
\everymath{\displaystyle}

\newcommand{\maruwaku}[1]{\begin{tikzpicture}[scale=0.7, baseline={([yshift=-22pt] current bounding box.north)}]
\filldraw[color=CUDBlue, line width=1pt, rounded corners=2pt] (-0.1,0)--(2.1,0)--(2.1,1.1)--(-0.1,1.1)--cycle;
\draw(1,0.5) node[white]{#1};
\end{tikzpicture}
}

\setbeamersize{text margin left=5mm,text margin right=5mm}
\fboxrule=0.8pt

\makeatletter
\newcommand{\hooklen}[2]{\settowidth{\@tempdima}{\(#1\)}
%\advance\@tempdima by.3ex
% ↑ 数式モードで式の前後に入るスペースを制御したかったが、
%    難しいのでやめた。段々難解なコードになっているのでやめた方がよい？
\hbox to\@tempdima{\hfil \(#2\)\hfil}}
%■等号付き不等号 by 美文書作成入門
\newcommand{\myleqq}{\mathrel{\mathpalette\gl@align<}}
\newcommand{\mygeqq}{\mathrel{\mathpalette\gl@align>}}
\newcommand{\gl@align}[2]{\lower.6ex\vbox{\baselineskip\z@skip\lineskip\z@
	\ialign{$\m@th#1\hfil##\hfil$\crcr#2\crcr=\crcr}}}
\makeatother

% カラーユニバーサルデザインを調べたつもりだがあまり自信がありません
% http://www.fukushihoken.metro.tokyo.jp/kiban/machizukuri/kanren/color.files/colorudguideline.pdf
% http://jfly.iam.u-tokyo.ac.jp/colorset/

% ■ アクセントカラー小面積を目立たせる高彩度色
\definecolor{CUDRed}{RGB}{255,75,0}
\definecolor{CUDGreen}{RGB}{3,175,122}
\definecolor{CUDBlue}{RGB}{0,90,255}
\definecolor{CUDCyan}{RGB}{77,196,255}
\definecolor{CUDMagenta}{RGB}{153,0,153}
\definecolor{CUDYellow}{RGB}{255,241,0}
\definecolor{CUDBrown}{RGB}{128,64,0}

\definecolor{CUDOrange}{RGB}{246,170,0}

% ■ ベースカラー広い面積の塗り分けに用いる低・中彩度色
\definecolor{CUDPink}{RGB}{255,202,191}
\definecolor{CUDBrightGreen}{RGB}{119,217,168}
\definecolor{CUDLime}{RGB}{216,242,85}
\definecolor{CUDCream}{RGB}{255,255,128}
\definecolor{CUDBrightCyan}{RGB}{191,228,255}
%\definecolor{CUD}{RGB}{}%

\setbeamercolor{CUDBrightGreen}{fg=black,bg=CUDBrightGreen!50}
\setbeamercolor{CUDBrightCyan}{fg=black,bg=CUDBrightCyan}
\setbeamercolor{CUDCream}{fg=black,bg=CUDCream!75}

%\setbeamercolor{frametitle}{use=structure,fg=white,bg=structure.fg}
%\setbeamercolor{palette primary}{use=structure,fg=white,bg=structure.fg}
%\setbeamercolor{frametitle}{bg=CUDOrange}
%\setbeamercolor{palette primary}{bg=CUDOrange}
%\setbeamercolor{frametitle}{bg=CUDCyan!25,fg=black}
%\setbeamercolor{palette primary}{bg=CUDCyan!25,fg=black}
%\setbeamercolor{frametitle}{bg=CUDCream!75,fg=black!80}
%\setbeamercolor{palette primary}{bg=CUDCream!75,fg=black!80}

\newcommand{\myfrac}[2]{\mathchoice%■ 修正 Oct 8,2023
	{%\fontsize{20}{20}\selectfont%	\displaystyle
		\frac{\raisebox{-.6ex}{\(#1\)}}{\raisebox{.4ex}{\(#2\)}}}%
	{\textstyle{\frac{#1}{#2}}}%	\textstyle
	{%\fontsize{13}{13}\selectfont%	\scriptstyle
		\frac{\raisebox{0ex}{\(#1\)}}{\raisebox{-.4ex}{\(#2\)}}}%
	{%\fontsize{11}{11}\selectfont%	\scriptscriptstyle
		\frac{\raisebox{0ex}{\(#1\)}}{\raisebox{-.4ex}{\(#2\)}}}%
	}


\begin{document}

\title{プレ高数学科}\author{gbb60166}

%■ 本文と数式モードの間に挿入される空白スペース
%\mathsurround=-1pt
%■ \( \log\log \)でlogとlogの間に挿入される細スペース（省略形は \,）
%\thinmuskip=3mu
%■ \( a+b \)で+の前後に挿入される中スペース（省略形は \>）
%\medmuskip=2mu plus 1mu minus 1mu
\medmuskip=1mu minus 1mu
%■ \( x=5 \)で=の前後に挿入される太スペース（省略形は \;）
%\thickmuskip=5mu plus 5mu
\thickmuskip=2mu plus 1mu
% mu は math units の省略で長さの単位

%\abovedisplayskip=6pt
%\belowdisplayskip=6pt
%■効果が出ていない
\abovedisplayshortskip=-12pt
\belowdisplayshortskip=9pt

%■ 新しいプラスマイナス記号を自作（TikZを利用）
\newcommand{\newpm}{\,\tikz[baseline=-0.1ex]
	\draw[line width=0.1ex,line cap=round](0,0)--++(0.6em,0)
					(0,0.8ex)--++(0.6em,0)
					(0.3em,0.25ex)--++(0,1.1ex);
	\,}

\newcommand{\tmyfrac}{\textstyle\myfrac}

\newcommand{\ArrowRightup}{\CID{12205}}
\newcommand{\ArrowRightdown}{\CID{12202}}
\newcommand{\UeTotsu}{\tikz\draw[thick](9pt,0)
	arc[start angle=0, end angle=180, radius=9pt];}
\newcommand{\ShitaTotsu}{\tikz\draw[thick](-9pt,9pt)
	arc[start angle=-180, end angle=0, radius=9pt];}
\newcommand{\RightUp}{\CID{16201}}
%\def\RightDown{\raisebox{-3pt}{\CID{16202}}}%	面倒くさいので微調整はやめた
\newcommand{\RightDown}{\CID{16202}}
\newcommand{\DownRight}{\raisebox{1zh}{\rotatebox{-90}{\CID{16201}}}}
%\def\UpRight{\raisebox{-3pt}{\rotatebox{90}{\CID{16202}}}}%	面倒くさいので微調整はやめた
\newcommand{\UpRight}{\rotatebox{90}{\CID{16202}}}


%■■■■■■■■■■■■■　テスト領域　■■■■■■■■■■■■■■



%\end{document}
%■■■■■■■■■■■■■　完成品　■■■■■■■■■■■■■■

%\begin{frame}[t]
%\frametitle{\( y=\tmyfrac{1}{x^2+1} \)の極値や凹凸などを調べグラフをかきなさい}
%今ではグラフ描画ソフトを使えばすぐにグラフはかけるが、紙とペンを使ってグラフがかけるようになることは重要だ。

%\bigskip
%紙とペンを使ってグラフがかけない人は、微分法が理解できていないはずだ。

%\bigskip
%希望する進路実現のために頑張るのか、能力の限界を悟って諦めるのか、あなた次第だ！
%\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\( y=\tmyfrac{1}{x^2+1} \)の極値や凹凸などを調べ
			グラフをかきなさい}
微分する。次の公式を使う。

\[ \left\{ \myfrac{f(x)}{g(x)} \right\}'
	=\myfrac{f'(x)g(x)-f(x)g'(x)}{\{g(x)\}^2 } \]

\medskip
この問題の場合は、上記公式の変形版である

\[ \left\{ \myfrac{1}{g(x)} \right\}'= \ - \ \myfrac{g'(x)}{\{g(x)\}^2 } \]

の方が適している。
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\( y=\tmyfrac{1}{x^2+1} \)の極値や凹凸などを調べ
			グラフをかきなさい}
\[ \text{公式}\quad \left\{ \myfrac{1}{g(x)} \right\}'
	= \ - \ \myfrac{g'(x)}{\{g(x)\}^2 } \quad\text{を使うと} \]

\bigskip
\[ y' = \myfrac{-2x}{(x^2+1)^2} \quad\text{となる} \]
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\( y'=0 \)の解は？}
\( y' = \myfrac{-2x}{(x^2+1)^2}=0 \) \quad を解きたい。

\bigskip
\( (x^2+1)^2>0 \)だから

\bigskip
\[ -2x=0 \quad\text{を解いて} \]

\[ x=0 \]
となる。
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\( y' = \tmyfrac{-2x}{(x^2+1)^2} \)が分かった}
\vspace*{-1ex}
さらにもう一度微分する。今度は次の公式を使う。

\[ \left\{ \myfrac{f(x)}{g(x)} \right\}'
	=\myfrac{f'(x)g(x)-f(x)g'(x)}{\{g(x)\}^2 } \]

\bigskip
\[ y'' = \myfrac{(-2x)' \cdot (x^2+1)^2-(-2x) \cdot \bigl((x^2+1)^2\bigr)'}
		{\bigl( (x^2+1)^2 \bigr)^2} \]

まず\quad  \( (-2x)'=-2 \) \quad はすぐ分かる。

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\( \bigl((x^2+1)^2\bigr)' \)を計算したい}

次に\( \bigl((x^2+1)^2\bigr)' \)を計算したい。

\bigskip
\( y=f(u), \hskip1em u=g(x) \)の合成関数\( y=f(g(x)) \)の導関数は
\( \myfrac{dy}{dx}=\myfrac{dy}{du}\cdot\myfrac{du}{dx} \)
%見えない支柱を入れる
\rule[0pt]{0pt}{4ex}
だから\\[2ex]

\( y=u^2, \hskip2em u=x^2+1 \)\quad と考えて
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\( \bigl((x^2+1)^2\bigr)' \)を計算したい}
\( y=u^2, \hskip2em u=x^2+1 \)\quad だから

\bigskip
\( \myfrac{dy}{du}=2u, \hskip2em \myfrac{du}{dx}=2x \)\quad となって

\begin{eqnarray*}
\myfrac{dy}{dx} &=& \myfrac{dy}{du}\cdot\myfrac{du}{dx}
	=2u \cdot 2x=2(x^2+1)\cdot 2x \\[2ex]
			&=& 4x(x^2+1)
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\( y' = \tmyfrac{-2x}{(x^2+1)^2} \)のとき\( y''= \)？}
\( \textcolor{CUDRed}{(-2x)'=-2}, \) \hskip2em
 \textcolor{CUDBlue}{\( \bigl((x^2+1)^2\bigr)'=4x(x^2+1) \)}\\
が分かったところで、元に戻って

\vspace*{-3ex}
\begin{eqnarray*}
y'' &=& \myfrac{(-2x)' \cdot (x^2+1)^2-(-2x) \cdot \bigl((x^2+1)^2\bigr)'}
		{\bigl( (x^2+1)^2 \bigr)^2}\\[2ex]
	&=& \myfrac{\textcolor{CUDRed}{-2}(x^2+1)^2-(-2x)\cdot
		\textcolor{CUDBlue}{4x(x^2+1)}}{\bigl( (x^2+1)^2 \bigr)^2}
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\( y' = \tmyfrac{-2x}{(x^2+1)^2} \)のとき\( y''= \)？}
\vspace*{-5ex}
\begin{eqnarray*}
y'' &=& \myfrac{-2(x^2+1)^2-(-2x)\cdot4x(x^2+1)}{\bigl( (x^2+1)^2 \bigr)^2}
								\\[2ex]
	&=& \myfrac{-2(x^2+1)^2+8x^2(x^2+1)}{(x^2+1)^4 }\\[2ex]
	&=& \myfrac{-2(x^2+1)+8x^2}{(x^2+1)^3 } 
		= \myfrac{-2x^2-2+8x^2}{(x^2+1)^3 }
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\( y''=0 \)を解くと？}
\vspace*{-4ex}
\begin{eqnarray*}
y'' &=& \myfrac{-2x^2-2+8x^2}{(x^2+1)^3 } \\
	&=& \myfrac{6x^2-2}{(x^2+1)^3 }
\end{eqnarray*}
が分かった。次に\quad \( y''=\myfrac{6x^2-2}{(x^2+1)^3 }=0 \)\quad を解く。
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\( y''=0 \)を解くと？}
\( \myfrac{6x^2-2}{(x^2+1)^3 }=0 \) \quad は \quad \( 6x^2-2=0 \)となって
\begin{eqnarray*}
\hspace*{18ex} 6x^2 &=& 2 \\
3x^2 &=& 1 \\
x^2 &=& \myfrac{\,1\,}{3} \\
x &=& \newpm\sqrt{\myfrac{\,1\,}{3}} =\newpm\myfrac{\,1\,}{\sqrt{3}}
	=\newpm\myfrac{\sqrt{3}}{3} \\
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\( y=\tmyfrac{1}{x^2+1} \hskip4em 
	y'=\tmyfrac{-2x}{(x^2+1)^2} \hskip4em 
	y''=\tmyfrac{6x^2-2}{(x^2+1)^3 } \)}

\( y'=0 \)\ の解は\ \( x=0 \)
\hfill
\( y''=0 \)\ の解は\ \( x=\newpm\myfrac{\sqrt{3}}{3} \)

\centerline{%
\begin{tabular}[t]{|c||c|c|c|c|c|c|c|} \hline
\(x\) & \(\cdots\) & \(-\tmyfrac{\sqrt{3}}{3}\) & \(\cdots\) & \makebox[1.5em]{\(0\)} & \(\cdots\) & \(\tmyfrac{\sqrt{3}}{3}\) & \(\cdots\) \\ \hline
\(y'\) & \temporal<2-3>{}{\(+\)}{\ArrowRightup} & \temporal<2-3>{}{\(+\)}{\ArrowRightup} & \temporal<2-3>{}{\(+\)}{\ArrowRightup} & \(0\)& %
	\temporal<2-3>{}{\(-\)}{\ArrowRightdown} & \temporal<2-3>{}{\(-\)}{\ArrowRightdown} & \temporal<2-3>{}{\(-\)}{\ArrowRightdown} \\ \hline
\(y''\) & \temporal<3-4>{}{\(+\)}{\ShitaTotsu} & \(0\) & \temporal<3-4>{}{\(-\)}{\UeTotsu} & \temporal<3-4>{}{\(-\)}{\UeTotsu} & %
	\temporal<3-4>{}{\(-\)}{\UeTotsu} & \(0\)& \temporal<3-4>{}{\(+\)}{\ShitaTotsu} \\ \hline
\(y\) & \visible<6->{\RightUp} & \(\) & \visible<6->{\UpRight} &  & \visible<6->{\RightDown} & \(\) & \visible<6->{\DownRight} \\ \hline
\end{tabular}%
}

\bigskip
\visible<7->{\( x=0,\ \newpm\myfrac{\sqrt{3}}{3} \)のときの\(y\)の値を求めると}

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\( y=\tmyfrac{1}{x^2+1} \hskip4em 
	y'=\tmyfrac{-2x}{(x^2+1)^2} \hskip4em
 y''=\tmyfrac{6x^2-2}{(x^2+1)^3 } \)}

\( x=\newpm\myfrac{\sqrt{3}}{3} \)のとき

\vskip-2ex
\begin{eqnarray*}
y &=& \myfrac{1}{x^2+1} =\myfrac{1}{\Bigl(\newpm\myfrac{\sqrt{3}}{3}\Bigr)^2+1}\\
	&=& \myfrac{1}{\ \myfrac{\,3\,}{9}+1 \ }
	=\myfrac{\,1\,}{\ \myfrac{\,1\,}{3}+1 \ }
	 =\myfrac{1}{\ \myfrac{\,4\,}{3}\ } =\myfrac{\,3\,}{4}
\end{eqnarray*}


\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\( y=\tmyfrac{1}{x^2+1} \hskip3em 
	y'=\tmyfrac{-2x}{(x^2+1)^2} \hskip3em 
	y''=\tmyfrac{6x^2-2}{(x^2+1)^3 } \)}

\( x=0 \)のとき

\vskip-2ex
\begin{eqnarray*}
y &=& \myfrac{1}{\ x^2+1 \ } =\myfrac{1}{\ 0^2+1 \ } =1
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\( y=\tmyfrac{1}{x^2+1} \hskip3em 
	y'=\tmyfrac{-2x}{(x^2+1)^2} \hskip3em 
	y''=\tmyfrac{6x^2-2}{(x^2+1)^3 } \)}

よって増減表は

\centerline{%
\begin{tabular}[t]{|c||c|c|c|c|c|c|c|} \hline
\(x\) & \(\cdots\) & \(-\tmyfrac{\sqrt{3}}{3}\) & \(\cdots\) & \makebox[1.5em]{\(0\)} & \(\cdots\) & \(\tmyfrac{\sqrt{3}}{3}\) & \(\cdots\) \\ \hline
\(y'\) & \ArrowRightup & \ArrowRightup & \ArrowRightup & \(0\)& %
	\ArrowRightdown & \ArrowRightdown & \ArrowRightdown \\ \hline
\(y''\) & \ShitaTotsu & \(0\) & \UeTotsu & \UeTotsu & %
	\UeTotsu & \(0\)& \ShitaTotsu \\ \hline
\(y\) & \RightUp & \(\tmyfrac{\,3\,}{4}\) & \UpRight & \(1\) & \RightDown & \(\tmyfrac{\,3\,}{4}\) & \DownRight \\ \hline
\end{tabular}%
}

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\( y=\tmyfrac{1}{x^2+1} \)の極値や凹凸などを調べ
			グラフをかきなさい}
さらに

\vskip-2ex
\[ \lim_{x\rightarrow\infty} \myfrac{1}{\ x^2+1 \ }
	=\lim_{x\rightarrow\infty}
	\myfrac{\myfrac{\,1\,}{\,x^2}}{\ 1+\myfrac{\,1\,}{\,x^2}\ }
	=\myfrac{0}{\ 1+0 \ } =0 \]

同様に

\vskip-2ex
\[ \lim_{x\rightarrow-\infty} \myfrac{1}{\ x^2+1 \ } =0 \]

となるので、グラフは
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\( y=\tmyfrac{1}{x^2+1} \)の極値や凹凸などを調べ
			グラフをかきなさい}

\vspace*{-1.5ex}
\begin{tikzpicture}[scale=1.3, domain=-5:5, thick, >=stealth]
\coordinate (A) at ($ ({-1/sqrt(3)},0) $);
\coordinate (B) at ($ ({-1/sqrt(3)},{3/4}) $);
\coordinate (C) at ($ ({1/sqrt(3)},{3/4}) $);
\coordinate (D) at ($ ({1/sqrt(3)},0) $);

\footnotesize
\draw[->](-5,0)--(5,0) node[right]{$x$};
\draw[->](0,-0.5)--(0,1.5) node[above]{$y$};
\tiny
\draw[thin, dashed] (A) node[below]{\( -\myfrac{\sqrt{3}}{3} \)}
	--(B)--(C)--(D) node[below]{\( \myfrac{\sqrt{3}}{3} \)};
\draw(0.5em,-1ex) node{O};
\draw(0.5em,0.65) node[fill=white, inner sep=0pt]{\( \myfrac{3}{4} \)};
\draw(0.5em,1) node[inner sep=0pt]{\( 1 \)};

\draw[CUDRed] plot[smooth] (\x,{1/((\x)^2+1)}) 
	node[above] {\footnotesize \( \displaystyle y=\myfrac{1}{x^2+1} \)};
\end{tikzpicture}

\vspace*{-2ex}
\centerline{%
\begin{tabular}[t]{|c||c|c|c|c|c|c|c|} \hline
\(x\) & \(\cdots\) & \(-\tmyfrac{\sqrt{3}}{3}\) & \(\cdots\) & \makebox[1.5em]{\(0\)} & \(\cdots\) & \(\tmyfrac{\sqrt{3}}{3}\) & \(\cdots\) \\ \hline
\(y'\) & \ArrowRightup & \ArrowRightup & \ArrowRightup & \(0\)& %
	\ArrowRightdown & \ArrowRightdown & \ArrowRightdown \\ \hline
\(y''\) & \ShitaTotsu & \(0\) & \UeTotsu & \UeTotsu & %
	\UeTotsu & \(0\)& \ShitaTotsu \\ \hline
\(y\) & \RightUp & \(\tmyfrac{\,3\,}{4}\) & \UpRight & \(1\) & \RightDown & \(\tmyfrac{\,3\,}{4}\) & \DownRight \\ \hline
\end{tabular}%
}

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%




\end{document}
