4

I'm modifying this code to have an arrow as in the picture. I'm tried but unable to reach my goal. Please help me modify it! Thank you so much!

\documentclass[12pt]{extreport}
\usepackage[left = 1in, right = 1in, top = 0.7in, bottom = 0.7in]{geometry}
\usepackage{graphicx,latexsym,mathtools}
\usepackage{tikz,tkz-tab}

\begin{document}

\begin{tikzpicture}
\tkzTab
[lgt=5,espcl=3] 
{$x$/1, $f_\alpha'(x)$/1, $f_\alpha(x)$/3} 
{$0$, $\alpha-1$, $x^*(\alpha)$, $+\infty$} 
{,+,0,-,-,-,} 
{-/ $0$, +/ $f_\alpha(\alpha-1)$, -/ $0$ , -/ $-\infty$} 
\end{tikzpicture}

\end{document}

enter image description here

7
  • 1
    Are you sure you're going to + infinity?
    – AndréC
    Feb 21, 2020 at 22:23
  • @AndréC, It's my bad. It should be - infinity.
    – Akira
    Feb 21, 2020 at 22:24
  • ok, please, now give a fully compilable code.
    – AndréC
    Feb 21, 2020 at 22:25
  • I've edited my code @AndréC.
    – Akira
    Feb 21, 2020 at 22:28
  • 1
    @Navier_Stokes - please see if the answer below meets your requirement or if anything more is required -- a second negative/descent/derivative is provided with the help of intermediate entries
    – js bibra
    Feb 22, 2020 at 5:33

3 Answers 3

5

I think plain TikZ is not so difficult as OP complained ^^ What does plain mean? SIMPLE! Change [yscale=.8,xscale=1.4] as you wish!

enter image description here

\documentclass[tikz,border=5mm]{standalone}
\begin{document}
\begin{tikzpicture}[yscale=.8,xscale=1.4]
\draw[shift={(-.5,.5)}] (0,0) rectangle +(8,-5) 
(0,-1)--+(0:8) (0,-2)--+(0:8) (1,0)--+(-90:5);
\path
(0,0)    node{$x$}          % <<< line 1
++(0:1)  node{$0$}
++(0:2)  node{$\alpha -1$}
++(0:2)  node{$x^*(\alpha)$}
++(0:2)  node{$+\infty$}
(0,-1)   node{$f'_{\alpha}(x)$}         % <<< line 2
++(0:2)  node{$+$}
++(0:1)  node{$0$}
++(0:1)  node{$-$}
++(0:2)  node{$-$}
(0,-3)   node{$f_{\alpha}(x)$}          % <<< line 3
++(0:1)  node (A) {$0$}
++(0:2)  +(90:1) node (B) {$f_{\alpha}(\alpha -1)$}
++(0:2)  node (C) {$0$}
++(0:2)  +(-90:1) node (D) {$-\infty$}
;
\foreach \p/\q in {A/B,B/C,C/D}
\draw[-stealth,magenta] (\p)--(\q);
\end{tikzpicture}
\end{document}
2
  • Thank you so much! You save my day :)
    – Akira
    Feb 22, 2020 at 9:01
  • You are welcome!
    – Black Mild
    Feb 22, 2020 at 9:04
5

I used the intermediate tab value command from page 33 of the manual

\tkzTabVal{2}{3}{0.5}{$x^*(\alpha)$}{0}

the {2}{3} refers to the column precedent and antecedent and 0.5 on the sloped line -- -- a second negative/descent/derivative is provided with the help of intermediate entries-- these intermediate entries are not shown with a column heading of - since the continuous down slope shows that anyway

The manual is at -- http://ftp.cc.uoc.gr/mirrors/CTAN/macros/latex/contrib/tkz/tkz-tab/doc/tkz-tab-screen.pdf

enter image description here

\documentclass[12pt]{extreport}
\usepackage[left = 1in, right = 1in, top = 0.7in, bottom 
= 0.7in]{geometry}
\usepackage{graphicx,latexsym,mathtools}
\usepackage{tikz,tkz-tab}

\begin{document}

\begin{tikzpicture}

\newcommand*{\va}{\colorbox{red!50} {$\scriptscriptstyle 
V_a$}}
\newcommand*{\vb}{\colorbox{blue!50} {$\scriptscriptstyle 
V_b$}}
\newcommand*{\vbo}{\colorbox{blue!50} 
{$\scriptscriptstyle 
V_{b1}$}}
\newcommand*{\vbt}{\colorbox{yellow!50} 
{$\scriptscriptstyle 
V_{b2}$}}
\newcommand*{\vc}{\colorbox{gray!50} {$\scriptscriptstyle 
V_c$}}
\newcommand*{\vd}{\colorbox{magenta!50} 
{$\scriptscriptstyle 
V_d$}}
\newcommand*{\ve}{\colorbox{orange!50} 
{$\scriptscriptstyle 
V_e$}}
\tkzTabInit[color,
colorT = yellow!20,
colorC = red!20,
colorL = green!20,
colorV = lightgray!20,
lgt = 3,
espcl = 4]%
{$x$/1, $f_\alpha'(x)$/1, $f_\alpha(x)$/3}%
{$0$, $\alpha-1$,$+\infty$} 
\tkzTabLine{ ,+,0,-,-,, }
\tkzTabVar{-/ $0$, +/ $f_\alpha(\alpha-1)$,-/ $-\infty$}%
\tkzTabVal{2}{3}{0.5}{$x^*(\alpha)$}{0}
\end{tikzpicture}
\end{document}
3
  • Could you please adjust your code to have both 0's on the same hight and and - infty below them?
    – Akira
    Feb 22, 2020 at 7:35
  • 1
    @Navier_Stokes -- then your table would be assymetrical/ unbalanced -- right side would be longer than left side -- the manual has no such parameter defined -- better to go for plain tikz -- see -- tex.stackexchange.com/a/447188/197451 -- and compare with plain tikz -- tex.stackexchange.com/a/528556/197451
    – js bibra
    Feb 22, 2020 at 8:25
  • Thank you so much! It seems to me that there is no easy solution.
    – Akira
    Feb 22, 2020 at 8:36
0

I have found other solutions too. Just post here for future reference.

\usetikzlibrary{calc,angles,quotes}
\begin{document}
\begin{tikzpicture}
\tkzTabInit[lgt=1.5,espcl=3,deltacl=0.5]
{$x$/.6, $f_{\alpha}'(x)$/.6, $f_{\alpha}(x)$/2.5}
{$0$,$\alpha-1$,$x^*(\alpha)$,$+\infty$}
\tkzTabLine{,+,z,-,-,-,}
\draw
($(N12)!0.5!(N13)$) node (A){$0$}
(N22) node[below] (B) {$f_{\alpha}(\alpha-1)$}
($(N32)!0.5!(N33)$) node (C) {$0$}
(N43) node[above] (D) {$-\infty$};
\draw[-stealth] (A)--(B);
\draw[-stealth] (B)--(C);
\draw[-stealth] (C)--(D);
\end{tikzpicture}

\begin{tikzpicture}[>=stealth,line join=round,line cap=round,font=\footnotesize,scale=1]
\tkzTabInit[nocadre=false,lgt=1.5,espcl=2,deltacl=0.5]{$x$/.7 ,$f'_\alpha(x)$/.7,$f_\alpha(x)$/2}
{ $0$ , $\alpha-1$ ,$x^*(\alpha)$ , $+\infty$}
\tkzTabLine{ , + , $0$ , - ,-, - , }
\draw
($(N13)!0.5!(N12)$) node(A){$0$}
(N22) node[below] (B){$f_\alpha(\alpha-1)$ }
($(N32)!0.5!(N33)$) node (C){$0$}
(N43) node[above](D){$-\infty$};
\draw[->=stealth] (A)--(B) ;
\draw[->=stealth](B)--(C)--(D);
\end{tikzpicture}

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