# How to modify the direction of this arrow?

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}


• Are you sure you're going to + infinity? – AndréC Feb 21 '20 at 22:23
• @AndréC, It's my bad. It should be - infinity. – LE Anh Dung Feb 21 '20 at 22:24
• ok, please, now give a fully compilable code. – AndréC Feb 21 '20 at 22:25
• I've edited my code @AndréC. – LE Anh Dung Feb 21 '20 at 22:28
• @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 '20 at 5:33

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!

\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}

• Thank you so much! You save my day :) – LE Anh Dung Feb 22 '20 at 9:01
• You are welcome! – Black Mild Feb 22 '20 at 9:04

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

\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}

• Could you please adjust your code to have both 0's on the same hight and and - infty below them? – LE Anh Dung Feb 22 '20 at 7:35
• @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 '20 at 8:25
• Thank you so much! It seems to me that there is no easy solution. – LE Anh Dung Feb 22 '20 at 8:36

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}