Why can't I plot (x+2)^{1/3} properly using pgfplots?

Question

I'm using the code

\documentclass{book}
\usepackage{pgfplots}
\begin{document}

\begin{tikzpicture}
\begin{axis}[
        xmin=-5,
        xmax=5,
        ymin=-5,
        ymax=5,
        axis x line=middle,
        axis y line=middle,
        samples=100,
        domain=-5:5,
    ]
        \addplot[thick](x,{(x+2)^(1/3)});
\end{axis} 
\end{tikzpicture}

\end{document}

to plot y= (x+2)^{1/3} which gives me this graph:
enter image description here

but it is not complete, because the right graph is:

enter image description here

How can I plot it like the second screenshot?

Possible duplicate: PGF Math Function to compute cube root or Pgfplots : can't plot some usual mathematical functions. — DJP, Jun 04 '15 at 21:18
f(x) is not real for x<2, because (x+2)^(1/3)=exp(1/3*ln(x+2)) and ln(x) is only defined for x>0. You'll get the complete graph (on the real part) with domain=-2:5. — anderstood, Jun 04 '15 at 21:34
@djp: Thanks for your links, but I couldn't solve my problem. — Sisabe, Jun 04 '15 at 21:37
From the code given in the second link, modify to (x+2)/abs(x+2)*abs(x+2)^(1/3) and the graph gets shifted as you asked. — DJP, Jun 04 '15 at 21:45
@anderstood: here we have 3 which is an odd number. Why the domain of (x+2)^(1/3) is [-2,\infty)? — Sisabe, Jun 04 '15 at 22:00
Well, f can be defined from R to R. For any real, the cube root is either a real or two conjugate complex numbers. You can do as suggested by DJP, but the reason why pgfplot does not work is, I guess, because it uses the definition x^p=exp(p*ln(x)). Maybe someone can confirm or not that pgfplots works this way. Anyway, that's probably not a coincidence that the graph started at x=2. — anderstood, Jun 04 '15 at 22:10

Why can't I plot (x+2)^{1/3} properly using pgfplots?

0 Answers0