How to draw the graph of this function? (very difficult) I'm trying to draw the graph of this function: 
$$f(x) =
\begin{cases}
|x|^{|x|},  & \text{if $x\neq 0$} \\
1, & \text{if $x=0$}  \\
\end{cases}$$
I divided the question in two parts, if $x\lt 0$ and $x\gt0$.
If $x\gt 0$
We have the first and second derivatives:
$f'(x)=e^{x\ln x}(\ln x+1)$
$f''(x)=e^{x\ln x}(\ln x+1)^2+\frac{e^{x\ln x}}{x}$
the only information I have is these derivatives are positive everywhere and never zero, so it's increasing, what else can I say about $f$?
I need help to draw this graph.
Thanks in advance. 
 A: Using LaTeX and my package xpicture it is very easy to do it. This is the LaTeX code:
\documentclass{article}
\usepackage{xpicture}
\begin{document}
\setlength{\unitlength}{1cm}
\PRODUCTfunction{\IDENTITYfunction}{\LOGfunction}{\f}
\COMPOSITIONfunction{\EXPfunction}{\f}{\F}
\begin{Picture}(-3.5,-0.5)(3.5,10.5)
\cartesiangrid(-3,0)(3,10)
\PlotFunction[10]\F{0.01}{2.5}
\changereferencesystem(0,0)(-1,0)(0,1)
\PlotFunction[20]\F{0.01}{2.5}
\end{Picture}
\end{document}

We define the function as $F(x)=\exp(x\log x)$, and plot it a positive interval.
Then we change of reference system to (0,0)(-1,0)(0,1) and replot the same function, because $F$ is a even function.
 This is the result:

A: First, this is an even function, since if $x>0$, we have $f(-x) = |-x|^{|-x|} = x^x = f(x)$. Thus we only need to concern ourselves with the graph for $x\ge 0$. As @Dario noted, $f'(x) = 0$ when $x = \frac{1}{e}$, at which point $f(x) = e^{-1/e}\approx 0.692$. For $0<x<\frac{1}{e}$, we have $f'(x) < 0$, while for $x>\frac{1}{e}$, $f'(x) > 0$. Thus $f$ decreases from $x=0$ to a minimum at $x=\frac{1}{e}$ and increases thereafter. For $x>0$, the second derivative is always positive (the first term is always nonnegative, and the second term is always positive), so the graph is concave up everywhere. A graph of $f(x)$ is

