Graph of $y=x^x$ for $x<0$ I have been wondering about the graph of $y=x^x$.  Most graphing calculators will quite happily graph it up to $0$, but after that they don't do anything else.  Basic calculation suggests that, while for some points ($x=-\frac 12$) there are no real number solutions, for others ($x=-1,-2,-3$ etc.) there are solutions.  Why, if at all, do the graphing calculators stop at $0$, and can anyone produce a graph of the real number solutions of $y=x^x$ past the $0$ point?
Thanks.
 A: Note that $f(x)=x^x = e^{x\ln(x)}.$ Since $e^x>0$ for $x\in\mathbb{R}$, there exists no real logarithm of negative real numbers.
However, by the Euler identity $e^{\pi i}=-1$, therefore you could say that "$\ln(-1)=\pi i$", which is a complex number. The problem with that is that the exponential function is periodic, i.e. $e^{(2k+1)\pi i}=-1$ for every odd number $2k+1$, $k\in\mathbb{Z}$. Therefore you could just as well say that "$\ln(-1)=3\pi i, -\pi i, \dots$". See multivalued function, complex logarithm.
The graph which you see in Wolfram Alpha gives you for negative real values the real and imaginary part corresponding to using the principal branch (one choice of values for the logarithm which is somehow canonical) of the logarithm.
A: A more direct answer is the reason your graphing calculator doesn't graph when $x<0$ is because there are infinite undefined "holes" and infinite defined points in the real plane. Even when you restrict the domain to $[-2,-1]$ this will still be the case. 
Note that for $x^x$ when $x<0$ if you calculate for the output of certain x-values (using the Texas I-85) you will have... 
$$x^x=\begin{cases} (-x)^x & x=\left\{ {2n\over 2m+1}\ |\ n, m \in \Bbb Z\right\}\frac{\text{even integer}}{\text{odd integer}}\\ -(-x)^{x} & x=\left\{ {2n+1\over 2m+1}\ |\ n, m \in \Bbb Z\right\}\frac{\text{odd integer}}{\text{odd integer}}\ \\ \text{undefined} & x=\left\{ {2n+1\over 2m}\ |\ n, m \in \Bbb Z\right\}\bigcup \left\{\mathbb{R}\setminus{\mathbb{Q}}\right\} \left(\frac{\text{odd integer}}{\text{even integer}},\text{irrational numbers}\right) \end{cases}$$
(Just remember to simplify fractions all the way until the denominator is a prime number (ex: $2/6\to1/3$))
This is because when we have $x^a$ it can only extend to the negative domain if $a$'s denominator is odd (ex: $x^{1/3},x^{2/3}$). 
Thus there are infinite undefined values from $[-2,-1]$ that are still (even/odd) when simplified. For example $(-3/2,-1/2)$ are undefined but so is $ (-19/10, -17/10, -15/10...-11/10)$ and $(-199/100, -197/100, -195/100,.....-101/100)$. This includes irrational numbers.
There is also infinite defined values. There are infinite defined values that have positive output and infinite defined values that have a negative output. For example there is $(-2,-4/3$), ($-2,-24/13,-22/13,-16/13...-14/13)$ and $(-2,-52/27,-50/27,-48/27,-46/27,-44/27...-28/27)$ that are still positive.
Then there is $(-5/3,-3/3)$, $(-25/13,-23/13,-21/13,-19/13..-13/13)$ and $(-53/27,-51/27,-49/27,-47/27,-45/27,-43/27...-27/27)$ that is negative.
Because the function is so "disconnected" with undefined holes and real numbers the graphing calculator still fails to register a graph of $x^x$ when $x<0$. 
Thus when you see $x^x$ with the three graphs in the piecewise definition note that I am hiding the infinite holes that exist for ${x}^{x}$.
Now since the outputs for the negative domain can be positive or negative we have two "trajectories". Thus we must graph $\left(-x\right)^{x}$ and $-\left(-x\right)^{x}$ with $x^x$.

However, if you want to graph $x^x$ to seem "more continuous" you can either $|x|^{x}$ or $\text{sgn}{\left(x\right)}|x|^{x}$.
A: Graph $y=\left\lvert x\right\rvert^x$ instead.
You can't raise negative real numbers to arbitrary powers and expect to get a real output. For instance $(-1/2)^{-1/2}$ is not real; although there are two nonreal complex numbers this could be interpreted to mean. Expressions like $(-\pi)^{-\pi}$ are even more problematic.
But $y=\left\lvert x\right\rvert^x$ gives an interesting, smooth curve (with a vertical tangent at $(0,1)$.)
