Understanding this graph The function is given as $f(x)=2x+3x^\left(2/3\right)$.  The enclosed graph is from wolfram..
My question is why doesn't $f(-1)=1$ show up to the left of the y axis.  I don't see why the domain of this graph is $x>=0.$
 A: First off, the source of the confusion is that WolframAlpha uses the principal root so that 
$$(-1)^{2/3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i,$$
as you can see like so: (-1)^(2/3). 
This is in no way an error on the part of WolframAlpha (as indicated in this comment on Jose's otherwise reasonable answer) but it is a simple design decision.
There are two simple ways to trigger the use of the real root, if you desire.
Getting the real root with a link
If you make your screenshot just a little bit larger, you see the following:

Notice that there is a link that allows you to "use the real-valued root". When you press that, you generate the graph that you expect:

Getting the real root with cbrt
If you know that you want to use the real root, you can use cbrt to denote that from the outset. Thus, cbrt(-1)^2 returns $1$, as you expect and 2x+3cbrt(x)^2 gives the graph that you'd expect.
A: Because when you write $x^{\frac23}$, what you have in mind is (I suppose) $\sqrt[3]{x^2}$. Yes, this is defined for every real number and it is another real number.
However, many computing systems “think” that$$x^{\frac23}=e^{\frac23\ln x}.$$ Since there are no (real) logarithms of negative real numbers, this ledas to a proble, which is reflected in the picture that you posted here.
