what is the domain of $\sqrt[3]{x}$ Find the Domain of $\sqrt[3]{x}$
Can we take negative numbers in domain?
Because $F(x)=x^3$ and $g(x)=\sqrt[3]{x}$ are inverse of each other so their graphs are symmetric about $y=x$. But my book gave graph of $\sqrt[3]{x}$ only in first quadrant? Any reasons
 A: Technically, the domain is integral part of the definition of the function so you don't really find it, it is supposed to be given. However, in this context, when the domain isn't given but asked, it is usually assumed that they want you to give the maximal or natural domain, i.e. the largest subset of $\mathbb{R}$ for which the function can be meaningfully defined.
Since $\sqrt[3]x$ exists for all real numbers $x$, its (maximal/natural) domain is $\mathbb{R}$.

Addendum: why would it make sense to consider a function on a domain smaller than its maximal domain? Think about function inverses. Clearly $x^2$ exists for all real numbers $x$ but the function
$$f : \mathbb{R} \to \mathbb{R} : x \mapsto x^2$$
isn't invertible because it isn't injective (nor surjective). Restricting the domain (and codomain) to $\mathbb{R}^+$ turns it into a function which is invertible:
$$g : \mathbb{R}^+ \to \mathbb{R}^+ : x \mapsto x^2 \quad \mbox{and} \quad g^{-1} : \mathbb{R}^+ \to \mathbb{R}^+ : x \mapsto \sqrt{x}$$
A: Yes, you can take the domain to be all real numbers, by your reasoning that $x^3$ is invertible.  That's fine!  There are also reasons besides differentiability that the authors might not have shown this: because $\sqrt[n]{x}$ can be unambiguously (by convention) defined for positive real numbers and 0, but for negative numbers, it only works for odd values of $n$.  
As soon as you start talking about roots of negative numbers, that opens up a can of worms. (Or a wondrous world of complex numbers and geometry!) Negative real numbers do have square roots over the complex numbers.  But if we want to consider complex numbers, then it makes sense to consider all roots of a number: suddenly $\sqrt[3]x$ can become ambiguous, and any nonzero number has $n$ different nth roots.  For a non-negative real number, we usually define $\sqrt[n]x$ as "the $n$th root that is also a non-negative real number" regardless of $n$, but then for negative real numbers, it's very awkward to say "$\sqrt[n]{x}$ is the negative real number that is an $n$th root of x, if $x$ is odd, and doesn't exist / we'll ignore it if $n$ is even."  So while I'm happy to use that convention and say $\sqrt[3]{-27} = -3$, perhaps your authors wanted to avoid that. :)
A: The cubic root of a real number $x$ is usually defined to be the real number $y$ such that
$$y^3=x.$$
By the observation that
$$(-y)^3=-y^3$$ the definition implies that
$$\sqrt[3]{-x}=-\sqrt[3]x$$ and the function is indeed defined for negatives (and is odd).

Anyway, neither the roots of even order nor the irrational powers of a negative number are defined in the reals, so that in some cases the negative arguments are somehow "ignored".
A: The natural domain of definiton of $\sqrt[3]{x}$ is indeed $\mathbb{R}$ as you've mentioned. However, the authors might have restricted the graph to the first quadrant in order to have this function differentiable as well (as $y=\sqrt[3]{x}$ has a vertical tangent at $x=0$).  
