Why is the domain of $x^2$ the set of all real numbers? I try to understand, why domain of $x^2$ is the set of all real numbers.
My doubts:
The domain of square root is not defined for negative numbers. Reason to that (If I am not wrong) is that function is supposed to have output with only one input leading to it. Therefore 9 cannot be square-rooted to -3 and 3, only to the positive number (in this case it's 3). Is not the situation with  $x^2$ the same? Should not a domain for it be limited to non-negative numbers? Otherwise, we are having an ambiguity of 3 and -3 leading to 9...
 A: You're right about one thing: the square root function is the inverse function to the squaring function, after the squaring function is restricted to a domain on which it's one-to-one.  That domain is nonnegative numbers.
However, I think you're overthinking this problem.  When a problem asks for the the domain of a function defined by an algebraic expression, the task is to calculate the entire subset of real numbers which can be substituted into the expression.  For instance, if the expression is $\frac{1}{1-x^2}$, you're supposed to notice that the denominator cannot be zero for this to make sense. So the domain must carve out any numbers which do that, namely $\pm 1$.  Therefore the domain is $\mathbb{R}\setminus\{-1,1\}$.
But the expression you're given is just $x^2$.  This is defined for all real numbers.  So the domain is $\mathbb{R}$.
A: The domain is $\Bbb R$, the set of all real numbers. The two inputs $\pm3$ both have output $9$, yes, but this doesn't stop either from being an input: $x^2$ still a function.
A: A function $f$ on a set $X$ is a subset of the cartesian product of $X$ with itself such that if $(x,y) \in f$ and $(x,z) \in f$ then $y=z$. The domain of $f$ is the set $d$ of left coordinates of ordered pairs of $f$. In short, the domain is something you can have the distinct privilege of specifying. 
I.e. if you have a function $f$ and a domain $d$ such that $d$ has more than, (for example, $10$ real numbers). Then you can create a new function $g$ by creating a new domain $d'$ by removing one element from $d$ and setting $g$ equal to the restriction of values of $f$ defined over $d'$.
In short, one domain for the function $f(x)=x^{2}$ is $\mathbb{R}$ because for each real number $x$ in the domain you get a unique real number, (namely $x^2$.)
You might opine that this is a "natural" domain for the function. But then again, we can specify the same equation defining the function $f$ and just restrict our domain set $d$ to be a new domain, $d = \{274848638463926284\} $ instead of all of $\mathbb{R}$ and the result is still a domain for $f(x)=x^{2}$. Although, the graph of this function is a single point in the plane-- which is rather boring compared to a parabola.
A: 
I try to understand, why domain of $x^2$ is the set of all real numbers.

The key idea is to distinguish these two things:


*

*the image of the function, i.e. the "output" of the function, and here you're right: it's not $\mathbb{R}$, but $\mathbb{R}^+$

*the domain of definition, i.e. the set of all authorized numbers as "input" of the function: here all real numbers can be squared. Can $2.317$ be squared? Yes. Can $-3$ be squared? Yes. Indeed: $(-3)^2 = -3 \times (-3) = 9$. Thus the domain of definition of the function $x \mapsto x^2$ is $\mathbb{R}$.
