why is the range of the function $\sqrt x$ the set of numbers $\geq 0$ I have tried looking for an explanation why the range of $$\sqrt x$$ is the set of non-negative real numbers, but I have found none.

I think the range should include negative real numbers because $$\sqrt 4= \pm 2 \ .$$

So why is it so?
Let me tell you first that
$\sqrt4$ is always $2$ not $\pm \,2$. Square root of a positive no. Is always positive.

Like $\sqrt{16}$ is just $4$, not $\pm \, 4$.

And we develop misconception from here. 
In school we are taught to solve ques like
$$x^2 = 4$$
Then we take root both side,

$$x = \sqrt4$$
And then we write, $x = \pm\,2$
Although the above solution is correct but the way it is done is wrong.

The right method is
$$x^2 = 4$$
Taking root both side 
$$\sqrt{x^2} = \sqrt4,$$
$$|\,X\,| = 2$$
Then we come to result that
$x= \pm\,2$
So in your $f(x) : \sqrt x$ , 
 x can't take negative values
And if it takes positive values then answer will be surely a positive number.

I hope you cleared your misconception today that $\sqrt4$ is just $2$.
A function is a relation from a domain to a codomain such that every element of the domain is related to exactly one element of the codomain.