Inclusivity of the domain of the derivative of a function with a vertical tangent

Question

My question stems from an argument between a teacher and myself. The argument came from a debate of whether the domain of the derivative of a function similar to $\sqrt x$

Anyways my question is when writing the domain of the derivative, with respect to the original function, would it be (0, +$\infty$) or [0,+$\infty$) when using interval notation?

My thoughts on this are because the derivative at 0 tends towards $\infty$ when approaching from the right meaning that there is no evaluable answer making it ambiguous and since the derivative is ambiguous at the point f'(0) then the domain cannot be inclusive of that point (for the same reason you can't be inclusive with infinity) As well many places will state that a vertical tangent is non differentiable and a derivative does not exist.

After searching for an answer to this both my teacher and I found some cases where people were arguing similar questions but none actually answered the question.

The question in class that we were specifically looking at was $y = x \sqrt(4-x)$ and we were looking at where the derivative is negative and the specific notation on the board that caused the debate was as follows: the interval where y'<0 fits the domain of ($\frac{8}{3}$,4)

Answer 1

$f'(a) = \lim_\limits{x\to a} \frac {f(x) - f(a)}{x-a}$

if $f:\mathbb R_{\ge0} \to \mathbb R_{\ge 0}; f(x) = \sqrt x$

That is $f(x)$ is defined over the non-negative real numbers.

Question 1: does $\lim_\limits{x\to 0} \sqrt{x}$ exist.

Since there are no negative values of $x$ in the domain, we do not consider the right hand limit when evaluating whether the limit exists. That is, in the definition of limits, we consider all $x$ in the domain of $f(x)$ in the neighborhood of $0$.

$\lim_\limits{x\to 0} \sqrt{x} = 0$

Using similar logic, I am inclined to say:

$f'(0) = \lim_\limits{x\to 0} \frac {\sqrt x}{x} = \infty $

Answer 2

The definition of derivative that I like says that the limit of the difference quotient must exist and be finite. So my answer would be $(0, \infty)$.

Others might allow "infinite" derivatives, but even then, there's a tricky question: for a function defined on a closed interval, does the limit in the difference quotient have to exist as a limit from both sides, or is one side sufficient at the endpoints? Again, this is a matter of definition. If you say that one-sided limits are OK whenever you're at an edge of a closed interval, then you've got a slight technical problem: every function on a degenerate interval like $[1,1]$ ends up being characterized as differentiable. And the restriction of a non-differentiable function (like $f(x) = |x|$) to such an interval (like $[0,0]$) ends up being differentiable there.

So which definition is "right"? Whichever one you've agreed to use with the people you're talking with. Your textbook, if it's a good one, will have a clear and unambiguous definition. On the other hand, if it's like most calculus books, it'll have a vague and somewhat opaque definition. (Sigh.) In that case, there's no right answer -- just a rotten text.

I'll repeat my usual suggestion here: get a copy of Spivak's Calculus, and smile at how very well-written it is.