We have $$f(x) = \frac{(\sqrt{x}-\sqrt{x-1} )}{( \sqrt{x}+\sqrt{x-1} )}$$ The domain of $f$ is: $$D_f = \{ x \in \mathbb{R} : (\sqrt{x}+\sqrt{x-1} \ne 0) \wedge (x \ge 0) \wedge (x-1 \ge 0) \}$$

• Let we consider the first inequality: $\sqrt{x}-\sqrt{x-1} \ne 0$
  To make the explanation clearer let we consider to negation: $$\sqrt{x}+\sqrt{x-1} = 0 \Leftrightarrow \sqrt{x-1} = -\sqrt{x}$$ Because $(\forall x \in \mathbb{R}): \sqrt{x} \ge 0 \Rightarrow \sqrt{x-1} = -\sqrt{x}$ is not solvable($\sqrt{x-1}$ can not be negative)
  The solution is $\emptyset$, because we considered the negation, so we must negate it again what result $\mathbb{R}$ Let $D_1$ denotes the first solution set, so $D_1 = \mathbb{R}$
• Now let consider the second inequality: $x \ge 0$
  This inequality is already solved. In analogue to the first case let $D_2$ denotes the second solution set, so $D_2 = [0,+\infty[$
• Now let consider the last inequality: $ x-1 \ge 0 $
  $ x-1 \ge 0 \Leftrightarrow x \ge 1 \Leftrightarrow D_3 = [1,+\infty[$
  The whole solution $$D_f= D_1 \cap D_2 \cap D_3$$ $$\Leftrightarrow D_f= \mathbb{R} \cap [0,+\infty[ \cap [1,+\infty[$$ $$\Leftrightarrow D_f= [1,+\infty[$$

Yes, absolutely, it should be defined. I had the same experience with my high-school teacher who actually got mad. I do not remember exactly, but apparently it is assumed (without stating) that the domain is the largest possible set on which the (rational, real, complex?) function is defined. I mean, something like $\sqrt{x-4}$ could als be defined on $[2014,\infty)$. So always ask for a specification of the domain.