Assuming that you are looking at a real function of a real variable you can determine the allowed domain as those values of t that produce a real result for g. In this case you need $t^2 + 6t \ge 0$ otherwise you are trying to get the square root of a negative number. Factorising $t(t+6) \ge 0$ with solutions $t \ge 0$ and $t \le -6$. So the allowed domain is $t \le -6$ and $t \ge 0$. The corresponding range is the values that g ranges over given this domain which can be seen to start from 0 (if $t = 0$ or $t=-6)$ and extend to $+\infty$

This changes if you allow g to be a complex function of a real variable, or a complex function of a complex variable.

The domain of a function is also often specified as a subset of the allowed domain, so you might have a function like g restriced by definition to a range $t \ge 0$.

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 $$D_f= D_1 \cap D_2 \cap D_3$$ $$\Leftrightarrow D_f= \mathbb{R} \cap [0,+\infty[ \cap [1,+\infty[$$ $$\Leftrightarrow D_f= [1,+\infty[$$

As you say, the domain of a function is part of its definition. The "definitions" of $f$ and $g$, in the context of the question, are not definitions at all, but merely equations. I would reformulate the question as: what is the largest domain in $\mathbb{R}$ for which this equation defines a function?