Given that [tex]f(x) = \sqrt{7-x}[/tex] and [tex]g(x) = \sqrt{x + 2}[/tex], we can say the following:
[tex]\Bigg(\dfrac{f}{g}\Bigg)(x) = \dfrac{f (x)}{g(x)} = \dfrac{\sqrt{7 - x}}{\sqrt{x+2}}[/tex]
Now, remember what happens if we have a negative square root: it becomes an imaginary number. We don't want this, so we want to make sure whatever is under a square root is greater than 0 (given we are talking about real numbers only).
Thus, let's set what is under both square roots to be greater than 0:
[tex]\sqrt{7 - x} \Rightarrow 7 - x \geq 0 \Rightarrow x \leq 7[/tex]
[tex]\sqrt{x + 2} \Rightarrow x + 2 \geq 0 \Rightarrow x \geq -2[/tex]
Since both of the square roots are in the same function, we want to take the union of the domains of the individual square roots to find the domain of the overall function.
[tex]x \leq 7 \,\,\cup x \geq -2 = \boxed{-2 \leq x \leq 7}[/tex]
Now, let's look back at the function entirely, which is:
[tex]\Bigg( \dfrac{f}{g} \Bigg)(x) = \dfrac{\sqrt{7 - x}}{\sqrt{x+2}}[/tex]
Since [tex]\sqrt{x + 2}[/tex] is on the bottom of the fraction, we must say that [tex]\sqrt{x + 2} \neq 0[/tex], since the denominator can't equal 0. Thus, we must exclude [tex]\sqrt{x + 2} = 0 \Rightarrow x + 2 = 0 \Rightarrow x = -2[/tex] from the domain.
Thus, our answer is Choice C, or [tex]\boxed{ \{ x | -2 < x \leq 7 \}}[/tex].
If you are wondering why the choices begin with the [tex]x |[/tex] symbol, it is because this is a way of representing that [tex]x[/tex] lies within a particular set.
