The problem asks to determine the domain and range of the function $f(x) = \frac{1}{\sqrt{x - 4}}$.

Domain:

For the function $f(x) = \frac{1}{\sqrt{x - 4}}$, we must ensure that:

1. The expression inside the square root, $x - 4$, must be non-negative, meaning $x - 4 \geq 0$. Therefore, $x \geq 4$.
2. The denominator cannot be zero, meaning $\sqrt{x - 4} \neq 0$, which implies $x - 4 > 0$ or $x > 4$.

Thus, the domain of the function is all $x$ such that $x > 4$.

$\text{Domain: } (4, \infty)$

Range:

For the range of the function:

1. The function $f(x) = \frac{1}{\sqrt{x - 4}}$ is a reciprocal of the square root function, which means $f(x)$ will always be positive (since both the square root and reciprocal are positive).
2. As $x \to 4^+$ (i.e., as $x$ approaches 4 from the right), $\sqrt{x - 4}$ approaches 0, making $f(x) \to \infty$.
3. As $x \to \infty$, $\sqrt{x - 4} \to \infty$, making $f(x) \to 0^+$ (but never reaching zero).

Thus, the range of the function is:

$\text{Range: } (0, \infty)$

Final Answer:

• Domain: $(4, \infty)$
• Range: $(0, \infty)$

Would you like further clarification on any step? Here are some related questions to extend your understanding:

1. What is the domain and range of $f(x) = \frac{1}{\sqrt{4 - x}}$?
2. How does the reciprocal nature of the function affect the range?
3. What happens if we remove the square root from the denominator?
4. Can the function $f(x) = \frac{1}{\sqrt{x-4}}$ be defined for negative values of $x$?
5. How does the function behave as $x \to \infty$?

Tip: Always check both the conditions for square roots and denominators when determining the domain of a function.