The function provided is:

$f(x) = \frac{\sqrt{x - 4}}{x^2 - 25}$

To find the domain and range of this function, let's analyze it step by step:

Domain:

The domain consists of all values of $x$ for which the function is defined.

1. Square Root Condition: The expression under the square root, $x - 4$, must be non-negative: $x - 4 \geq 0 \implies x \geq 4.$

2. Denominator Condition: The denominator $x^2 - 25$ cannot be zero: $x^2 - 25 \neq 0 \implies x \neq \pm 5.$

So, the function is defined for $x \geq 4$, except when $x = 5$. Therefore, the domain is: $x \in [4, 5) \cup (5, \infty).$

Range:

For the range, we need to consider how the function behaves for the allowed values of $x$.

1. As $x \to 4^+$, $\sqrt{x - 4}$ approaches 0, so $f(x)$ tends to 0 (but does not actually reach 0 since $x^2 - 25$ is non-zero).

2. As $x$ increases beyond 5, both $\sqrt{x - 4}$ and $x^2 - 25$ increase. However, since the denominator grows faster than the numerator, $f(x)$ approaches 0 from above as $x \to \infty$.

Given these behaviors, the range of $f(x)$ is: $f(x) \in (0, \infty).$

Conclusion:

• Domain: $[4, 5) \cup (5, \infty)$
• Range: $(0, \infty)$

Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. How does the discontinuity at $x = 5$ affect the graph of the function?
2. What is the behavior of the function near the critical points $x = 4$ and $x = 5$?
3. How would you determine if this function is continuous or differentiable over its domain?
4. Can you find the asymptotes of this function?
5. How does the graph of this function compare to simpler rational functions?

Tip: When determining the domain of a function involving square roots and rational expressions, always check for both non-negative conditions under the root and restrictions due to denominators.