The task involves calculating the domain of the given functions. Let's analyze each function step by step.

a) $\frac{3}{x^2 - 25}$

The domain of a rational function is all real numbers except where the denominator is zero. For this function: $x^2 - 25 = 0$ Factorize: $(x - 5)(x + 5) = 0$ The values $x = 5$ and $x = -5$ make the denominator zero.

Domain: $x \in \mathbb{R} \setminus \{-5, 5\}$

b) $\frac{4 + x}{\sqrt{x}}$

The domain is restricted by:

1. The denominator $\sqrt{x}$ must not be zero: $x > 0$ (since $\sqrt{x}$ is undefined for $x \leq 0$).
2. The numerator $4 + x$ has no restrictions.

Domain: $x > 0$

c) $\frac{4 - x}{x^2 - x - 20}$

The domain is restricted by the denominator: $x^2 - x - 20 = 0$ Factorize: $(x - 5)(x + 4) = 0$ The values $x = 5$ and $x = -4$ make the denominator zero.

Domain: $x \in \mathbb{R} \setminus \{-4, 5\}$

Final Answer

1. $a(x): x \in \mathbb{R} \setminus \{-5, 5\}$
2. $b(x): x > 0$
3. $c(x): x \in \mathbb{R} \setminus \{-4, 5\}$

Would you like detailed explanations for any specific part?

Related Questions:

1. How do you handle square roots in determining domains?
2. What happens to the domain if the numerator is zero?
3. Can you generalize domain restrictions for rational functions?
4. How do absolute values affect the domain of a function?
5. Why is the square root undefined for negative numbers?

Tip:

Always check for zeros in the denominator and ensure the argument of square roots is non-negative when determining domains!