To find the domain of the function

$f(x) = \frac{\sqrt{7 - 5x}}{x^2 - 64},$

we need to consider the constraints imposed by both the square root in the numerator and the division by the denominator.

Step-by-Step Analysis:

1. Square Root Constraint: The expression under the square root, $7 - 5x$, must be non-negative (since we cannot take the square root of a negative number in the real number system):

   $7 - 5x \geq 0 \implies x \leq \frac{7}{5}.$

2. Denominator Constraint: The denominator $x^2 - 64$ cannot be zero (since division by zero is undefined):

   $x^2 - 64 \neq 0 \implies x \neq \pm 8.$

3. Combining Constraints: We combine the constraints from both the numerator and the denominator:

   • $x \leq \frac{7}{5}$
   • $x \neq -8, 8$

   The first constraint, $x \leq \frac{7}{5}$, limits the domain to all $x$ values less than or equal to $\frac{7}{5}$. The second constraint removes $x = -8$ and $x = 8$. Since $x = 8$ is already greater than $\frac{7}{5}$, it is not within the domain specified by the first constraint. However, $x = -8$ is less than $\frac{7}{5}$, so we must exclude it from the domain.

4. Interval Notation: The domain of $f(x)$ in interval notation is:

   $(-\infty, -8) \cup (-8, \frac{7}{5}].$

Conclusion:

The domain of the function $f(x) = \frac{\sqrt{7 - 5x}}{x^2 - 64}$ in interval notation is:

$\boxed{(-\infty, -8) \cup (-8, \frac{7}{5}]}.$

Would you like more details or have any questions?

Here are some related questions you might find useful:

1. How do you solve inequalities involving square roots?
2. What is the domain of a function, and how is it determined?
3. How do you handle rational functions when finding the domain?
4. What happens to the graph of a function when there are restrictions on its domain?
5. How would the domain change if there were different expressions under the square root or in the denominator?

Tip: Always check both numerator and denominator constraints when finding the domain of a function that involves both a square root and division!