To find the domain of the function:

$f(x) = \sqrt{\frac{x - 1}{x + 4}},$

we need to ensure two things:

1. The expression inside the square root must be non-negative (since the square root of a negative number is not defined in the set of real numbers).

   $\frac{x - 1}{x + 4} \geq 0$

2. The denominator $x + 4$ cannot be zero (since division by zero is undefined).

Step 1: Solve the inequality $\frac{x - 1}{x + 4} \geq 0$

To solve this inequality, we analyze when the expression $\frac{x - 1}{x + 4}$ changes sign. We check the critical points where the numerator and denominator are zero:

• $x - 1 = 0 \implies x = 1$,
• $x + 4 = 0 \implies x = -4$.

These critical points divide the number line into intervals. We test the sign of $\frac{x - 1}{x + 4}$ in each interval:

• For $x < -4$, say $x = -5$: $\frac{-5 - 1}{-5 + 4} = \frac{-6}{-1} = 6 > 0.$

• For $-4 < x < 1$, say $x = 0$: $\frac{0 - 1}{0 + 4} = \frac{-1}{4} = -\frac{1}{4} < 0.$

• For $x > 1$, say $x = 2$: $\frac{2 - 1}{2 + 4} = \frac{1}{6} > 0.$

Step 2: Include critical points

• At $x = 1$, $\frac{1 - 1}{1 + 4} = 0$, which is valid because the square root of 0 is 0.
• At $x = -4$, the expression is undefined because the denominator becomes zero.

Conclusion

The function $f(x)$ is defined where $\frac{x - 1}{x + 4} \geq 0$. Based on the sign analysis, the solution to the inequality is:

$x \in (-\infty, -4) \cup [1, \infty)$

Thus, the domain of $f(x)$ is:

$(-\infty, -4) \cup [1, \infty)$

Would you like a more detailed explanation or further examples?

Here are some related questions:

1. How do you solve more complex rational inequalities?
2. What happens to the domain if there is no square root in the function?
3. How can we determine the range of the function?
4. What is the graph of this function?
5. Can we apply this method to other root-based functions?

Tip: Always look for critical points where the numerator or denominator changes sign when solving rational inequalities!