To find the domain of the function $f(x) = \frac{3x}{\sqrt{x + 5}}$, we need to determine the values of $x$ for which the function is defined. There are two conditions to consider:

1. The denominator cannot be zero. So, we need to ensure that $\sqrt{x + 5} \neq 0$.
2. The expression inside the square root must be non-negative, since the square root of a negative number is undefined for real numbers. So, $x + 5 \geq 0$.

Step-by-step:

1. Denominator condition:
   $\sqrt{x + 5} \neq 0 \quad \Rightarrow \quad x + 5 \neq 0 \quad \Rightarrow \quad x \neq -5$

2. Non-negative square root condition:
   $x + 5 \geq 0 \quad \Rightarrow \quad x \geq -5$

Conclusion:

The domain of the function $f(x) = \frac{3x}{\sqrt{x + 5}}$ is all real numbers $x$ such that $x > -5$. In interval notation, the domain is: $(-5, \infty)$

Let me know if you'd like more details or have any questions!

Related Questions:

1. What happens if we have a different function, like $f(x) = \frac{3}{\sqrt{x + 2}}$?
2. How would we handle domain restrictions for rational functions?
3. What is the domain of a function involving cube roots?
4. How can we find the range of functions with square roots?
5. Can a function have more than one domain restriction?

Tip:

When dealing with square roots in the denominator, always check both for non-negativity (inside the root) and that the denominator does not become zero.