Find the domain of f(x).
f, of, x, equals, square root of, 9, x, plus, 45, end square root
f(x)= 
9x+45

​


To find the domain of the function \( f(x) = \sqrt{9x + 45} \), we need to determine where the expression inside the square root is non-negative. This is because the square root function is only defined for values that are greater than or equal to zero.

### Steps:

1. Set the expression inside the square root greater than or equal to zero:

\[
9x + 45 \geq 0
\]

2. Solve for \(x\):

\[
9x \geq -45
\]
\[
x \geq -5
\]

### Conclusion:
The domain of \( f(x) = \sqrt{9x + 45} \) is \( x \geq -5 \). In interval notation, this is:

\[
\boxed{[-5, \infty)}
\]

Would you like further details or have any questions about this?

Here are 5 related questions:
1. How do you find the domain of a rational function?
2. What happens to the domain if the square root function has a negative value inside?
3. How does the domain of \( f(x) = \sqrt{x} \) compare to other square root functions?
4. Can the domain of a function change based on transformations (shifts, stretches)?
5. What is the general rule for the domain of even-root functions?

**Tip:** Always check for values that make the expression inside the square root negative, as they are excluded from the domain.

Learn how to find the domain of the function f(x) = √(9x + 45) by determining where the expression inside the square root is non-negative. This problem explores algebraic concepts like square root functions and domain determination, suitable for students in Grades 8-10.

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