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To find the domain of the function \( f(x) = \sqrt{32 - 8x} \), we need to ensure that the expression inside the square root is non-negative. This is because the square root of a negative number is not defined in the set of real numbers.

### Steps:
1. Set the expression inside the square root to be greater than or equal to zero:
   \[
   32 - 8x \geq 0
   \]
2. Solve for \(x\):
   \[
   32 \geq 8x
   \]
   Divide both sides by 8:
   \[
   4 \geq x \quad \text{or} \quad x \leq 4
   \]

Thus, the domain of the function is all values of \(x\) such that \(x \leq 4\).

### Interval Notation:
The domain is \( (-\infty, 4] \).

Would you like further details on this process or have any questions?

Here are 5 follow-up questions:
1. What would the domain be if the square root expression was \( \sqrt{32 + 8x} \)?
2. How does the domain change if the function had a cube root instead of a square root?
3. What happens to the domain if there is an additional term outside the square root, like \( f(x) = 5 + \sqrt{32 - 8x} \)?
4. What is the domain of a function without any square roots?
5. How would the domain change if the square root was over a fraction?

### Tip:
Always check the inside of the square root for non-negative values when finding domains of functions involving square roots.

Find the domain of the function f(x) = sqrt(32 - 8x). Write your answer using interval notation.

Learn how to find the domain of the function f(x) = sqrt(32 - 8x). This problem involves understanding the square root function and determining the values of x that make the expression inside the square root non-negative. Suitable for students in Grades 9-11.

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