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.