Let's analyze the given problem.

The function is: $f(x) = -8\sqrt{2x + 1} - 9$

We need to determine its domain and range in interval notation.

Step 1: Determine the Domain

The domain is determined by the values of $x$ for which the function is defined. Since there is a square root, the expression inside the square root ($2x + 1$) must be non-negative. Thus: $2x + 1 \geq 0$

Solve for $x$: $x \geq -\frac{1}{2}$

So, the domain is: $\boxed{[-\frac{1}{2}, \infty)}$

Step 2: Determine the Range

The range is determined by the possible values of $f(x)$. First, consider the square root term, $\sqrt{2x + 1}$:

• Since $x \geq -\frac{1}{2}$, the smallest value of $2x + 1$ is $0$, meaning $\sqrt{2x + 1} \geq 0$.

Now substitute the behavior of $\sqrt{2x + 1}$ into the function: $f(x) = -8\sqrt{2x + 1} - 9$

1. When $x = -\frac{1}{2}$, $\sqrt{2x + 1} = 0$, so: $f\left(-\frac{1}{2}\right) = -8(0) - 9 = -9$

2. As $x$ increases, $\sqrt{2x + 1}$ increases, which means $-8\sqrt{2x + 1}$ decreases. Thus, $f(x)$ decreases without bound.

Hence, the range is: $\boxed{(-\infty, -9]}$

Final Answer:

• Domain in interval notation: $\boxed{[-\frac{1}{2}, \infty)}$
• Range in interval notation: $\boxed{(-\infty, -9]}$

Would you like a detailed explanation of any step?
Here are some follow-up questions to consider:

1. What happens to the domain if the square root is replaced by a cube root?
2. How would the range change if the coefficient of the square root was positive instead of negative?
3. Can a domain ever be all real numbers if a square root is present in the function?
4. How does the subtraction of $-9$ affect the range of the function?
5. What happens to the domain if we add a restriction such as $x < 10$?

Tip: Always check constraints such as square roots or divisions to determine the domain!