To determine the domain and range of the function $f(x) = \sqrt{-3 - 8x}$, we need to follow these steps:

Domain

Since the function involves a square root, the expression inside the square root must be non-negative to avoid complex numbers. Thus, we require: $-3 - 8x \geq 0$ Solving for $x$:

1. Add 3 to both sides: $-8x \geq 3$
2. Divide by -8 (note that dividing by a negative reverses the inequality): $x \leq -\frac{3}{8}$

So, the domain of $x$ is $(-\infty, -\frac{3}{8}]$.

Range

For the range, we analyze the values that $f(x)$ can take based on the square root function.

Since $f(x) = \sqrt{-3 - 8x}$, the square root function always yields non-negative values, starting from 0 (when $x = -\frac{3}{8}$) and increasing as $x$ decreases further.

Thus, the range of $f(x)$ is $[0, \infty)$.

Final Answer

• Domain: $x \in (-\infty, -\frac{3}{8}]$
• Range: $y \in [0, \infty)$

Would you like further details on any part of this solution? Here are some related questions to deepen your understanding:

1. How would the domain change if the function were $f(x) = \sqrt{3 + 8x}$?
2. What would be the range if the square root were replaced by a cube root?
3. How do we determine the domain and range for functions without square roots?
4. Can you explain why dividing by a negative number reverses the inequality sign?
5. How does the presence of a negative coefficient affect the domain of functions with square roots?

Tip: When working with square roots, remember that the expression under the square root must be non-negative to keep the function within real numbers.