To determine the domain and range of the function \( f(x) = -3 - \sqrt{4x - 12} \), we need to analyze the requirements for the function to be defined and the values it can take.
### Domain
1. The term inside the square root \(\sqrt{4x - 12}\) must be non-negative for \(f(x)\) to be a real-valued function. This gives us the condition:
[tex]\[
4x - 12 \geq 0
\][/tex]
2. Solving this inequality:
[tex]\[
4x \geq 12 \implies x \geq 3
\][/tex]
Therefore, the domain of the function is:
[tex]\[
x \geq 3
\][/tex]
### Range
1. For the function \( f(x) = -3 - \sqrt{4x - 12} \):
- The term \(\sqrt{4x - 12}\) represents the square root of a non-negative number, so \(\sqrt{4x - 12} \geq 0\).
- Consequently, \(-\sqrt{4x - 12}\) will be less than or equal to zero, making it non-positive.
2. Thus, \(-3 - \sqrt{4x - 12}\) will also be less than or equal to \(-3\), because we subtract a non-positive number (or zero) from \(-3\):
[tex]\[
f(x) \leq -3
\][/tex]
3. As \( x \) increases from \( 3 \) and onwards:
- When \( x = 3 \), \( \sqrt{4(3) - 12} = \sqrt{0} = 0 \) so \( f(3) = -3 - 0 = -3 \).
- As \( x \) increases further, \( 4x - 12 \) increases, making \(\sqrt{4x - 12}\) increase. Consequently, \(-\sqrt{4x - 12}\) becomes more negative, making \( f(x) \) decrease further from \(-3\).
Hence, the range of the function \( f(x) \) is:
[tex]\[
f(x) \leq -3
\][/tex]
### Conclusion
Summarizing the above findings, the domain of the function is \( x \geq 3 \) and the range is \( f(x) \leq -3 \).
The correct answer from the given choices is:
- The domain is [tex]\( x \geq 3 \)[/tex], and the range is [tex]\( f(x) \leq -3 \)[/tex].
