What are the domain and range of the real-valued function f(x) = -3 + sqrt 4x-12?
A)The domain is x>3, and the range is f(x)<-3.
B)The domain is x<3, and the range is f(x)<-3.
C)The domain is x>3, and the range is f(x)>-3.
D)The domain is x>3, and the range is all real numbers.
The domain of the function is determined by the restrictions on the square root within the function. Since the square root of a real number must be non-negative, the expression under the square root, 4x - 12, must be greater than or equal to 0:
4x - 12 ≥ 0
4x ≥ 12
x ≥ 3
Therefore, the domain of the function is x ≥ 3.
The range of the function can be found by analyzing the behavior of the function as x approaches infinity and negative infinity. As x approaches negative infinity, the sqrt(4x - 12) approaches sqrt(-∞), which is undefined. Therefore, the function approaches -3. As x approaches positive infinity, the sqrt(4x - 12) approaches infinity, making the function approach infinity as well. The range of the function is all real numbers greater than -3.
Therefore, the correct answer is D) The domain is x ≥ 3, and the range is all real numbers.