The domain of the function is the set of all real numbers that make the expression under the square root non-negative. In this case, the expression under the square root must be greater than or equal to zero, so we have:
4x - 12 ≥ 0
4x ≥ 12
x ≥ 3
Therefore, the domain of the function is x ≥ 3.
The range of the function is the set of all possible values the function can take. The function f(x) = -3 + sqrt(4x-12) has its minimum value at x = 3, where the square root term becomes zero. As x increases from 3, the function also increases. Therefore, the range of the function is all real numbers greater than or equal to -3.
In interval notation, the domain is [3, ∞) and the range is [-3, ∞).
What are the domain and range of the real-valued function f(x) = -3 + sqrt 4x-12?
1 answer
1 answer