Question

What are the domain and range of the real-valued function f(x) = - 3+ √ (4x - 12)?-O The domain is x <= 3, and the range is f(x) => - 3.The domain is x => 3, and the range is f(x) = -3.O The domain is x => 3, and the range is all real numbers.O The domain is x => 3, and the range is f(x) => – 3.

Answer 1

ANSWER:

4th option: The domain is x ≥ 3, and the range is f(x) ≥ – 3.

STEP-BY-STEP EXPLANATION:

We have the following equation:

$f\mleft(x\mright)=-3+\sqrt{4x-12}$

The domain is the range of x values and the range is the range of y values.

There can be no negative values inside the square root therefore:

$\begin{gathered} 4x-12=0 \\ 4x=12 \\ x=\frac{12}{4}=3 \\ \text{ Therefore:} \\ \text{ Domain: }x\ge\:3\: \end{gathered}$

Since the minimum number that the root can take is 0, the range would be:

$\text{ Range: }f\left(x\right)\ge\:-3$

Therefore, the correct answer is the 4th option: The domain is x ≥ 3, and the range is f(x) ≥ – 3.