Let's determine the domain and range of the function [tex]\( g(x) = \sqrt{x-3} \)[/tex].
### 1. Determining the Domain:
The domain of a function is the set of all possible input values [tex]\( x \)[/tex] for which the function is defined.
For the function [tex]\( g(x) = \sqrt{x-3} \)[/tex] to be defined, the expression inside the square root must be non-negative because the square root function is only defined for non-negative numbers. This gives us the inequality:
[tex]\[ x - 3 \geq 0 \][/tex]
Solving this inequality:
[tex]\[ x \geq 3 \][/tex]
The domain of the function [tex]\( g(x) = \sqrt{x-3} \)[/tex] is all [tex]\( x \)[/tex] values that are greater than or equal to 3. In interval notation, this is:
[tex]\[ [3, \infty) \][/tex]
### 2. Determining the Range:
The range of a function is the set of all possible output values [tex]\( y \)[/tex] that [tex]\( g(x) \)[/tex] can take.
Since [tex]\( g(x) = \sqrt{x-3} \)[/tex], we see that [tex]\( g(x) \)[/tex] outputs the square root of a non-negative number. The square root function produces non-negative values. Therefore, the smallest value [tex]\( g(x) \)[/tex] can take is 0 (when [tex]\( x = 3 \)[/tex]), and there is no upper limit because as [tex]\( x \)[/tex] increases, [tex]\( g(x) \)[/tex] also increases without bound. Hence, the range is all non-negative real numbers.
In interval notation, the range is:
[tex]\[ [0, \infty) \][/tex]
### Conclusion:
The domain of [tex]\( g(x) = \sqrt{x-3} \)[/tex] is [tex]\( [3, \infty) \)[/tex] and the range is [tex]\( [0, \infty) \)[/tex].
Given the choices:
- D: [tex]\( [3, \infty) \)[/tex] and R: [tex]\( [0, \infty) \)[/tex]
- D: [tex]\( [-3, \infty) \)[/tex] and R: [tex]\( [0, \infty) \)[/tex]
- D: [tex]\( (-3, \infty) \)[/tex] and R: [tex]\( (-\infty, 0) \)[/tex]
- D: [tex]\( (3, \infty) \)[/tex] and R: [tex]\( (-\infty, 0) \)[/tex]
The correct answer is:
[tex]\[ \text{D: } [3, \infty) \text{ and R: } [0, \infty) \][/tex]
