To determine the domain and range of the function [tex]\( g(x) = \sqrt{x-3} \)[/tex], let's go through the following steps:
1. Domain Determination:
- The domain of a function refers to all the possible input values (x-values) that make the function defined.
- For the function [tex]\( g(x) = \sqrt{x-3} \)[/tex], we are dealing with a square root. The expression inside the square root, [tex]\( x-3 \)[/tex], must be non-negative because the square root of a negative number is not a real number.
- This means we need [tex]\( x-3 \geq 0 \)[/tex].
- Solving for [tex]\( x \)[/tex], we get [tex]\( x \geq 3 \)[/tex].
- Therefore, the domain of [tex]\( g(x) \)[/tex] is all values of [tex]\( x \)[/tex] that are greater than or equal to 3, which we can write in interval notation as [tex]\( [3, \infty) \)[/tex].
2. Range Determination:
- The range of a function refers to all possible output values (y-values).
- For [tex]\( g(x) = \sqrt{x-3} \)[/tex], we should consider how the square root function behaves.
- Since the square root function only yields non-negative results (i.e., [tex]\(\sqrt{y}\)[/tex] is always [tex]\( \geq 0 \)[/tex] for real [tex]\( y \)[/tex]), the output [tex]\( g(x) \)[/tex] can only be non-negative.
- The smallest value inside the square root occurs when [tex]\( x = 3 \)[/tex], giving [tex]\( \sqrt{3-3} = \sqrt{0} = 0 \)[/tex].
- As [tex]\( x \)[/tex] increases beyond 3, the value of [tex]\( \sqrt{x-3} \)[/tex] will increase without bound.
- Hence, the range of [tex]\( g(x) \)[/tex] is all non-negative real numbers, which can be written in interval notation as [tex]\( [0, \infty) \)[/tex].
Given these steps, we can see:
- The domain [tex]\( D \)[/tex] is [tex]\( [3, \infty) \)[/tex]
- The range [tex]\( R \)[/tex] is [tex]\( [0, \infty) \)[/tex]
Thus, the correct answer is:
- [tex]\( D:[3, \infty) \)[/tex] and [tex]\( R:[0, \infty) \)[/tex]
