Answer :
To determine the range of the function [tex]\( f(x) = \sqrt{x-2} + 4 \)[/tex], we need to analyze how this function behaves and what values it can take.
1. Domain Analysis:
- The function inside the square root, [tex]\( \sqrt{x-2} \)[/tex], requires that [tex]\( x-2 \geq 0 \)[/tex] because the square root of a negative number is not defined in the set of real numbers.
- Hence, [tex]\( x \geq 2 \)[/tex].
- So, the domain of [tex]\( f(x) \)[/tex] is [tex]\( [2, \infty) \)[/tex].
2. Evaluating [tex]\( f(x) \)[/tex] at the boundary of the domain:
- When [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = \sqrt{2-2} + 4 = \sqrt{0} + 4 = 0 + 4 = 4 \][/tex]
Thus, at [tex]\( x = 2 \)[/tex], [tex]\( f(x) = 4 \)[/tex].
3. Behavior of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] increases:
- As [tex]\( x \)[/tex] increases from 2 onwards, [tex]\( x-2 \)[/tex] is always non-negative, and the square root function increases as [tex]\( x \)[/tex] increases.
- Therefore, [tex]\( \sqrt{x-2} \)[/tex] also increases.
- For any [tex]\( x \geq 2 \)[/tex], [tex]\( \sqrt{x-2} \)[/tex] will produce values starting from 0 up to positive infinity as [tex]\( x \)[/tex] grows larger.
- Adding 4 to [tex]\( \sqrt{x-2} \)[/tex] means [tex]\( f(x) \)[/tex] starts at 4 and increases without bound as [tex]\( x \rightarrow \infty \)[/tex].
4. Conclusion about the range:
- The minimum value of [tex]\( f(x) \)[/tex] is 4 (when [tex]\( x = 2 \)[/tex]).
- As [tex]\( x \)[/tex] grows, [tex]\( f(x) \)[/tex] increases and can take any value greater than or equal to 4.
- Therefore, the range of the function is [tex]\( [4, \infty) \)[/tex].
Looking at the choices provided:
- A. [tex]\([2, \infty)\)[/tex]: This is incorrect; this represents the domain of the function, not the range.
- B. [tex]\(\{-2, \infty\)[/tex]: This is incorrectly formatted and cannot be correct.
- C. [tex]\((4, *)\)[/tex]: This represents an incorrect interval notation and does not specify a lower bound correctly nor an upper bound.
- D. [tex]\((-4,-)\)[/tex]: This does not make sense given the context and is not related to the behavior of this function.
The correct answer is:
[tex]\[ [4, \infty) \][/tex]
1. Domain Analysis:
- The function inside the square root, [tex]\( \sqrt{x-2} \)[/tex], requires that [tex]\( x-2 \geq 0 \)[/tex] because the square root of a negative number is not defined in the set of real numbers.
- Hence, [tex]\( x \geq 2 \)[/tex].
- So, the domain of [tex]\( f(x) \)[/tex] is [tex]\( [2, \infty) \)[/tex].
2. Evaluating [tex]\( f(x) \)[/tex] at the boundary of the domain:
- When [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = \sqrt{2-2} + 4 = \sqrt{0} + 4 = 0 + 4 = 4 \][/tex]
Thus, at [tex]\( x = 2 \)[/tex], [tex]\( f(x) = 4 \)[/tex].
3. Behavior of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] increases:
- As [tex]\( x \)[/tex] increases from 2 onwards, [tex]\( x-2 \)[/tex] is always non-negative, and the square root function increases as [tex]\( x \)[/tex] increases.
- Therefore, [tex]\( \sqrt{x-2} \)[/tex] also increases.
- For any [tex]\( x \geq 2 \)[/tex], [tex]\( \sqrt{x-2} \)[/tex] will produce values starting from 0 up to positive infinity as [tex]\( x \)[/tex] grows larger.
- Adding 4 to [tex]\( \sqrt{x-2} \)[/tex] means [tex]\( f(x) \)[/tex] starts at 4 and increases without bound as [tex]\( x \rightarrow \infty \)[/tex].
4. Conclusion about the range:
- The minimum value of [tex]\( f(x) \)[/tex] is 4 (when [tex]\( x = 2 \)[/tex]).
- As [tex]\( x \)[/tex] grows, [tex]\( f(x) \)[/tex] increases and can take any value greater than or equal to 4.
- Therefore, the range of the function is [tex]\( [4, \infty) \)[/tex].
Looking at the choices provided:
- A. [tex]\([2, \infty)\)[/tex]: This is incorrect; this represents the domain of the function, not the range.
- B. [tex]\(\{-2, \infty\)[/tex]: This is incorrectly formatted and cannot be correct.
- C. [tex]\((4, *)\)[/tex]: This represents an incorrect interval notation and does not specify a lower bound correctly nor an upper bound.
- D. [tex]\((-4,-)\)[/tex]: This does not make sense given the context and is not related to the behavior of this function.
The correct answer is:
[tex]\[ [4, \infty) \][/tex]