Let's analyze the given function [tex]\( f(x) = -2 \sqrt{x-7} + 1 \)[/tex] to determine if -6 is in its domain and/or range.
1. Domain Analysis:
- The expression inside the square root, [tex]\(x-7\)[/tex], must be greater than or equal to zero since we cannot take the square root of a negative number in the real number system.
- Thus, [tex]\(x - 7 \geq 0\)[/tex], which simplifies to [tex]\(x \geq 7\)[/tex].
- Therefore, the domain of [tex]\(f(x)\)[/tex] is all real numbers [tex]\(x\)[/tex] such that [tex]\(x \geq 7\)[/tex].
2. Range Analysis:
- Now, we need to determine the range of [tex]\(f(x)\)[/tex].
- The expression [tex]\(\sqrt{x-7}\)[/tex] is always non-negative (i.e., [tex]\(\sqrt{x-7} \geq 0\)[/tex]).
- When [tex]\(x = 7\)[/tex], [tex]\(\sqrt{x-7} = \sqrt{0} = 0\)[/tex].
- As [tex]\(x\)[/tex] becomes larger, [tex]\(\sqrt{x-7}\)[/tex] also becomes larger, approaching infinity as [tex]\(x\)[/tex] approaches infinity.
- The function [tex]\(f(x) = -2 \sqrt{x-7} + 1\)[/tex]:
- When [tex]\(x = 7\)[/tex], we have [tex]\(f(7) = -2 \cdot 0 + 1 = 1\)[/tex].
- As [tex]\(x\)[/tex] increases, [tex]\(\sqrt{x-7}\)[/tex] increases, so [tex]\(-2\sqrt{x-7}\)[/tex] decreases (becomes more negative), causing [tex]\(f(x)\)[/tex] to decrease.
- Thus, the function values range from [tex]\(1\)[/tex] downward to [tex]\(-\infty\)[/tex], or [tex]\((-\infty, 1]\)[/tex].
3. Evaluating -6:
- Domain: Since the domain of [tex]\(f(x)\)[/tex] is [tex]\(x \geq 7\)[/tex], [tex]\(-6\)[/tex] is not included in the domain.
- Range: Since the range of [tex]\(f(x)\)[/tex] is [tex]\((-\infty, 1]\)[/tex], [tex]\(-6\)[/tex] is included in the range.
Given this analysis, the statement that best describes [tex]\(f(x)\)[/tex] in relation to [tex]\(-6\)[/tex] is:
"−6 is not in the domain of [tex]\(f(x)\)[/tex] but is in the range of [tex]\(f(x)\)[/tex]."
