Answer :
To determine which statement best describes the function [tex]\( f(x) = -2 \sqrt{x-7} + 1 \)[/tex] in relation to the value [tex]\(-6\)[/tex], let's carefully analyze the domain and range of the function and the given value.
1. Domain Analysis:
The domain of [tex]\( f(x) \)[/tex] is determined by the expression inside the square root [tex]\(\sqrt{x-7}\)[/tex]. For the square root to be defined, the expression inside must be non-negative:
[tex]\[ x - 7 \geq 0 \implies x \geq 7 \][/tex]
Therefore, the domain of [tex]\( f(x) \)[/tex] is [tex]\( [7, \infty) \)[/tex].
2. Range Analysis:
To find the range of [tex]\( f(x) \)[/tex], we evaluate the bounds of the function:
- At the lower boundary of the domain ([tex]\( x = 7 \)[/tex]), the function value is:
[tex]\[ f(7) = -2 \sqrt{7-7} + 1 = -2 \cdot 0 + 1 = 1 \][/tex]
- As [tex]\( x \to \infty \)[/tex], [tex]\( \sqrt{x-7} \to \infty \)[/tex] and hence [tex]\( -2 \sqrt{x-7} \to -\infty \)[/tex], making [tex]\( f(x) \)[/tex] approach [tex]\( -\infty \)[/tex].
Therefore, the range is [tex]\( (-\infty, 1] \)[/tex].
3. Checking [tex]\(-6\)[/tex]:
- Domain Check:
[tex]\(-6 < 7\)[/tex], which means [tex]\(-6\)[/tex] is not in the domain of [tex]\( f(x) \)[/tex].
- Range Check:
We need to check if [tex]\(-6\)[/tex] can be an output of [tex]\( f(x) \)[/tex]. Thus, we solve for [tex]\( x \)[/tex] when [tex]\( f(x) = -6 \)[/tex]:
[tex]\[ -6 = -2 \sqrt{x-7} + 1 \][/tex]
Rearranging the equation:
[tex]\[ -6 - 1 = -2 \sqrt{x-7} \implies -7 = -2 \sqrt{x-7} \implies 7 = 2 \sqrt{x-7} \implies \sqrt{x-7} = \frac{7}{2} \implies x-7 = \left( \frac{7}{2} \right)^2 = \frac{49}{4} \implies x = 7 + \frac{49}{4} = \frac{77}{4} \][/tex]
Since [tex]\( x = \frac{77}{4} = 19.25 \geq 7 \)[/tex], [tex]\(-6\)[/tex] is indeed in the range of [tex]\( f(x) \)[/tex].
Combining the results:
- [tex]\(-6\)[/tex] is not in the domain of [tex]\( f(x) \)[/tex].
- [tex]\(-6\)[/tex] is in the range of [tex]\( f(x) \)[/tex].
Therefore, the best statement describing the function [tex]\( f(x) = -2 \sqrt{x-7} + 1 \)[/tex] is:
-6 is not in the domain of [tex]\( f(x) \)[/tex] but is in the range of [tex]\( f(x) \)[/tex].
1. Domain Analysis:
The domain of [tex]\( f(x) \)[/tex] is determined by the expression inside the square root [tex]\(\sqrt{x-7}\)[/tex]. For the square root to be defined, the expression inside must be non-negative:
[tex]\[ x - 7 \geq 0 \implies x \geq 7 \][/tex]
Therefore, the domain of [tex]\( f(x) \)[/tex] is [tex]\( [7, \infty) \)[/tex].
2. Range Analysis:
To find the range of [tex]\( f(x) \)[/tex], we evaluate the bounds of the function:
- At the lower boundary of the domain ([tex]\( x = 7 \)[/tex]), the function value is:
[tex]\[ f(7) = -2 \sqrt{7-7} + 1 = -2 \cdot 0 + 1 = 1 \][/tex]
- As [tex]\( x \to \infty \)[/tex], [tex]\( \sqrt{x-7} \to \infty \)[/tex] and hence [tex]\( -2 \sqrt{x-7} \to -\infty \)[/tex], making [tex]\( f(x) \)[/tex] approach [tex]\( -\infty \)[/tex].
Therefore, the range is [tex]\( (-\infty, 1] \)[/tex].
3. Checking [tex]\(-6\)[/tex]:
- Domain Check:
[tex]\(-6 < 7\)[/tex], which means [tex]\(-6\)[/tex] is not in the domain of [tex]\( f(x) \)[/tex].
- Range Check:
We need to check if [tex]\(-6\)[/tex] can be an output of [tex]\( f(x) \)[/tex]. Thus, we solve for [tex]\( x \)[/tex] when [tex]\( f(x) = -6 \)[/tex]:
[tex]\[ -6 = -2 \sqrt{x-7} + 1 \][/tex]
Rearranging the equation:
[tex]\[ -6 - 1 = -2 \sqrt{x-7} \implies -7 = -2 \sqrt{x-7} \implies 7 = 2 \sqrt{x-7} \implies \sqrt{x-7} = \frac{7}{2} \implies x-7 = \left( \frac{7}{2} \right)^2 = \frac{49}{4} \implies x = 7 + \frac{49}{4} = \frac{77}{4} \][/tex]
Since [tex]\( x = \frac{77}{4} = 19.25 \geq 7 \)[/tex], [tex]\(-6\)[/tex] is indeed in the range of [tex]\( f(x) \)[/tex].
Combining the results:
- [tex]\(-6\)[/tex] is not in the domain of [tex]\( f(x) \)[/tex].
- [tex]\(-6\)[/tex] is in the range of [tex]\( f(x) \)[/tex].
Therefore, the best statement describing the function [tex]\( f(x) = -2 \sqrt{x-7} + 1 \)[/tex] is:
-6 is not in the domain of [tex]\( f(x) \)[/tex] but is in the range of [tex]\( f(x) \)[/tex].
