which statement best describes $f(x)=-2sqrt{x - 7}+1$?\n-6 is in the domain of $f(x)$ but not in the range…

Explanation:

Step1: Find the domain of (f(x))

For the square - root function (\sqrt{x - 7}), the expression inside the square - root must be non - negative. So (x-7\geq0), which gives (x\geq7). Since (- 6<7), (-6) is not in the domain of (f(x)).

Step2: Find the range of (f(x))

The square - root function (y = \sqrt{x - 7}) has a range of (y\geq0). Then (y=-2\sqrt{x - 7}) has a range of (y\leq0) (because of the negative coefficient (-2)). And (f(x)=-2\sqrt{x - 7}+1) has a range of (y\leq1). Let (f(x)=-6), then (-6=-2\sqrt{x - 7}+1). Rearranging gives (-2\sqrt{x - 7}=-7), or (\sqrt{x - 7}=\frac{7}{2}). Squaring both sides gives (x - 7=\frac{49}{4}), and (x=\frac{49}{4}+7=\frac{49 + 28}{4}=\frac{77}{4}=19.25\geq7). So (-6) is in the range of (f(x)).

Answer:

-6 is not in the domain of (f(x)) but is in the range of (f(x)).