Question

which statement best describes $f(x) = -2\sqrt{x - 7} + 1$?

• $-6$ is in the domain of $f(x)$ and in the range of $f(x)$.
• $-6$ is neither in the domain of $f(x)$ nor in the range of $f(x)$.
• $-6$ is not in the domain of $f(x)$ but is in the range of $f(x)$.
• $-6$ is in the domain of $f(x)$ but not in the range of $f(x)$.

Explanation:

Step1: Find the domain of \( f(x) = -2\sqrt{x - 7} + 1 \)

For the square root function \( \sqrt{x - 7} \), the expression inside the square root must be non - negative. So we set up the inequality \( x - 7\geq0 \). Solving for \( x \), we add 7 to both sides of the inequality, getting \( x\geq7 \). Since \( - 6<7 \), \( -6 \) is not in the domain of \( f(x) \).

Step2: Find the range of \( f(x) = -2\sqrt{x - 7} + 1 \)

We know that the square root function \( \sqrt{x - 7}\geq0 \) for all \( x \) in its domain (\( x\geq7 \)). Multiply both sides of the inequality \( \sqrt{x - 7}\geq0 \) by \( - 2 \). When we multiply an inequality by a negative number, the direction of the inequality sign changes. So we have \( - 2\sqrt{x - 7}\leq0 \). Then add 1 to both sides of the inequality: \( - 2\sqrt{x - 7}+1\leq1 \). Let's check if \( y = - 6 \) can be in the range. We set \( -2\sqrt{x - 7}+1=-6 \). Subtract 1 from both sides: \( -2\sqrt{x - 7}=-7 \). Divide both sides by \( - 2 \): \( \sqrt{x - 7}=\frac{7}{2} \). Squaring both sides (since both sides are non - negative), we get \( x - 7=\frac{49}{4} \), then \( x=\frac{49}{4}+7=\frac{49 + 28}{4}=\frac{77}{4}=19.25\geq7 \), which is in the domain of \( f(x) \). 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 \(f(x)\) (the third option among the given boxes).