Question

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

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

Explanation:

Step1: Check Domain of \( f(x) \)

The function is \( f(x) = -2\sqrt{x - 7} + 1 \). For the square root, the expression inside must be non - negative: \( x - 7\geq0 \), so \( x\geq7 \). Since \( - 6<7 \), \( -6 \) is not in the domain.

Step2: Check Range of \( f(x) \)

Let's set \( y=-2\sqrt{x - 7}+1 \). Solve for \( x \):
\( y - 1=-2\sqrt{x - 7} \)
\( \frac{1 - y}{2}=\sqrt{x - 7} \)
For the square root to be real, \( \frac{1 - y}{2}\geq0 \), so \( 1 - y\geq0 \), \( y\leq1 \). Also, the square root \( \sqrt{x - 7}\geq0 \), so \( - 2\sqrt{x - 7}\leq0 \), \( -2\sqrt{x - 7}+1\leq1 \). Let's see if \( y = - 6 \) is possible:
Set \( -6=-2\sqrt{x - 7}+1 \)
\( -7=-2\sqrt{x - 7} \)
\( \sqrt{x - 7}=\frac{7}{2} \)
\( x - 7=\frac{49}{4} \)
\( x=\frac{49}{4}+7=\frac{49 + 28}{4}=\frac{77}{4}=19.25\geq7 \)
So \( -6 \) is in the range.

Answer:

\( -6 \) is not in the domain of \( f(x) \) but is in the range of \( f(x) \) (the option with this statement, e.g., if the options are labeled as above, the second option: \( -6 \) is not in the domain of \( f(x) \) but is in the range of \( f(x) \))