Question

The domain of the function $\sqrt{\frac{x - 7}{9 - x}}$ is

• A. $(7, 9)$
• B. $[7, 9)$
• C. $[7, 9]$
• D. $(7, 9]$

Hint:

For rational inequalities $\frac{x-a}{x-b} \le 0$, the solution is the bounded interval between roots $[a,b]$. Always check the denominator root separately to ensure it is excluded with an open parenthesis.

Answer 1

The Correct Option is B

To find the domain of the function \(\sqrt{\frac{x - 7}{9 - x}}\), we need to ensure that the expression inside the square root is non-negative, as the square root is defined only for non-negative numbers.

The function given is \(\sqrt{\frac{x - 7}{9 - x}}\). For the expression inside the square root to be non-negative, we have:

• \(\frac{x - 7}{9 - x} \geq 0\)

This inequality can be satisfied in two cases:

1. Both \(x - 7 \geq 0\) and \(9 - x \geq 0\) are satisfied.
2. Both \(x - 7 \leq 0\) and \(9 - x \leq 0\) are satisfied.

Let us look at each case:

1. Case 1: \(x - 7 \geq 0\) and \(9 - x \geq 0\)

• From \(x - 7 \geq 0\), we get \(x \geq 7\).
• From \(9 - x \geq 0\), we get \(x \leq 9\).

Thus, in this case, \(x\) must satisfy: \(7 \leq x \leq 9\).

2. Case 2: \(x - 7 \leq 0\) and \(9 - x \leq 0\)

• From \(x - 7 \leq 0\), we get \(x \leq 7\).
• From \(9 - x \leq 0\), we get \(x \geq 9\).

This scenario is impossible because \(x\) cannot simultaneously be less than or equal to 7 and greater than or equal to 9.

Therefore, the only feasible interval for \(x\) is \(7 \leq x < 9\), where:

• \(x = 7\) is allowed because the fraction becomes zero, which is non-negative.
• \(x = 9\) is not allowed because it makes the denominator zero, leading to undefined behavior.

Thus, the domain of the function is \([7, 9)\).

Correct Answer: \([7, 9)\)