Domain of the function `f(x) = log(sqrt(x-4)+sqrt(6-x))`

Options

A. [4,6]
B. `(-oo,6)`
C. (2,3)
D. None of these

Answer

To find the domain of the function \( f(x) = \log(\sqrt{x-4} + \sqrt{6-x}) \), we need to ensure that the argument of the logarithm is positive. This means we need to analyze the expression \( \sqrt{x-4} + \sqrt{6-x} \).

### Step 1: Determine the conditions for the square roots
1. The term \( \sqrt{x-4} \) is defined when \( x - 4 \geq 0 \), which simplifies to:
\[ x \geq 4 \]

2. The term \( \sqrt{6-x} \) is defined when \( 6 - x \geq 0 \), which simplifies to:
\[ x \leq 6 \]

### Step 2: Combine the inequalities
Now, we need to combine the two inequalities:
- From \( \sqrt{x-4} \): \( x \geq 4 \)
- From \( \sqrt{6-x} \): \( x \leq 6 \)

This gives us the combined condition:
\[ 4 \leq x \leq 6 \]

### Step 3: Ensure the argument of the logarithm is positive
Next, we need to ensure that the argument of the logarithm \( \sqrt{x-4} + \sqrt{6-x} \) is greater than 0:
- Since both \( \sqrt{x-4} \) and \( \sqrt{6-x} \) are non-negative for \( x \) in the interval \([4, 6]\), we check the endpoints:
- At \( x = 4 \):
\[ \sqrt{4-4} + \sqrt{6-4} = 0 + \sqrt{2} > 0 \]
- At \( x = 6 \):
\[ \sqrt{6-4} + \sqrt{6-6} = \sqrt{2} + 0 > 0 \]

Since the argument is positive at both endpoints and throughout the interval, we conclude that the logarithm is defined.

### Final Domain
Thus, the domain of the function \( f(x) \) is:
\[ [4, 6] \]