The domain of definition of function
`f(x)=log(sqrt(x^(2)-5x-24)-x-2)`, is

Options

A. `(-infty,-3]`
B. `(-infty,-3] cup [8, infty)`
C. `(-infty,(-28)/9)`
D. None of these

Answer

To find the domain of the function \( f(x) = \log(\sqrt{x^2 - 5x - 24} - x - 2) \), we need to ensure two conditions are satisfied:

1. The expression inside the square root must be non-negative.
2. The expression inside the logarithm must be positive.

### Step 1: Ensure the expression inside the square root is non-negative
We start with the inequality:
\[ x^2 - 5x - 24 \geq 0 \]
To solve this quadratic inequality, we first find the roots of the equation \( x^2 - 5x - 24 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot (-24)}}{2 \cdot 1} = \frac{5 \pm \sqrt{25 + 96}}{2} = \frac{5 \pm \sqrt{121}}{2} = \frac{5 \pm 11}{2} \]
This gives us the roots:
\[ x = \frac{16}{2} = 8 \quad \text{and} \quad x = \frac{-6}{2} = -3 \]

Next, we analyze the sign of the quadratic expression \( x^2 - 5x - 24 \) in the intervals determined by the roots:
- For \( x < -3 \): Choose \( x = -4 \) → \( (-4)^2 - 5(-4) - 24 = 16 + 20 - 24 = 12 \) (positive)
- For \( -3 < x < 8 \): Choose \( x = 0 \) → \( 0^2 - 5(0) - 24 = -24 \) (negative)
- For \( x > 8 \): Choose \( x = 9 \) → \( 9^2 - 5(9) - 24 = 81 - 45 - 24 = 12 \) (positive)

Thus, the solution to \( x^2 - 5x - 24 \geq 0 \) is:
\[ (-\infty, -3] \cup [8, \infty) \]

### Step 2: Ensure the expression inside the logarithm is positive
Next, we need to ensure:
\[ \sqrt{x^2 - 5x - 24} - x - 2 > 0 \]
This can be rewritten as:
\[ \sqrt{x^2 - 5x - 24} > x + 2 \]
Squaring both sides (noting that both sides must be non-negative):
\[ x^2 - 5x - 24 > (x + 2)^2 \]
Expanding the right-hand side:
\[ x^2 - 5x - 24 > x^2 + 4x + 4 \]
Cancelling \( x^2 \) from both sides:
\[ -5x - 24 > 4x + 4 \]
Rearranging gives:
\[ -5x - 4x > 4 + 24 \implies -9x > 28 \implies x < -\frac{28}{9} \]

### Step 3: Find the intersection of the intervals
Now we have two conditions:
1. \( x \in (-\infty, -3] \cup [8, \infty) \)
2. \( x < -\frac{28}{9} \)

To find the intersection:
- The interval \( (-\infty, -3] \) intersects with \( x < -\frac{28}{9} \) to give \( (-\infty, -\frac{28}{9}) \) since \( -\frac{28}{9} \approx -3.11 \) which is less than -3.
- The interval \( [8, \infty) \) does not intersect with \( x < -\frac{28}{9} \).

Thus, the final domain of the function \( f(x) \) is:
\[ \boxed{(-\infty, -\frac{28}{9})} \]