The domain of the real valued function ` sqrt((x+2) (5-x) ) - (1)/( sqrt(x^2 -4))` is

Options

A. `[1,2] uu [2,5]`
B. `(2,5]`
C. `[3,4]`
D. `[3,5]`

Answer

To find the domain of the function \( f(x) = \sqrt{(x+2)(5-x)} - \frac{1}{\sqrt{x^2 - 4}} \), we need to ensure that both parts of the function are defined and yield real values. This involves addressing the conditions for the square roots and the denominator.

### Step 1: Analyze the first part \( \sqrt{(x+2)(5-x)} \)

For the square root to be defined, the expression inside must be non-negative:

\[ (x+2)(5-x) \geq 0 \]

**Finding the critical points:**
Set the expression equal to zero:

\[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \]
\[ 5 - x = 0 \quad \Rightarrow \quad x = 5 \]

**Testing intervals:**
We will test the intervals determined by the critical points \( -2 \) and \( 5 \):

1. **Interval \( (-\infty, -2) \)**: Choose \( x = -3 \)
\[ (-3 + 2)(5 - (-3)) = (-1)(8) < 0 \quad \text{(not valid)} \]

2. **Interval \( (-2, 5) \)**: Choose \( x = 0 \)
\[ (0 + 2)(5 - 0) = (2)(5) > 0 \quad \text{(valid)} \]

3. **Interval \( (5, \infty) \)**: Choose \( x = 6 \)
\[ (6 + 2)(5 - 6) = (8)(-1) < 0 \quad \text{(not valid)} \]

**Conclusion for the first part:**
The valid interval for \( (x+2)(5-x) \geq 0 \) is:

\[ [-2, 5] \]

### Step 2: Analyze the second part \( \frac{1}{\sqrt{x^2 - 4}} \)

For this part, the expression inside the square root must be positive (since it is in the denominator):

\[ x^2 - 4 > 0 \]

**Finding the critical points:**
Set the expression equal to zero:

\[ x^2 - 4 = 0 \quad \Rightarrow \quad x = \pm 2 \]

**Testing intervals:**
We will test the intervals determined by the critical points \( -2 \) and \( 2 \):

1. **Interval \( (-\infty, -2) \)**: Choose \( x = -3 \)
\[ (-3)^2 - 4 = 9 - 4 = 5 > 0 \quad \text{(valid)} \]

2. **Interval \( (-2, 2) \)**: Choose \( x = 0 \)
\[ 0^2 - 4 = -4 < 0 \quad \text{(not valid)} \]

3. **Interval \( (2, \infty) \)**: Choose \( x = 3 \)
\[ 3^2 - 4 = 9 - 4 = 5 > 0 \quad \text{(valid)} \]

**Conclusion for the second part:**
The valid intervals for \( x^2 - 4 > 0 \) are:

\[ (-\infty, -2) \cup (2, \infty) \]

### Step 3: Find the intersection of the two intervals

Now, we need to find the intersection of the intervals from both parts:

1. From the first part: \( [-2, 5] \)
2. From the second part: \( (-\infty, -2) \cup (2, \infty) \)

**Finding the intersection:**
- The intersection with \( (-\infty, -2) \) gives us no valid values since \( -2 \) is not included.
- The intersection with \( (2, \infty) \) gives us \( (2, 5] \).

### Final Domain

Thus, the domain of the function is:

\[ (2, 5] \]