The largest set of real values of `x` for which `f(x)=sqrt((x + 2)(5-x))-1/sqrt(x^2-4)`- is a real function is

Options

A. `[1, 2) cup (2, 5]`
B. (2, 5]
C. [3, 4]
D. `[(1)/(sqrt(2)), 1]`

Answer

To find the largest set of real values of \( x \) for which the function

\[ f(x) = \sqrt{(x + 2)(5 - x)} - \frac{1}{\sqrt{x^2 - 4}} \]

is defined, we need to analyze the conditions under which both terms of the function are real.

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

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

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

To solve this inequality, we find the roots of the equation:

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

Now we can test the intervals defined by these roots:

1. \( (-\infty, -2) \)
2. \( [-2, 5] \)
3. \( (5, \infty) \)

Testing these intervals:

- For \( x < -2 \) (e.g., \( x = -3 \)):
\[ (-3 + 2)(5 - (-3)) = (-1)(8) < 0 \quad \text{(not valid)} \]

- For \( -2 \leq x \leq 5 \) (e.g., \( x = 0 \)):
\[ (0 + 2)(5 - 0) = (2)(5) = 10 \geq 0 \quad \text{(valid)} \]

- For \( x > 5 \) (e.g., \( x = 6 \)):
\[ (6 + 2)(5 - 6) = (8)(-1) < 0 \quad \text{(not valid)} \]

Thus, the first term is valid for:

\[ -2 \leq x \leq 5 \]

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

For this term to be defined, the expression under the square root must be positive:

\[ x^2 - 4 > 0 \]

This can be factored as:

\[ (x - 2)(x + 2) > 0 \]

Finding the roots:

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

Testing the intervals defined by these roots:

1. \( (-\infty, -2) \)
2. \( (-2, 2) \)
3. \( (2, \infty) \)

Testing these intervals:

- For \( x < -2 \) (e.g., \( x = -3 \)):
\[ (-3 - 2)(-3 + 2) = (-5)(-1) > 0 \quad \text{(valid)} \]

- For \( -2 < x < 2 \) (e.g., \( x = 0 \)):
\[ (0 - 2)(0 + 2) = (-2)(2) < 0 \quad \text{(not valid)} \]

- For \( x > 2 \) (e.g., \( x = 3 \)):
\[ (3 - 2)(3 + 2) = (1)(5) > 0 \quad \text{(valid)} \]

Thus, the second term is valid for:

\[ x < -2 \quad \text{or} \quad x > 2 \]

### Step 3: Combine the intervals

Now we need to find the intersection of the two sets of valid \( x \):

1. From the first term: \( -2 \leq x \leq 5 \)
2. From the second term: \( x < -2 \) or \( x > 2 \)

The intersection of these conditions gives us:

- For \( x < -2 \): This does not intersect with \( -2 \leq x \leq 5 \).
- For \( x > 2 \): The intersection with \( -2 \leq x \leq 5 \) is \( 2 < x \leq 5 \).

### Final Result

Thus, the largest set of real values of \( x \) for which \( f(x) \) is defined is:

\[ (2, 5] \]