The domain of the function
` sqrt(x^2 - 5x +6 ) + sqrt(2x + 8- x^2)` is

Options

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

Answer

To find the domain of the function \( f(x) = \sqrt{x^2 - 5x + 6} + \sqrt{2x + 8 - x^2} \), we need to ensure that both square root expressions are defined and non-negative. This means we need to solve the inequalities for both expressions under the square roots.

### Step 1: Solve the first inequality \( x^2 - 5x + 6 \geq 0 \)

1. **Factor the quadratic**:
\[ x^2 - 5x + 6 = (x - 2)(x - 3) \]

2. **Set the factors to zero** to find the critical points:
\[ (x - 2)(x - 3) = 0 \implies x = 2 \text{ and } x = 3 \]

3. **Determine the intervals**:
The critical points divide the number line into intervals: \( (-\infty, 2) \), \( (2, 3) \), and \( (3, \infty) \).

4. **Test each interval**:
- For \( x < 2 \) (e.g., \( x = 0 \)):
\[ (0 - 2)(0 - 3) = 6 \geq 0 \quad \text{(True)} \]
- For \( 2 < x < 3 \) (e.g., \( x = 2.5 \)):
\[ (2.5 - 2)(2.5 - 3) = -0.25 < 0 \quad \text{(False)} \]
- For \( x > 3 \) (e.g., \( x = 4 \)):
\[ (4 - 2)(4 - 3) = 2 > 0 \quad \text{(True)} \]

5. **Combine the results**:
The solution for the first inequality is:
\[ x \in (-\infty, 2] \cup [3, \infty) \]

### Step 2: Solve the second inequality \( 2x + 8 - x^2 \geq 0 \)

1. **Rearrange the inequality**:
\[ -x^2 + 2x + 8 \geq 0 \implies x^2 - 2x - 8 \leq 0 \]

2. **Factor the quadratic**:
\[ x^2 - 2x - 8 = (x - 4)(x + 2) \]

3. **Set the factors to zero** to find the critical points:
\[ (x - 4)(x + 2) = 0 \implies x = 4 \text{ and } x = -2 \]

4. **Determine the intervals**:
The critical points divide the number line into intervals: \( (-\infty, -2) \), \( (-2, 4) \), and \( (4, \infty) \).

5. **Test each interval**:
- For \( x < -2 \) (e.g., \( x = -3 \)):
\[ (-3 - 4)(-3 + 2) = 7 > 0 \quad \text{(False)} \]
- For \( -2 < x < 4 \) (e.g., \( x = 0 \)):
\[ (0 - 4)(0 + 2) = -8 < 0 \quad \text{(True)} \]
- For \( x > 4 \) (e.g., \( x = 5 \)):
\[ (5 - 4)(5 + 2) = 7 > 0 \quad \text{(False)} \]

6. **Combine the results**:
The solution for the second inequality is:
\[ x \in [-2, 4] \]

### Step 3: Find the intersection of both solutions

Now, we need to find the intersection of the two sets:
1. From the first inequality: \( (-\infty, 2] \cup [3, \infty) \)
2. From the second inequality: \( [-2, 4] \)

**Intersection**:
- The first part \( (-\infty, 2] \) intersects with \( [-2, 4] \) gives \( [-2, 2] \).
- The second part \( [3, \infty) \) intersects with \( [-2, 4] \) gives \( [3, 4] \).

Thus, the final domain of the function is:
\[ \text{Domain} = [-2, 2] \cup [3, 4] \]