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

Answer

To find the domain of the function \( f(x) = \sqrt{x^2 - 5x + 6} + \sqrt{2x + 8 - x^2} \), we need to determine the values of \( x \) for which both square root expressions are defined and non-negative.

### Step 1: Determine the domain of \( f_1(x) = \sqrt{x^2 - 5x + 6} \)

1. The expression inside the square root must be non-negative:
\[ x^2 - 5x + 6 \geq 0 \]

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

3. Find the critical points where the expression equals zero:
\[ (x - 2)(x - 3) = 0 \implies x = 2 \text{ and } x = 3 \]

4. Analyze the sign of the expression on the number line:
- Test intervals: \( (-\infty, 2) \), \( (2, 3) \), \( (3, \infty) \)
- For \( x < 2 \): both factors are negative, so the product is positive.
- For \( 2 < x < 3 \): one factor is negative and the other is positive, so the product is negative.
- For \( x > 3 \): both factors are positive, so the product is positive.

5. Therefore, the solution to \( (x - 2)(x - 3) \geq 0 \) is:
\[ x \in (-\infty, 2] \cup [3, \infty) \]

### Step 2: Determine the domain of \( f_2(x) = \sqrt{2x + 8 - x^2} \)

1. The expression inside the square root must be non-negative:
\[ 2x + 8 - x^2 \geq 0 \]

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

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

4. Find the critical points where the expression equals zero:
\[ (x - 4)(x + 2) = 0 \implies x = -2 \text{ and } x = 4 \]

5. Analyze the sign of the expression on the number line:
- Test intervals: \( (-\infty, -2) \), \( (-2, 4) \), \( (4, \infty) \)
- For \( x < -2 \): both factors are negative, so the product is positive.
- For \( -2 < x < 4 \): one factor is negative and the other is positive, so the product is negative.
- For \( x > 4 \): both factors are positive, so the product is positive.

6. Therefore, the solution to \( (x - 4)(x + 2) \leq 0 \) is:
\[ x \in [-2, 4] \]

### Step 3: Find the intersection of the domains

Now we need to find the intersection of the two domains:
- Domain of \( f_1(x) \): \( (-\infty, 2] \cup [3, \infty) \)
- Domain of \( f_2(x) \): \( [-2, 4] \)

The intersection is:
- From \( (-\infty, 2] \) and \( [-2, 4] \): \( [-2, 2] \)
- From \( [3, \infty) \) and \( [-2, 4] \): \( [3, 4] \)

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