If `f(x)=sqrt((3x-6)/(x+2))+root(4)((x^(4)-5x^3+6x^(2))(1-x^(2)))`, find complex values of x , for which f(x) is real

Answer

To find the complex values of \( x \) for which \( f(x) \) is real, we need to analyze the function given:

\[ f(x) = \sqrt{\frac{3x-6}{x+2}} + \sqrt[4]{(x^4 - 5x^3 + 6x^2)(1 - x^2)} \]

### Step 1: Analyze the first term \( \sqrt{\frac{3x-6}{x+2}} \)

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

\[ \frac{3x-6}{x+2} \geq 0 \]

This inequality can be solved by finding the critical points where the numerator and denominator are zero:

- **Numerator**: \( 3x - 6 = 0 \) implies \( x = 2 \)
- **Denominator**: \( x + 2 = 0 \) implies \( x = -2 \)

Now, we will test the intervals determined by these points: \( (-\infty, -2) \), \( (-2, 2) \), and \( (2, \infty) \).

1. **Interval \( (-\infty, -2) \)**: Choose \( x = -3 \):
\[ \frac{3(-3)-6}{-3+2} = \frac{-9-6}{-1} = \frac{-15}{-1} = 15 \quad (\text{positive}) \]

2. **Interval \( (-2, 2) \)**: Choose \( x = 0 \):
\[ \frac{3(0)-6}{0+2} = \frac{-6}{2} = -3 \quad (\text{negative}) \]

3. **Interval \( (2, \infty) \)**: Choose \( x = 3 \):
\[ \frac{3(3)-6}{3+2} = \frac{9-6}{5} = \frac{3}{5} \quad (\text{positive}) \]

Thus, the first term is real for \( x \in (-\infty, -2) \cup (2, \infty) \).

### Step 2: Analyze the second term \( \sqrt[4]{(x^4 - 5x^3 + 6x^2)(1 - x^2)} \)

For the fourth root to be real, the expression inside must be non-negative:

\[ (x^4 - 5x^3 + 6x^2)(1 - x^2) \geq 0 \]

#### Part A: Analyze \( 1 - x^2 \)

The expression \( 1 - x^2 \) is non-negative when:

\[ -1 \leq x \leq 1 \]

#### Part B: Analyze \( x^4 - 5x^3 + 6x^2 \)

Factor out \( x^2 \):

\[ x^2(x^2 - 5x + 6) = x^2(x-2)(x-3) \]

The roots are \( x = 0, 2, 3 \). The sign of this expression can be analyzed over the intervals determined by these roots: \( (-\infty, 0) \), \( (0, 2) \), \( (2, 3) \), and \( (3, \infty) \).

1. **Interval \( (-\infty, 0) \)**: Choose \( x = -1 \):
\[ (-1)^2(-1-2)(-1-3) = 1 \cdot (-3) \cdot (-4) = 12 \quad (\text{positive}) \]

2. **Interval \( (0, 2) \)**: Choose \( x = 1 \):
\[ 1^2(1-2)(1-3) = 1 \cdot (-1) \cdot (-2) = 2 \quad (\text{positive}) \]

3. **Interval \( (2, 3) \)**: Choose \( x = 2.5 \):
\[ (2.5)^2(2.5-2)(2.5-3) = 6.25 \cdot 0.5 \cdot (-0.5) = -1.5625 \quad (\text{negative}) \]

4. **Interval \( (3, \infty) \)**: Choose \( x = 4 \):
\[ 4^2(4-2)(4-3) = 16 \cdot 2 \cdot 1 = 32 \quad (\text{positive}) \]

Thus, \( x^4 - 5x^3 + 6x^2 \geq 0 \) for \( x \in (-\infty, 0] \cup [2, 3] \cup [3, \infty) \).

### Step 3: Combine conditions

The conditions for \( f(x) \) to be real are:

1. From the first term: \( x \in (-\infty, -2) \cup (2, \infty) \)
2. From the second term: \( x \in (-\infty, 0] \cup [2, 3] \cup [3, \infty) \)

The intersection of these conditions gives us:

- For \( (-\infty, -2) \): valid
- For \( (2, \infty) \): valid only for \( x \in (2, 3) \cup (3, \infty) \)

### Final Result

Thus, the complex values of \( x \) for which \( f(x) \) is real are:

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