Determine the range and domain of the function : `f(x)=(3+x^(2))/(2-x)`

Answer

To determine the domain and range of the function \( f(x) = \frac{3 + x^2}{2 - x} \), we will follow a systematic approach.

### Step 1: Finding the Domain

1. **Identify where the function is undefined**: The function is undefined where the denominator is zero. Therefore, we need to solve the equation:
\[ 2 - x = 0 \]
This gives:
\[ x = 2 \]

2. **State the domain**: Since the function is undefined at \( x = 2 \), the domain of \( f(x) \) is all real numbers except 2. In interval notation, this is expressed as:
\[ \text{Domain} = (-\infty, 2) \cup (2, \infty) \]

### Step 2: Finding the Range

1. **Set the function equal to \( y \)**:
\[ y = \frac{3 + x^2}{2 - x} \]

2. **Rearrange the equation**: Multiply both sides by \( (2 - x) \) (noting that \( 2 - x \neq 0 \)):
\[ y(2 - x) = 3 + x^2 \]
Rearranging gives us:
\[ x^2 + xy - 2y + 3 = 0 \]

3. **Identify this as a quadratic in \( x \)**: The equation can be treated as a quadratic equation in \( x \):
\[ x^2 + xy + (3 - 2y) = 0 \]

4. **Use the discriminant to find the range**: For \( x \) to have real solutions, the discriminant must be non-negative:
\[ D = b^2 - 4ac = y^2 - 4(1)(3 - 2y) \geq 0 \]
Simplifying this gives:
\[ D = y^2 + 8y - 12 \geq 0 \]

5. **Solve the quadratic inequality**: We can find the roots of the quadratic equation \( y^2 + 8y - 12 = 0 \) using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{D}}{2a} = \frac{-8 \pm \sqrt{(8)^2 - 4(1)(-12)}}{2(1)} = \frac{-8 \pm \sqrt{64 + 48}}{2} = \frac{-8 \pm \sqrt{112}}{2} \]
Simplifying \( \sqrt{112} = 4\sqrt{7} \):
\[ y = \frac{-8 \pm 4\sqrt{7}}{2} = -4 \pm 2\sqrt{7} \]
Thus, the roots are:
\[ y_1 = -4 - 2\sqrt{7}, \quad y_2 = -4 + 2\sqrt{7} \]

6. **Determine the intervals**: The quadratic \( y^2 + 8y - 12 \) opens upwards (as the coefficient of \( y^2 \) is positive). Therefore, the function is positive outside the roots:
\[ y \in (-\infty, -4 - 2\sqrt{7}) \cup (-4 + 2\sqrt{7}, \infty) \]

### Final Answer

- **Domain**: \( (-\infty, 2) \cup (2, \infty) \)
- **Range**: \( (-\infty, -4 - 2\sqrt{7}) \cup (-4 + 2\sqrt{7}, \infty) \)