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 these steps:

### Step 1: Determine the Domain
The domain of a function is the set of all possible input values (x-values) for which the function is defined. The function \( f(x) \) is undefined when the denominator is zero.

1. Set the denominator equal to zero:
\[ 2 - x = 0 \]
Solving for \( x \):
\[ x = 2 \]

2. Therefore, the function is undefined at \( x = 2 \). The domain will be all real numbers except \( 2 \):
\[ \text{Domain} = \mathbb{R} \setminus \{2\} \]

### Step 2: Determine the Range
To find the range, we will express \( f(x) \) in terms of \( y \) and solve for \( x \):

1. Replace \( f(x) \) with \( y \):
\[ y = \frac{3 + x^2}{2 - x} \]

2. Cross-multiply to eliminate the fraction:
\[ y(2 - x) = 3 + x^2 \]
Expanding this gives:
\[ 2y - yx = 3 + x^2 \]

3. Rearranging the equation to form a standard quadratic equation in \( x \):
\[ x^2 + yx + (3 - 2y) = 0 \]

4. For this quadratic equation to have real solutions for \( x \), the discriminant must be non-negative:
\[ b^2 - 4ac \geq 0 \]
Here, \( a = 1 \), \( b = y \), and \( c = 3 - 2y \):
\[ y^2 - 4(1)(3 - 2y) \geq 0 \]
Simplifying this:
\[ y^2 - 12 + 8y \geq 0 \]
Rearranging gives:
\[ y^2 + 8y - 12 \geq 0 \]

5. Now, we can factor or use the quadratic formula to find the roots:
\[ y = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1} \]
\[ y = \frac{-8 \pm \sqrt{64 + 48}}{2} \]
\[ y = \frac{-8 \pm \sqrt{112}}{2} \]
\[ y = \frac{-8 \pm 4\sqrt{7}}{2} \]
\[ y = -4 \pm 2\sqrt{7} \]

6. The roots are:
\[ y_1 = -4 - 2\sqrt{7}, \quad y_2 = -4 + 2\sqrt{7} \]

7. The quadratic \( y^2 + 8y - 12 \) opens upwards (since the coefficient of \( y^2 \) is positive). Therefore, the inequality \( y^2 + 8y - 12 \geq 0 \) holds for:
\[ y \leq -4 - 2\sqrt{7} \quad \text{or} \quad y \geq -4 + 2\sqrt{7} \]

### Final Results
- **Domain**: \( \mathbb{R} \setminus \{2\} \)
- **Range**: \( (-\infty, -4 - 2\sqrt{7}] \cup [-4 + 2\sqrt{7}, \infty) \)