Find the Domain & Range of the following functions :
(b) `f(x)=(x)/(1-x^2)`

Answer

To find the domain and range of the function \( f(x) = \frac{x}{1 - x^2} \), we will follow a systematic approach.

### Step 1: Finding the Domain

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

2. **Solve for \( x \)**:
\[ x^2 = 1 \]
Taking the square root of both sides gives:
\[ x = \pm 1 \]

3. **Determine the domain**: The function is defined for all real numbers except \( x = 1 \) and \( x = -1 \). Thus, the domain of \( f(x) \) is:
\[ \text{Domain} = \mathbb{R} \setminus \{-1, 1\} = (-\infty, -1) \cup (-1, 1) \cup (1, \infty) \]

### Step 2: Finding the Range

1. **Set the function equal to \( y \)**:
\[ y = \frac{x}{1 - x^2} \]
Rearranging gives:
\[ y(1 - x^2) = x \]
This can be rewritten as:
\[ y - yx^2 - x = 0 \]
Rearranging further:
\[ yx^2 + x - y = 0 \]

2. **Identify this as a quadratic equation in \( x \)**: The equation can be expressed in standard quadratic form:
\[ ax^2 + bx + c = 0 \]
where \( a = y \), \( b = 1 \), and \( c = -y \).

3. **Apply the discriminant condition**: For the quadratic equation to have real solutions for \( x \), the discriminant must be non-negative:
\[ D = b^2 - 4ac = 1^2 - 4(y)(-y) = 1 + 4y^2 \]
We need:
\[ 1 + 4y^2 \geq 0 \]
Since \( 4y^2 \) is always non-negative, \( 1 + 4y^2 \) is always greater than or equal to 1. Therefore, this condition is satisfied for all real numbers \( y \).

4. **Determine the range**: Since the discriminant is always non-negative, \( y \) can take any real value. Thus, the range of \( f(x) \) is:
\[ \text{Range} = \mathbb{R} = (-\infty, \infty) \]

### Final Answer

- **Domain**: \( (-\infty, -1) \cup (-1, 1) \cup (1, \infty) \)
- **Range**: \( (-\infty, \infty) \)