`f(x)=x/(1+x^(2))`. Find domain and range of f(x).

Answer

To find the domain and range of the function \( f(x) = \frac{x}{1 + x^2} \), we will follow these steps:

### Step 1: Finding the Domain

1. **Identify the denominator**: The function is defined as long as the denominator is not equal to zero. Thus, we need to ensure that \( 1 + x^2 \neq 0 \).

2. **Set the denominator to zero**:
\[ 1 + x^2 = 0 \]
This simplifies to:
\[ x^2 = -1 \]
Since \( x^2 \) cannot be negative for any real number \( x \), there are no real solutions to this equation.

3. **Conclusion about the domain**: Since the denominator is never zero for any real number \( x \), the domain of \( f(x) \) is all real numbers:
\[ \text{Domain: } x \in \mathbb{R} \]

### Step 2: Finding the Range

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

2. **Identify coefficients for the quadratic equation**:
The equation can be expressed in standard quadratic form \( ax^2 + bx + c = 0 \) where:
- \( a = y \)
- \( b = -1 \)
- \( c = y \)

3. **Use the discriminant**: For the quadratic equation to have real solutions, the discriminant must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Substituting the coefficients:
\[ (-1)^2 - 4(y)(y) \geq 0 \]
This simplifies to:
\[ 1 - 4y^2 \geq 0 \]

4. **Solve the inequality**:
Rearranging gives:
\[ 4y^2 \leq 1 \]
Dividing by 4:
\[ y^2 \leq \frac{1}{4} \]
Taking the square root:
\[ -\frac{1}{2} \leq y \leq \frac{1}{2} \]

5. **Conclusion about the range**: Therefore, the range of \( f(x) \) is:
\[ \text{Range: } \left[-\frac{1}{2}, \frac{1}{2}\right] \]

### Final Answer
- **Domain**: \( x \in \mathbb{R} \)
- **Range**: \( \left[-\frac{1}{2}, \frac{1}{2}\right] \)