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

Explanation

To find the domain and range of the function \( f(x) = \frac{x}{1 + x^2} \), 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. Since \( f(x) \) is a rational function, it is defined wherever the denominator is not equal to zero.

1. **Identify the denominator**: The denominator of our function is \( 1 + x^2 \).
2. **Set the denominator to zero**: We need to find when \( 1 + x^2 = 0 \).
\[ 1 + x^2 = 0 \implies x^2 = -1 \]
Since \( x^2 \) is always non-negative for real numbers, \( 1 + x^2 \) can never be zero. Therefore, there are no restrictions on \( x \).

**Conclusion for Domain**: The domain of \( f(x) \) is all real numbers, which can be expressed in interval notation as:
\[ \text{Domain} = (-\infty, \infty) \]

### Step 2: Determine the Range
To find the range, we will express \( y \) in terms of \( x \) and analyze the resulting quadratic equation.

1. **Set \( y = f(x) \)**:
\[ y = \frac{x}{1 + x^2} \]
2. **Cross-multiply to eliminate the fraction**:
\[ y(1 + x^2) = x \implies y + yx^2 = x \implies yx^2 - x + y = 0 \]
3. **Rearrange into standard quadratic form**:
\[ yx^2 - x + y = 0 \]
Here, \( a = y \), \( b = -1 \), and \( c = y \).

4. **Use the discriminant to find conditions for real roots**:
The discriminant \( D \) must be greater than or equal to zero for the quadratic to have real solutions.
\[ D = b^2 - 4ac = (-1)^2 - 4(y)(y) = 1 - 4y^2 \]
Set the discriminant \( D \geq 0 \):
\[ 1 - 4y^2 \geq 0 \]
\[ 4y^2 \leq 1 \implies y^2 \leq \frac{1}{4} \]
5. **Take the square root**:
\[ -\frac{1}{2} \leq y \leq \frac{1}{2} \]

**Conclusion for Range**: The range of \( f(x) \) is:
\[ \text{Range} = \left[-\frac{1}{2}, \frac{1}{2}\right] \]

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