Find the domain and range of the function
`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 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.

1. The denominator of the function is \( 1 + x^2 \).
2. We need to ensure that the denominator is not equal to zero:
\[ 1 + x^2 \neq 0 \]
3. Since \( x^2 \) is always non-negative (i.e., \( x^2 \geq 0 \)), the minimum value of \( 1 + x^2 \) occurs when \( x = 0 \):
\[ 1 + 0^2 = 1 \]
4. Therefore, \( 1 + x^2 \) is always greater than zero for all real numbers \( x \). This means the function is defined for all real numbers.

**Conclusion for Domain:**
\[ \text{Domain} = \mathbb{R} \quad \text{(all real numbers)} \]

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

1. Set \( f(x) = y \):
\[ y = \frac{x}{1 + x^2} \]
2. Rearranging gives:
\[ y(1 + x^2) = x \]
\[ yx^2 - x + y = 0 \]
3. This is a quadratic equation in \( x \):
\[ yx^2 - x + y = 0 \]
4. For this quadratic equation to have real roots, the discriminant must be greater than or equal to zero:
\[ D = b^2 - 4ac \]
Here, \( a = y \), \( b = -1 \), and \( c = y \):
\[ D = (-1)^2 - 4(y)(y) = 1 - 4y^2 \]
5. Set the discriminant \( D \geq 0 \):
\[ 1 - 4y^2 \geq 0 \]
\[ 4y^2 \leq 1 \]
\[ y^2 \leq \frac{1}{4} \]
6. Taking the square root gives:
\[ -\frac{1}{2} \leq y \leq \frac{1}{2} \]

**Conclusion for Range:**
\[ \text{Range} = \left[-\frac{1}{2}, \frac{1}{2}\right] \]

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