Find the domain and range of the following functions:
`(x^(2))/(1+x^(2))`

Answer

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

### Step 1: Determine the Domain
The function \( f(x) \) is a rational function, which is defined as the ratio of two polynomials. The domain of a rational function is all real numbers except where the denominator is equal to zero.

1. Identify the denominator: \( 1 + x^2 \).
2. Set the denominator equal to zero and solve for \( x \):
\[ 1 + x^2 = 0 \]
\[ x^2 = -1 \]
Since \( x^2 \) is always non-negative, \( 1 + x^2 \) can never be zero. Therefore, there are no restrictions on \( x \).

**Domain**: All real numbers, \( \mathbb{R} \).

### Step 2: Determine the Range
Next, we will find the range of the function \( f(x) \).

1. Rewrite the function:
\[ f(x) = \frac{x^2}{1+x^2} = 1 - \frac{1}{1+x^2} \]
2. Let \( y = f(x) \):
\[ y = 1 - \frac{1}{1+x^2} \]
3. Rearranging gives:
\[ \frac{1}{1+x^2} = 1 - y \]
4. Cross-multiplying:
\[ 1 = (1 - y)(1 + x^2) \]
\[ 1 = 1 + x^2 - y - yx^2 \]
\[ yx^2 - x^2 = y \]
\[ x^2(y - 1) = y \]
\[ x^2 = \frac{y}{1 - y} \]
5. Since \( x^2 \) is always non-negative, we have:
\[ \frac{y}{1 - y} \geq 0 \]
This implies \( y \geq 0 \) and \( 1 - y > 0 \) (or \( y < 1 \)). Therefore, \( y \) must lie in the interval \( [0, 1) \).

**Range**: \( [0, 1) \).

### Final Answer
- **Domain**: \( \mathbb{R} \) (all real numbers)
- **Range**: \( [0, 1) \)