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 domain of a function consists of all the values of \( x \) for which the function is defined.

1. **Identify the denominator**: The denominator of the function is \( 1 + x^2 \).
2. **Check for restrictions**: The expression \( 1 + x^2 \) is always greater than 0 for all real numbers \( x \) because \( x^2 \) is always non-negative and adding 1 ensures it is positive.
3. **Conclusion**: Since there are no restrictions on \( x \), the domain of \( f(x) \) is all real numbers.

**Domain**: \( \mathbb{R} \) (all real numbers)

### Step 2: Determine the Range
To find the range, we need to analyze the output values of the function \( f(x) \).

1. **Set \( y = f(x) \)**: Let \( y = \frac{x^2}{1 + x^2} \).
2. **Rearrange the equation**: Multiply both sides by \( 1 + x^2 \):
\[ y(1 + x^2) = x^2 \]
This simplifies to:
\[ y + yx^2 = x^2 \]
Rearranging gives:
\[ x^2(1 - y) = y \]
Thus, we have:
\[ x^2 = \frac{y}{1 - y} \]
3. **Analyze the expression**: For \( x^2 \) to be non-negative, the right-hand side \( \frac{y}{1 - y} \) must also be non-negative. This leads to two conditions:
- \( y \geq 0 \) (numerator must be non-negative)
- \( 1 - y > 0 \) (denominator must be positive), which implies \( y < 1 \).
4. **Combine the conditions**: Therefore, the range of \( y \) is:
\[ 0 \leq y < 1 \]
This means \( y \) can take values from 0 to just below 1.

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

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