The domain and range of the function `f(x)=(x^2 )/((1+x^2))` are ....... and ........ respectively.

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 is the set of all possible input values (x-values) that will not cause any undefined behavior, such as division by zero.

1. **Identify the denominator**: In our function, the denominator is \( 1 + x^2 \).
2. **Set the denominator to zero**: We need to find when \( 1 + x^2 = 0 \).
\[ 1 + x^2 = 0 \implies x^2 = -1 \]
3. **Solve for x**: The equation \( x^2 = -1 \) has no real solutions, as the square of a real number cannot be negative. The solutions are complex numbers \( x = i \) and \( x = -i \), which are not in the set of real numbers.
4. **Conclusion about the domain**: Since there are no restrictions on x from the real number set, the domain of \( f(x) \) is all real numbers:
\[ \text{Domain} = (-\infty, +\infty) \]

### Step 2: Determine the Range
The range of a function is the set of all possible output values (y-values).

1. **Set the function equal to y**: Let \( y = f(x) = \frac{x^2}{1 + x^2} \).
2. **Rearrange the equation**: Multiply both sides by \( 1 + x^2 \):
\[ y(1 + x^2) = x^2 \implies y + yx^2 = x^2 \]
Rearranging gives:
\[ yx^2 - x^2 + y = 0 \implies (y - 1)x^2 + y = 0 \]
3. **Identify coefficients**: This is a quadratic equation in \( x \):
\[ a = y - 1, \quad b = 0, \quad c = y \]
4. **Use the discriminant**: For the quadratic to have real solutions, the discriminant must be non-negative:
\[ D = b^2 - 4ac = 0^2 - 4(y - 1)(y) \geq 0 \]
Simplifying gives:
\[ -4(y^2 - y) \geq 0 \implies 4(y^2 - y) \leq 0 \]
Factoring gives:
\[ 4y(y - 1) \leq 0 \]
5. **Determine the intervals**: The critical points are \( y = 0 \) and \( y = 1 \). We test intervals:
- For \( y < 0 \): \( 4y(y - 1) > 0 \) (not valid)
- For \( 0 < y < 1 \): \( 4y(y - 1) < 0 \) (valid)
- For \( y = 0 \) or \( y = 1 \): \( 4y(y - 1) = 0 \) (valid)
- For \( y > 1 \): \( 4y(y - 1) > 0 \) (not valid)
6. **Conclusion about the range**: Therefore, the range of \( f(x) \) is:
\[ \text{Range} = [0, 1] \]

### Final Answer
- **Domain**: \( (-\infty, +\infty) \)
- **Range**: \( [0, 1] \)