Find the range of the following functions.
`y=(x^(2))/(1+x^(2))`

Answer

To find the range of the function \( y = \frac{x^2}{1 + x^2} \), we will analyze the behavior of the function step by step.

### Step 1: Understand the function
The function is given as:
\[ y = \frac{x^2}{1 + x^2} \]
Here, \( x^2 \) is always non-negative since it is a square term. Therefore, \( x^2 \geq 0 \).

### Step 2: Analyze the numerator and denominator
- The numerator \( x^2 \) ranges from \( 0 \) to \( \infty \) (i.e., \( 0 \leq x^2 < \infty \)).
- The denominator \( 1 + x^2 \) is always positive and ranges from \( 1 \) to \( \infty \) (i.e., \( 1 \leq 1 + x^2 < \infty \)).

### Step 3: Determine the range of \( y \)
Since \( y \) is a fraction where the numerator is non-negative and the denominator is always greater than or equal to \( 1 \), we can analyze the limits of \( y \):

1. **When \( x = 0 \)**:
\[ y = \frac{0^2}{1 + 0^2} = 0 \]

2. **As \( x \) approaches \( \infty \)**:
\[ y = \frac{x^2}{1 + x^2} \approx \frac{x^2}{x^2} = 1 \]
However, \( y \) never actually reaches \( 1 \) because the numerator will always be slightly less than the denominator as \( x \) increases.

### Step 4: Conclusion about the range
From the above analysis, we can conclude that:
- The minimum value of \( y \) is \( 0 \) (when \( x = 0 \)).
- The maximum value of \( y \) approaches \( 1 \) but never actually reaches it.

Thus, the range of the function \( y = \frac{x^2}{1 + x^2} \) is:
\[ \text{Range} = [0, 1) \]