Classify the following function `f(x)` defined in `RtoR` as injective, surjective, both or none.
`f(x)=(x^(2))/(1+x^(2))`

Explanation

To classify the function \( f(x) = \frac{x^2}{1 + x^2} \) defined from \( \mathbb{R} \) to \( \mathbb{R} \) as injective, surjective, both, or none, we will analyze its properties step by step.

### Step 1: Check for Injectivity
A function is injective (one-to-one) if different inputs produce different outputs. In mathematical terms, if \( f(a) = f(b) \) implies \( a = b \), then the function is injective.

1. Let's assume \( f(a) = f(b) \):
\[ \frac{a^2}{1 + a^2} = \frac{b^2}{1 + b^2} \]
2. Cross-multiplying gives:
\[ a^2(1 + b^2) = b^2(1 + a^2) \]
3. Expanding both sides:
\[ a^2 + a^2b^2 = b^2 + a^2b^2 \]
4. Simplifying leads to:
\[ a^2 = b^2 \]
5. This implies \( a = b \) or \( a = -b \). Since both \( a \) and \( b \) can yield the same output for different inputs (e.g., \( f(1) = f(-1) \)), the function is not injective.

### Step 2: Check for Surjectivity
A function is surjective (onto) if every element in the codomain has a pre-image in the domain. We need to check if the range of \( f(x) \) covers all real numbers.

1. The function \( f(x) = \frac{x^2}{1 + x^2} \) is always non-negative because both the numerator and denominator are non-negative.
2. The minimum value of \( f(x) \) occurs at \( x = 0 \):
\[ f(0) = \frac{0^2}{1 + 0^2} = 0 \]
3. The maximum value occurs as \( x \to \infty \):
\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{x^2}{1 + x^2} = 1 \]
4. Therefore, the range of \( f(x) \) is \( [0, 1) \).
5. The codomain is \( \mathbb{R} \), which includes negative numbers and values greater than or equal to 1. Since the range \( [0, 1) \) does not cover all of \( \mathbb{R} \), the function is not surjective.

### Conclusion
Since the function \( f(x) \) is neither injective nor surjective, we classify it as **none**.

### Summary of Steps:
1. **Check injectivity**: Found that \( f(a) = f(b) \) can yield \( a \neq b \), hence not injective.
2. **Check surjectivity**: Found that the range \( [0, 1) \) does not cover all of \( \mathbb{R} \), hence not surjective.