Find the domain and range of `f (x)= x^2 //(1+x^2)`(x real). Is the function one-to-one ?

Answer

To find the domain and range of the function \( f(x) = \frac{x^2}{1 + x^2} \) and to determine if it is one-to-one, we can follow these steps:

### Step 1: Determine the Domain
The domain of a function is the set of all possible input values (x-values) for which the function is defined.

For the function \( f(x) = \frac{x^2}{1 + x^2} \):
- The denominator \( 1 + x^2 \) is always positive for all real numbers \( x \) since \( x^2 \) is non-negative and adding 1 ensures it is never zero.
- Therefore, there are no restrictions on the values of \( x \).

**Domain:** \( (-\infty, \infty) \) or all real numbers \( \mathbb{R} \).

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

To find the range of \( f(x) \):
1. Notice that \( f(x) \) is always non-negative because both the numerator and denominator are non-negative.
2. We can analyze the behavior of \( f(x) \) as \( x \) approaches certain values:
- As \( x \to 0 \), \( f(0) = \frac{0^2}{1 + 0^2} = 0 \).
- As \( x \to \infty \) or \( x \to -\infty \), \( f(x) = \frac{x^2}{1 + x^2} \) approaches \( 1 \) since the \( x^2 \) terms dominate.

3. To confirm that \( f(x) \) never actually reaches \( 1 \), we can set up the inequality:
\[ \frac{x^2}{1 + x^2} < 1 \]
Simplifying gives:
\[ x^2 < 1 + x^2 \quad \text{(which is always true)} \]
Hence, \( f(x) \) can take values from \( 0 \) up to but not including \( 1 \).

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

### Step 3: Check if the Function is One-to-One
A function is one-to-one if different inputs produce different outputs. To check if \( f(x) \) is one-to-one, we can look for distinct \( x_1 \) and \( x_2 \) such that \( f(x_1) = f(x_2) \).

Consider:
- Let \( x_1 = 2 \) and \( x_2 = -2 \):
\[ f(2) = \frac{2^2}{1 + 2^2} = \frac{4}{5} \]
\[ f(-2) = \frac{(-2)^2}{1 + (-2)^2} = \frac{4}{5} \]
Thus, \( f(2) = f(-2) \) but \( 2 \neq -2 \).

Since we found two different inputs that yield the same output, the function is not one-to-one.

**Conclusion:**
- **Domain:** \( (-\infty, \infty) \)
- **Range:** \( [0, 1) \)
- **Is the function one-to-one?** No.