If `f: RvecR` is given by `f(x)=(x^2-4)/(x^2+1)` , identify the type of function.

Explanation

To identify the type of function \( f: \mathbb{R} \to \mathbb{R} \) given by

\[ f(x) = \frac{x^2 - 4}{x^2 + 1}, \]

we will follow these steps:

### Step 1: Check if the function is even or odd
To determine if the function is even, we need to check if \( f(-x) = f(x) \).

\[ f(-x) = \frac{(-x)^2 - 4}{(-x)^2 + 1} = \frac{x^2 - 4}{x^2 + 1} = f(x). \]

Since \( f(-x) = f(x) \), the function is **even**.

### Step 2: Determine if the function is one-one or many-one
A function is one-one if \( f(a) = f(b) \) implies \( a = b \). To check this, we can analyze the function:

Assume \( f(a) = f(b) \):

\[ \frac{a^2 - 4}{a^2 + 1} = \frac{b^2 - 4}{b^2 + 1}. \]

Cross-multiplying gives:

\[ (a^2 - 4)(b^2 + 1) = (b^2 - 4)(a^2 + 1). \]

This simplifies to:

\[ a^2b^2 + a^2 - 4b^2 - 4 = b^2a^2 + b^2 - 4a^2 - 4. \]

Rearranging terms leads to:

\[ a^2 - 4b^2 = b^2 - 4a^2. \]

This implies:

\[ 5a^2 = 5b^2 \implies a^2 = b^2 \implies a = b \text{ or } a = -b. \]

Since \( a \) and \( b \) can be different (for example, \( a = 2 \) and \( b = -2 \)), the function is **many-one**.

### Step 3: Determine the range of the function
To find the range, we can rewrite the function:

\[ f(x) = 1 - \frac{5}{x^2 + 1}. \]

The term \( x^2 + 1 \) is always positive and has a minimum value of 1 when \( x = 0 \). Thus, the maximum value of \( f(x) \) occurs as \( x^2 \) approaches infinity, making \( f(x) \) approach 1.

The minimum value occurs when \( x = 0 \):

\[ f(0) = \frac{0^2 - 4}{0^2 + 1} = -4. \]

Thus, the range of \( f(x) \) is:

\[ [-4, 1). \]

### Step 4: Determine if the function is onto or into
A function is onto if its range equals its codomain. Since the codomain is \( \mathbb{R} \) and the range is \( [-4, 1) \), the function is **into**.

### Conclusion
The function \( f(x) = \frac{x^2 - 4}{x^2 + 1} \) is an **even**, **many-one**, and **into** function.

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