Find whether the following function are one-one or many -one & into or onto if `f:DtoR` where `D` is its domain
`f(x)=x+1/x,xepsilon(0,oo)`

Explanation

To determine whether the function \( f(x) = x + \frac{1}{x} \) defined on the domain \( D = (0, \infty) \) is one-one or many-one, and whether it is into or onto, we will follow these steps:

### Step 1: Check if the function is one-one or many-one

To check if the function is one-one, we can differentiate it and analyze the sign of the derivative.

1. **Differentiate the function**:
\[ f'(x) = \frac{d}{dx}\left(x + \frac{1}{x}\right) = 1 - \frac{1}{x^2} \]

2. **Set the derivative equal to zero to find critical points**:
\[ 1 - \frac{1}{x^2} = 0 \implies \frac{1}{x^2} = 1 \implies x^2 = 1 \implies x = 1 \]

3. **Analyze the sign of the derivative**:
- For \( x < 1 \) (in the interval \( (0, 1) \)):
\[ f'(x) < 0 \quad \text{(function is decreasing)} \]
- For \( x > 1 \):
\[ f'(x) > 0 \quad \text{(function is increasing)} \]

Since the function decreases on \( (0, 1) \) and increases on \( (1, \infty) \), it has a minimum at \( x = 1 \). Therefore, the function is not one-one because it takes the same value for two different inputs (for example, \( f(0.5) = f(2) \)).

### Step 2: Determine if the function is into or onto

To check if the function is into or onto, we need to find the range of the function.

1. **Find the range of the function**:
\[ y = x + \frac{1}{x} \]
Rearranging gives:
\[ x^2 - yx + 1 = 0 \]

2. **Use the discriminant to find the conditions for real \( x \)**:
The discriminant \( D \) of the quadratic equation must be non-negative for \( y \) to be in the range:
\[ D = b^2 - 4ac = (-y)^2 - 4(1)(1) = y^2 - 4 \]
For real solutions:
\[ y^2 - 4 \geq 0 \implies y^2 \geq 4 \implies |y| \geq 2 \]
This implies:
\[ y \geq 2 \quad \text{or} \quad y \leq -2 \]

3. **Consider the domain**:
Since \( x \) is in \( (0, \infty) \), \( f(x) \) can only take positive values. Thus, we only consider:
\[ y \geq 2 \]

### Conclusion

- The range of \( f(x) \) is \( [2, \infty) \).
- The codomain is \( \mathbb{R} \).

Since the range \( [2, \infty) \) does not cover all of \( \mathbb{R} \), the function is **into** but not onto.

### Final Answer:
- The function \( f(x) = x + \frac{1}{x} \) is **many-one** and **into**.