A real valued function is given by `f (x) = (x^(2)+x+2)/(x^(2)+x +1)` , find its domain and range .

Answer

To find the domain and range of the function \( f(x) = \frac{x^2 + x + 2}{x^2 + x + 1} \), we will follow these steps:

### Step 1: Determine the Domain
The domain of a function consists of all the values of \( x \) for which the function is defined. For a rational function, we need to ensure that the denominator is not equal to zero.

1. **Set the denominator equal to zero:**
\[ x^2 + x + 1 = 0 \]

2. **Calculate the discriminant (D):**
The discriminant \( D \) is given by the formula:
\[ D = b^2 - 4ac \]
Here, \( a = 1 \), \( b = 1 \), and \( c = 1 \):
\[ D = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \]

3. **Interpret the discriminant:**
Since the discriminant is negative (\( D < 0 \)), the quadratic equation \( x^2 + x + 1 = 0 \) has no real roots. This means that the denominator is never zero for any real value of \( x \).

4. **Conclusion about the domain:**
Therefore, the domain of \( f(x) \) is all real numbers:
\[ \text{Domain: } (-\infty, \infty) \]

### Step 2: Determine the Range
To find the range of the function, we will express \( f(x) \) in a different form and analyze it.

1. **Rewrite the function:**
\[ f(x) = \frac{x^2 + x + 2}{x^2 + x + 1} = 1 + \frac{1}{x^2 + x + 1} \]

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

3. **Rearranging gives:**
\[ y - 1 = \frac{1}{x^2 + x + 1} \]

4. **Cross-multiply:**
\[ (y - 1)(x^2 + x + 1) = 1 \]
This leads to the quadratic equation:
\[ x^2(y - 1) + x(y - 1) + (1 - y) = 0 \]

5. **Find the discriminant of this quadratic:**
The discriminant \( D \) must be non-negative for \( x \) to have real solutions:
\[ D = (y - 1)^2 - 4(y - 1)(1 - y) \geq 0 \]

6. **Simplifying the discriminant:**
\[ D = (y - 1)^2 - 4(y - 1)(1 - y) = (y - 1)^2 + 4(y - 1)(y - 1) \]
\[ = (y - 1)^2 + 4(y - 1)^2 = 5(y - 1)^2 \]
Since \( 5(y - 1)^2 \geq 0 \), the discriminant is always non-negative.

7. **Finding the limits of \( y \):**
As \( x^2 + x + 1 \) is always positive, the term \( \frac{1}{x^2 + x + 1} \) will be positive and approaches 0 as \( x \) approaches \( \pm \infty \). Thus:
\[ y \to 1 \text{ as } x \to \pm \infty \]
The minimum value of \( y \) is just above 1, and it can take values approaching \( \infty \) when \( x^2 + x + 1 \) approaches 0 (which does not happen since it is always positive).

8. **Conclusion about the range:**
Therefore, the range of \( f(x) \) is:
\[ \text{Range: } (1, \infty) \]

### Final Answer:
- **Domain:** \( (-\infty, \infty) \)
- **Range:** \( (1, \infty) \)