Determine the range and domain of the function : `f(x)=(x+1)/(x^(2)+x+1)`

Answer

To determine the domain and range of the function \( f(x) = \frac{x+1}{x^2 + x + 1} \), we will 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 rational functions, we need to ensure that the denominator is not equal to zero.

1. **Identify the denominator**: The denominator of our function is \( x^2 + x + 1 \).
2. **Set the denominator to zero and solve**:
\[ x^2 + x + 1 = 0 \]
To find the roots, we calculate the discriminant \( D \):
\[ D = b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \]
Since the discriminant is negative (\( D < 0 \)), the quadratic equation has no real roots. This means that \( x^2 + x + 1 \) is always positive for all real numbers.

3. **Conclusion on the domain**: Since the denominator is never zero, the function is defined for all real numbers.
\[ \text{Domain: } x \in \mathbb{R} \]

### Step 2: Determine the Range
To find the range, we will express \( f(x) \) in terms of \( y \) and solve for \( y \).

1. **Set \( f(x) = y \)**:
\[ y = \frac{x+1}{x^2 + x + 1} \]
2. **Cross-multiply**:
\[ y(x^2 + x + 1) = x + 1 \]
Rearranging gives:
\[ yx^2 + yx + y - x - 1 = 0 \]
This can be rewritten as:
\[ yx^2 + (y - 1)x + (y - 1) = 0 \]
3. **Identify coefficients**:
The equation is a quadratic in \( x \) with coefficients:
- \( a = y \)
- \( b = y - 1 \)
- \( c = y - 1 \)

4. **Use the discriminant for real solutions**:
For the quadratic to have real solutions, the discriminant must be non-negative:
\[ D = (y - 1)^2 - 4y(y - 1) \geq 0 \]
Simplifying the discriminant:
\[ D = (y - 1)^2 - 4y^2 + 4y = -3y^2 + 6y - 1 \geq 0 \]
Rearranging gives:
\[ 3y^2 - 6y + 1 \leq 0 \]

5. **Solve the quadratic inequality**:
First, find the roots of the quadratic equation \( 3y^2 - 6y + 1 = 0 \):
\[ D' = (-6)^2 - 4 \cdot 3 \cdot 1 = 36 - 12 = 24 \]
The roots are:
\[ y = \frac{6 \pm \sqrt{24}}{2 \cdot 3} = \frac{6 \pm 2\sqrt{6}}{6} = 1 \pm \frac{\sqrt{6}}{3} \]
Let \( y_1 = 1 - \frac{\sqrt{6}}{3} \) and \( y_2 = 1 + \frac{\sqrt{6}}{3} \).

6. **Determine the range**:
The quadratic \( 3y^2 - 6y + 1 \) opens upwards (since the coefficient of \( y^2 \) is positive), so the values of \( y \) for which the quadratic is less than or equal to zero are between the roots:
\[ \text{Range: } \left[ 1 - \frac{\sqrt{6}}{3}, 1 + \frac{\sqrt{6}}{3} \right] \]

### Final Answer
- **Domain**: \( x \in \mathbb{R} \)
- **Range**: \( \left[ 1 - \frac{\sqrt{6}}{3}, 1 + \frac{\sqrt{6}}{3} \right] \)