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.

1. **Identify the denominator**: The function is defined as long as the denominator is not zero.
\[ x^2 + x + 1 \neq 0 \]
2. **Check for real roots**: To find if the denominator can be zero, we can check the discriminant of the quadratic equation \( x^2 + x + 1 = 0 \).
\[ \text{Discriminant} = b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \]
Since the discriminant is negative, the quadratic equation has no real roots. Therefore, the denominator is never zero for any real number \( x \).

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

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

1. **Set the function equal to \( y \)**:
\[ y = \frac{x+1}{x^2 + x + 1} \]
2. **Cross-multiply** to eliminate the fraction:
\[ y(x^2 + x + 1) = x + 1 \]
Rearranging gives:
\[ yx^2 + yx + y - x - 1 = 0 \]
This is a quadratic equation in \( x \):
\[ yx^2 + (y - 1)x + (y - 1) = 0 \]
3. **Use the discriminant** to find conditions for \( y \) such that \( x \) has real solutions:
\[ D = (y - 1)^2 - 4y(y - 1) \geq 0 \]
Simplifying the discriminant:
\[ D = (y - 1)^2 - 4y^2 + 4y = -3y^2 + 6y - 1 \geq 0 \]
4. **Solve the quadratic inequality**:
- First, find the roots of the equation \( -3y^2 + 6y - 1 = 0 \) using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-6 \pm \sqrt{6^2 - 4 \cdot (-3) \cdot (-1)}}{2 \cdot (-3)} = \frac{-6 \pm \sqrt{36 - 12}}{-6} = \frac{-6 \pm \sqrt{24}}{-6} \]
Simplifying gives:
\[ y = \frac{-6 \pm 2\sqrt{6}}{-6} = 1 \mp \frac{\sqrt{6}}{3} \]
Thus, the roots are:
\[ y_1 = 1 - \frac{\sqrt{6}}{3}, \quad y_2 = 1 + \frac{\sqrt{6}}{3} \]

5. **Determine the intervals**: The quadratic opens downwards (since the coefficient of \( y^2 \) is negative). Thus, the range of \( y \) for which the discriminant is non-negative is:
\[ y \in \left[1 - \frac{\sqrt{6}}{3}, 1 + \frac{\sqrt{6}}{3}\right] \]

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