Let `f(x) = (x^(2) + 4x + 1)/( x^(2) + x + 1), x inR`
If `m le f(x) le M AA x in R`, then `(1)/(2)` (M - m) = _______

Answer

To solve the problem, we need to find the minimum and maximum values of the function \( f(x) = \frac{x^2 + 4x + 1}{x^2 + x + 1} \) and then calculate \( \frac{1}{2}(M - m) \), where \( m \) is the minimum value and \( M \) is the maximum value of \( f(x) \).

### Step-by-step Solution:

1. **Identify the Function**:
\[ f(x) = \frac{x^2 + 4x + 1}{x^2 + x + 1} \]

2. **Set Up the Quadratic Equation**:
We want to analyze the behavior of \( f(x) \). To do this, we can set \( f(x) = y \):
\[ y = \frac{x^2 + 4x + 1}{x^2 + x + 1} \]
Rearranging gives:
\[ y(x^2 + x + 1) = x^2 + 4x + 1 \]
This leads to:
\[ (y - 1)x^2 + (y - 4)x + (y - 1) = 0 \]

3. **Determine the Discriminant**:
For the roots of this quadratic equation to be real, the discriminant must be non-negative:
\[ D = (y - 4)^2 - 4(y - 1)(y - 1) \geq 0 \]
Expanding this gives:
\[ D = (y - 4)^2 - 4(y - 1)^2 \]

4. **Simplify the Discriminant**:
Expanding both terms:
\[ D = (y^2 - 8y + 16) - 4(y^2 - 2y + 1) \]
\[ D = y^2 - 8y + 16 - 4y^2 + 8y - 4 \]
\[ D = -3y^2 + 12 \geq 0 \]

5. **Solve the Inequality**:
Rearranging gives:
\[ 3y^2 \leq 12 \implies y^2 \leq 4 \implies -2 \leq y \leq 2 \]
Thus, the range of \( f(x) \) is:
\[ m = -2, \quad M = 2 \]

6. **Calculate \( \frac{1}{2}(M - m) \)**:
Now we can substitute the values of \( m \) and \( M \):
\[ \frac{1}{2}(M - m) = \frac{1}{2}(2 - (-2)) = \frac{1}{2}(2 + 2) = \frac{1}{2}(4) = 2 \]

### Final Answer:
\[ \frac{1}{2}(M - m) = 2 \]