Find the greatest and least values of
`(x+2)/(2x^(2)+3x+6) AA x in R`

Answer

To find the greatest and least values of the function \( y = \frac{x + 2}{2x^2 + 3x + 6} \), we will follow these steps:

### Step 1: Define the Function
Let \( y = \frac{x + 2}{2x^2 + 3x + 6} \).

### Step 2: Analyze the Denominator
We need to check the discriminant of the denominator \( 2x^2 + 3x + 6 \) to ensure it does not equal zero. The discriminant \( D \) is given by:
\[ D = B^2 - 4AC = 3^2 - 4 \cdot 2 \cdot 6 = 9 - 48 = -39 \]
Since \( D < 0 \), the quadratic \( 2x^2 + 3x + 6 \) does not have real roots and is always positive.

### Step 3: Set Up the Equation
We can express \( y \) in terms of \( x \):
\[ y(2x^2 + 3x + 6) = x + 2 \]
Rearranging gives:
\[ 2yx^2 + (3y - 1)x + (6y - 2) = 0 \]

### Step 4: Find the Conditions for Real Roots
For \( x \) to have real values, the discriminant of this quadratic must be non-negative:
\[ (3y - 1)^2 - 4(2y)(6y - 2) \geq 0 \]
Calculating the discriminant:
\[ (3y - 1)^2 - 8y(6y - 2) \geq 0 \]
Expanding gives:
\[ 9y^2 - 6y + 1 - (48y^2 - 16y) \geq 0 \]
Combining like terms:
\[ -39y^2 + 10y + 1 \geq 0 \]
Multiplying through by -1 (which reverses the inequality):
\[ 39y^2 - 10y - 1 \leq 0 \]

### Step 5: Solve the Quadratic Inequality
To solve \( 39y^2 - 10y - 1 = 0 \), we use the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 39 \cdot (-1)}}{2 \cdot 39} \]
Calculating the discriminant:
\[ 100 + 156 = 256 \]
Thus,
\[ y = \frac{10 \pm 16}{78} \]
Calculating the two roots:
1. \( y = \frac{26}{78} = \frac{1}{3} \)
2. \( y = \frac{-6}{78} = -\frac{1}{13} \)

### Step 6: Determine the Range
The quadratic \( 39y^2 - 10y - 1 \) opens upwards (since the coefficient of \( y^2 \) is positive). Therefore, the values of \( y \) that satisfy \( 39y^2 - 10y - 1 \leq 0 \) are between the roots:
\[ -\frac{1}{13} \leq y \leq \frac{1}{3} \]

### Final Answer
Thus, the greatest value of \( y \) is \( \frac{1}{3} \) and the least value is \( -\frac{1}{13} \).