If x be real, find the maximum value of `((x+2))/((2x^(2)+3x+6))`

Answer

To find the maximum value of the expression \(\frac{x + 2}{2x^2 + 3x + 6}\), we can follow these steps:

### Step 1: Identify the function
Let \(y = \frac{x + 2}{2x^2 + 3x + 6}\). We want to maximize \(y\).

### Step 2: Find the critical points
To find the maximum value, we can use the method of calculus. We will differentiate \(y\) with respect to \(x\) and set the derivative equal to zero.

Using the quotient rule:
\[ y' = \frac{(2x^2 + 3x + 6)(1) - (x + 2)(4x + 3)}{(2x^2 + 3x + 6)^2} \]

### Step 3: Set the derivative to zero
Setting the numerator of the derivative to zero:
\[ (2x^2 + 3x + 6) - (x + 2)(4x + 3) = 0 \]
Expanding the equation:
\[ 2x^2 + 3x + 6 - (4x^2 + 3x + 8) = 0 \]
Simplifying:
\[ -2x^2 - 2 = 0 \]
This simplifies to:
\[ x^2 + 1 = 0 \]
This equation has no real solutions, indicating that the maximum occurs at the endpoints or at a critical point.

### Step 4: Analyze the quadratic function
The denominator \(2x^2 + 3x + 6\) is a quadratic function. Since the coefficient of \(x^2\) is positive, this function has a minimum value. We can find the vertex of this quadratic function to determine its minimum value.

The vertex \(x\) is given by:
\[ x = -\frac{b}{2a} = -\frac{3}{2 \cdot 2} = -\frac{3}{4} \]

### Step 5: Calculate the minimum value of the denominator
Substituting \(x = -\frac{3}{4}\) into the denominator:
\[ 2\left(-\frac{3}{4}\right)^2 + 3\left(-\frac{3}{4}\right) + 6 \]
Calculating:
\[ = 2 \cdot \frac{9}{16} - \frac{9}{4} + 6 \]
\[ = \frac{18}{16} - \frac{36}{16} + \frac{96}{16} \]
\[ = \frac{18 - 36 + 96}{16} = \frac{78}{16} = \frac{39}{8} \]

### Step 6: Calculate the maximum value of the expression
Now substituting \(x = -\frac{3}{4}\) into the numerator:
\[ x + 2 = -\frac{3}{4} + 2 = \frac{5}{4} \]
Thus, the maximum value of \(y\) is:
\[ y = \frac{\frac{5}{4}}{\frac{39}{8}} = \frac{5}{4} \cdot \frac{8}{39} = \frac{10}{39} \]

### Final Answer
The maximum value of \(\frac{x + 2}{2x^2 + 3x + 6}\) is \(\frac{10}{39}\).

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