If 'x' be real, find the maximum and minimum value of : `y=(x+2)/(2x^(2)+3x+6)`

Answer

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

### Step 1: Identify the function
We start with the function:
\[ y = \frac{x + 2}{2x^2 + 3x + 6} \]

### Step 2: Find the derivative
To find the maximum and minimum values, we need to find the critical points by taking the derivative of \( y \) with respect to \( x \) and setting it to zero. We will use the quotient rule for differentiation:
\[ y' = \frac{(2x^2 + 3x + 6)(1) - (x + 2)(4x + 3)}{(2x^2 + 3x + 6)^2} \]

### Step 3: Set the derivative to zero
Setting the numerator of the derivative equal to zero gives us:
\[ (2x^2 + 3x + 6) - (x + 2)(4x + 3) = 0 \]

### Step 4: Simplify the equation
Now we will simplify the equation:
\[ 2x^2 + 3x + 6 - (4x^2 + 3x + 8x + 6) = 0 \]
This simplifies to:
\[ 2x^2 + 3x + 6 - 4x^2 - 11x - 6 = 0 \]
\[ -2x^2 - 8x = 0 \]
Factoring out \(-2x\):
\[ -2x(x + 4) = 0 \]

### Step 5: Solve for critical points
Setting each factor to zero gives us:
\[ x = 0 \quad \text{or} \quad x = -4 \]

### Step 6: Evaluate the function at critical points
Now we will evaluate \( y \) at these critical points:
1. For \( x = 0 \):
\[ y(0) = \frac{0 + 2}{2(0)^2 + 3(0) + 6} = \frac{2}{6} = \frac{1}{3} \]
2. For \( x = -4 \):
\[ y(-4) = \frac{-4 + 2}{2(-4)^2 + 3(-4) + 6} = \frac{-2}{32 - 12 + 6} = \frac{-2}{26} = -\frac{1}{13} \]

### Step 7: Analyze the behavior of the function
Next, we need to check the limits as \( x \to \infty \) and \( x \to -\infty \):
- As \( x \to \infty \), \( y \to 0 \).
- As \( x \to -\infty \), \( y \to 0 \).

### Step 8: Determine maximum and minimum values
From our evaluations:
- The maximum value of \( y \) occurs at \( x = 0 \) and is \( \frac{1}{3} \).
- The minimum value of \( y \) occurs at \( x = -4 \) and is \( -\frac{1}{13} \).

### Final Result
Thus, the maximum value of \( y \) is \( \frac{1}{3} \) and the minimum value of \( y \) is \( -\frac{1}{13} \).

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