The difference of maximum and minimum value of `(x^(2)+4x+9)/(x^(2)+9)` is

Options

A. `1//3`
B. `2//3`
C. `-2//3`
D. `4//3`

Explanation

To solve the problem of finding the difference between the maximum and minimum values of the function \( y = \frac{x^2 + 4x + 9}{x^2 + 9} \), we will follow these steps:

### Step 1: Set up the equation
We start by letting \( y = \frac{x^2 + 4x + 9}{x^2 + 9} \).

### Step 2: Cross-multiply
Cross-multiplying gives us:
\[ y(x^2 + 9) = x^2 + 4x + 9 \]
This simplifies to:
\[ yx^2 + 9y = x^2 + 4x + 9 \]

### Step 3: Rearrange the equation
Rearranging the equation, we get:
\[ yx^2 - x^2 + 4x + 9 - 9y = 0 \]
This can be rewritten as:
\[ (1 - y)x^2 + 4x + (9 - 9y) = 0 \]

### Step 4: Apply the discriminant condition
For the quadratic equation to have real solutions, the discriminant must be non-negative. The discriminant \( D \) is given by:
\[ D = b^2 - 4ac \]
Here, \( a = 1 - y \), \( b = 4 \), and \( c = 9 - 9y \). Thus,
\[ D = 4^2 - 4(1 - y)(9 - 9y) \geq 0 \]
This simplifies to:
\[ 16 - 4(1 - y)(9 - 9y) \geq 0 \]

### Step 5: Expand and simplify the discriminant
Expanding the discriminant:
\[ D = 16 - 4[(9 - 9y) - (9y - 9y^2)] \]
\[ = 16 - 4[9 - 9y - 9y + 9y^2] \]
\[ = 16 - 36 + 36y - 36y^2 \geq 0 \]
\[ = -36y^2 + 36y - 20 \geq 0 \]

### Step 6: Factor the quadratic inequality
Dividing through by -4 (and reversing the inequality):
\[ 9y^2 - 9y + 5 \leq 0 \]
To find the roots, we use the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 9 \cdot 5}}{2 \cdot 9} \]
Calculating the discriminant:
\[ = \frac{9 \pm \sqrt{81 - 180}}{18} = \frac{9 \pm \sqrt{-99}}{18} \]
Since the discriminant is negative, the quadratic does not cross the x-axis, and we check the sign of the quadratic. Since \( a > 0 \), it opens upwards, meaning it is always positive.

### Step 7: Find the range of \( y \)
Returning to our earlier discriminant condition:
\[ 4 - 9(y - 1)^2 \geq 0 \]
This leads to:
\[ 9(y - 1)^2 \leq 4 \]
Taking square roots:
\[ |y - 1| \leq \frac{2}{3} \]
Thus, we have:
\[ 1 - \frac{2}{3} \leq y \leq 1 + \frac{2}{3} \]
This simplifies to:
\[ \frac{1}{3} \leq y \leq \frac{5}{3} \]

### Step 8: Calculate the difference between maximum and minimum values
The maximum value of \( y \) is \( \frac{5}{3} \) and the minimum value is \( \frac{1}{3} \). Therefore, the difference is:
\[ \frac{5}{3} - \frac{1}{3} = \frac{4}{3} \]

### Conclusion
The difference of the maximum and minimum values of the function is:
\[ \boxed{\frac{4}{3}} \]