Greatest value of the expression `(8)/( 9x^(2) - 6x + 5)` is

Options

A. 2
B. 5
C. `8(1)/(3)`
D. `9. 2`

Answer

To find the greatest value of the expression \( \frac{8}{9x^2 - 6x + 5} \), we will follow these steps:

### Step 1: Define the expression
Let \( y = \frac{8}{9x^2 - 6x + 5} \).

### Step 2: Rearrange the equation
Rearranging gives us:
\[ y(9x^2 - 6x + 5) = 8 \]
This can be rewritten as:
\[ 9yx^2 - 6yx + (5y - 8) = 0 \]

### Step 3: Identify coefficients
In this quadratic equation \( ax^2 + bx + c = 0 \):
- \( a = 9y \)
- \( b = -6y \)
- \( c = 5y - 8 \)

### Step 4: Use the discriminant
For the roots of the quadratic to be real, the discriminant must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Substituting the coefficients:
\[ (-6y)^2 - 4(9y)(5y - 8) \geq 0 \]
This simplifies to:
\[ 36y^2 - 36y(5y - 8) \geq 0 \]

### Step 5: Simplify the discriminant
Expanding the equation:
\[ 36y^2 - 180y^2 + 288y \geq 0 \]
Combining like terms:
\[ -144y^2 + 288y \geq 0 \]
Dividing through by -144 (and reversing the inequality):
\[ y^2 - 2y \leq 0 \]

### Step 6: Factor the inequality
Factoring gives:
\[ y(y - 2) \leq 0 \]

### Step 7: Find critical points
The critical points are \( y = 0 \) and \( y = 2 \).

### Step 8: Test intervals
We test the intervals determined by the critical points:
- For \( y < 0 \): \( y(y - 2) > 0 \)
- For \( 0 < y < 2 \): \( y(y - 2) < 0 \)
- For \( y > 2 \): \( y(y - 2) > 0 \)

### Step 9: Determine the range
The solution to \( y(y - 2) \leq 0 \) gives us:
\[ 0 \leq y \leq 2 \]

### Step 10: Conclusion
The greatest value of \( y \) is \( 2 \).

Thus, the greatest value of the expression \( \frac{8}{9x^2 - 6x + 5} \) is \( \boxed{2} \).

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