<p>If <em>x</em> is real, the maximum value of <span>\(\dfrac{3x^2+9x+17}{3x^2+9x+7}\)</span> is</p>
Step-by-Step Solution
Key Concept: Substitute y = 3x² + 9x + 7 to convert the fraction into a simpler form, then use the constraint that the discriminant must be non-negative for real x to find the maximum value.
<p><strong>Step 1:</strong> Let y = 3x² + 9x + 7. Then the numerator is 3x² + 9x + 17 = (3x² + 9x + 7) + 10 = y + 10.</p><p><strong>Step 2:</strong> So the expression becomes: f(x) = (y + 10)/y = 1 + 10/y</p><p><strong>Step 3:</strong> To maximize this, we need to minimize y = 3x² + 9x + 7. Complete the square:</p><p>y = 3(x² + 3x) + 7 = 3(x + 3/2)² - 27/4 + 7 = 3(x + 3/2)² + 1/4</p><p><strong>Step 4:</strong> The minimum value of y is 1/4, occurring when x = -3/2.</p><p><strong>Step 5:</strong> Therefore, the maximum value of f(x) = 1 + 10/(1/4) = 1 + 40 = <strong>41</strong></p>
Correct Answer: B