Applications of Derivatives
Local maxima and minima
Grade 12
Question:
<p>Given function is \( f(x) = \dfrac{x}{2} + \dfrac{2}{x} \). Which of the following is correct?</p>
<p>\( f(x) \) has a local minima at \( x = 2 \)</p>
<p>\( f(x) \) has a local maxima at \( x = 2 \)</p>
<p>\( f(x) \) has no local extrema</p>
<p>\( f(x) \) has a local minima at \( x = -2 \)</p>
Step-by-Step Solution
Key Concept: Find critical points by setting f'(x) = 0, then use the second derivative test to distinguish between local minima and maxima. For rational functions of this form, the critical point corresponds to an extremum that can be verified using AM-GM inequality.
<p><strong>Step 1:</strong> Find the derivative: f'(x) = 1/2 - 2/x²</p><p><strong>Step 2:</strong> Set f'(x) = 0: 1/2 - 2/x² = 0 ⟹ x² = 4 ⟹ x = ±2 (since domain is x ≠ 0)</p><p><strong>Step 3:</strong> Find the second derivative: f''(x) = 4/x³</p><p><strong>Step 4:</strong> At x = 2: f''(2) = 4/8 = 1/2 > 0, so x = 2 is a local minimum. At x = -2: f''(-2) = -1/2 < 0, so x = -2 is a local maximum</p><p><strong>Step 5:</strong> Calculate values: f(2) = 1 + 1 = 2 (local minimum), f(-2) = -1 - 1 = -2 (local maximum)</p><p><strong>Verification:</strong> By AM-GM inequality: for x > 0, (x/2 + 2/x)/2 ≥ √(x/2 · 2/x) = 1, so f(x) ≥ 2 with equality at x = 2</p><p>∴ Answer: A</p>
Correct Answer: A