Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p>\(f(x) = \dfrac{(x-2)(x-1)}{(x-3)}\), \(\forall x > 3\). The minimum value of \(f(x)\) is equal to</p>
<p>(1) \(3 + 2\sqrt{2}\)</p>
<p>(2) \(3 + 2\sqrt{3}\)</p>
<p>(3) \(3\sqrt{2} + 2\)</p>
<p>(4) \(3\sqrt{2} - 2\)</p>
Step-by-Step Solution
Key Concept: Use calculus to find critical points by setting f'(x) = 0, then verify it's a minimum using the second derivative test or analyzing sign changes of f'(x).
<p><strong>Step 1:</strong> Find f'(x) using the quotient rule.</p><p>f(x) = [(x-2)(x-1)]/(x-3)</p><p>Numerator: (x-2)(x-1) = x² - 3x + 2</p><p>Using quotient rule: f'(x) = [(2x-3)(x-3) - (x²-3x+2)(1)]/(x-3)²</p><p><strong>Step 2:</strong> Simplify the numerator of f'(x).</p><p>= [(2x-3)(x-3) - (x²-3x+2)]/(x-3)²</p><p>= [2x² - 6x - 3x + 9 - x² + 3x - 2]/(x-3)²</p><p>= [x² - 6x + 7]/(x-3)²</p><p><strong>Step 3:</strong> Set f'(x) = 0 to find critical points.</p><p>x² - 6x + 7 = 0</p><p>x = [6 ± √(36-28)]/2 = [6 ± √8]/2 = [6 ± 2√2]/2 = 3 ± √2</p><p><strong>Step 4:</strong> Check which critical point lies in domain x > 3.</p><p>x = 3 + √2 ≈ 4.414 > 3 ✓ (valid)</p><p>x = 3 - √2 ≈ 1.586 < 3 ✗ (invalid)</p><p><strong>Step 5:</strong> Verify this is a minimum and calculate f(3 + √2).</p><p>At x = 3 + √2: f(x) = [(3+√2-2)(3+√2-1)]/[(3+√2-3)]</p><p>= [(1+√2)(2+√2)]/√2</p><p>= [2 + √2 + 2√2 + 2]/√2 = [4 + 3√2]/√2</p><p>= 4/√2 + 3 = 2√2 + 3</p><p>∴ Minimum value = 3 + 2√2 or equivalently (3 + 2√2)</p>
Correct Answer: A