Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p>Let \(f(x) = x^2 + \dfrac{1}{x^2}\) and \(g(x) = x - \dfrac{1}{x}\), \(x \in \mathbb{R} - \{-1, 0, 1\}\). If \(h(x) = \dfrac{f(x)}{g(x)}\), then the local minimum value of \(h(x)\) is</p>
<p>\(-3\)</p>
<p>\(-2\sqrt{2}\)</p>
<p>\(2\sqrt{2}\)</p>
<p>3</p>
Step-by-Step Solution
Key Concept: Express h(x) by factoring the numerator using the algebraic identity a² + b² = (a-b)² + 2ab, which reveals that f(x) = g(x)² + 2. This transforms h(x) into g(x) + 2/g(x), a form amenable to AM-GM inequality or calculus optimization.
<p><strong>Step 1: Recognize the algebraic relationship</strong></p><p>Note that g(x) = x - 1/x, so g(x)² = x² - 2 + 1/x²</p><p>Therefore: f(x) = x² + 1/x² = g(x)² + 2</p><p><strong>Step 2: Simplify h(x)</strong></p><p>h(x) = f(x)/g(x) = (g(x)² + 2)/g(x) = g(x) + 2/g(x)</p><p><strong>Step 3: Apply calculus to find critical points</strong></p><p>Let u = g(x), then h = u + 2/u</p><p>dh/du = 1 - 2/u²</p><p>Setting dh/du = 0: u² = 2, so u = ±√2</p><p><strong>Step 4: Determine local minimum using second derivative</strong></p><p>d²h/du² = 4/u³</p><p>When u = √2: d²h/du² = 4/(2√2) > 0 (local minimum)</p><p>When u = -√2: d²h/du² = 4/(-2√2) < 0 (local maximum)</p><p><strong>Step 5: Calculate the local minimum value</strong></p><p>At u = √2: h = √2 + 2/√2 = √2 + √2 = 2√2</p><p>∴ <strong>Answer: C (local minimum value = 2√2)</strong></p>
Correct Answer: C