Applications of Derivatives
Local maxima and minima
Grade 12
Question:
<p>Given <br>\(f(x) = x^2 + \dfrac{1}{x^2}\) and \(g(x) = x - \dfrac{1}{x}\),<br>\(h(x) = \dfrac{f(x)}{g(x)}\). The local maximum and local minimum values of \(h(x)\) are respectively:</p>
<p>\(-2\sqrt{2}\) and \(2\sqrt{2}\)</p>
<p>\(2\sqrt{2}\) and \(-2\sqrt{2}\)</p>
<p>\(2\sqrt{2}\) and \(2\sqrt{2}\)</p>
<p>\(-2\sqrt{2}\) and \(-2\sqrt{2}\)</p>
Step-by-Step Solution
Key Concept: Recognize that h(x) = f(x)/g(x) can be simplified using algebraic factorization before differentiation. Note that f(x) = (g(x))² + 2, which transforms the problem into finding extrema of a simpler function.
<p><strong>Step 1:</strong> Simplify h(x) by expressing f(x) in terms of g(x).</p><p>Note that g(x) = x - 1/x, so [g(x)]² = x² - 2 + 1/x² = f(x) - 2</p><p>Therefore: f(x) = [g(x)]² + 2</p><p><strong>Step 2:</strong> Substitute into h(x):</p><p>h(x) = f(x)/g(x) = ([g(x)]² + 2)/g(x) = g(x) + 2/g(x)</p><p>Let u = g(x) = x - 1/x, then h(x) = u + 2/u</p><p><strong>Step 3:</strong> Find critical points by differentiating with respect to u:</p><p>dh/du = 1 - 2/u² = 0</p><p>u² = 2 → u = ±√2</p><p><strong>Step 4:</strong> Determine nature of extrema using second derivative:</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:</strong> Calculate extrema values:</p><p>At u = √2: h = √2 + 2/√2 = √2 + √2 = 2√2 (local minimum)</p><p>At u = -√2: h = -√2 + 2/(-√2) = -√2 - √2 = -2√2 (local maximum)</p><p>∴ Local maximum = -2√2, Local minimum = 2√2</p>
Correct Answer: A