<p>The minimum value of the function \(f(x) = x^{3/2} + x^{-3/2} - 4\left(x + \dfrac{1}{x}\right)\) for all permissible real \(x\), is:</p>
Step-by-Step Solution
Key Concept: Substitute u = √x + 1/√x to reduce the problem to a single variable, then use calculus to find the minimum of the transformed function. The constraint u ≥ 2 (from AM-GM) ensures you find the global minimum over the permissible domain.
<p><strong>Step 1:</strong> Note that f(x) is defined for x > 0 (permissible real x means positive x).</p><p><strong>Step 2:</strong> Let u = √x + 1/√x. By AM-GM, u ≥ 2.</p><p><strong>Step 3:</strong> Square u: u² = x + 2 + 1/x, so x + 1/x = u² - 2.</p><p><strong>Step 4:</strong> Express f(x) in terms of u. Note that x^(3/2) + x^(-3/2) = (√x)³ + (1/√x)³ = (√x + 1/√x)³ - 3(√x + 1/√x) = u³ - 3u.</p><p><strong>Step 5:</strong> Therefore, f(x) = u³ - 3u - 4(u² - 2) = u³ - 4u² - 3u + 8.</p><p><strong>Step 6:</strong> Let g(u) = u³ - 4u² - 3u + 8 for u ≥ 2. Find g'(u) = 3u² - 8u - 3 = (3u + 1)(u - 3).</p><p><strong>Step 7:</strong> Critical point in [2, ∞) is at u = 3 (since 3u + 1 = 0 gives u = -1/3, outside domain).</p><p><strong>Step 8:</strong> g(3) = 27 - 36 - 9 + 8 = -10. Check g(2) = 8 - 16 - 6 + 8 = -6. Since g'(u) < 0 for 2 ≤ u < 3 and g'(u) > 0 for u > 3, u = 3 gives the minimum.</p><p>∴ <strong>Answer: -10</strong></p>
Correct Answer: B