<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 y = x^(1/2) + x^(-1/2) to transform the problem into a single-variable optimization, recognizing that x^(3/2) + x^(-3/2) can be expressed using the identity (a+b)³ = a³ + b³ + 3ab(a+b).
<p><strong>Step 1:</strong> Note that x must be positive for f(x) to be defined in real numbers (due to x^(1/2) and x^(-1/2)).</p><p><strong>Step 2:</strong> Let y = x^(1/2) + x^(-1/2). Then y ≥ 2 by AM-GM inequality (since x^(1/2) · x^(-1/2) = 1).</p><p><strong>Step 3:</strong> Find x^(3/2) + x^(-3/2) using the identity: (x^(1/2) + x^(-1/2))³ = x^(3/2) + x^(-3/2) + 3(x^(1/2) + x^(-1/2))</p><p>Therefore: x^(3/2) + x^(-3/2) = y³ - 3y</p><p><strong>Step 4:</strong> Also note that x + 1/x = (x^(1/2))² + (x^(-1/2))² = (x^(1/2) + x^(-1/2))² - 2 = y² - 2</p><p><strong>Step 5:</strong> Substitute into f(x): f = (y³ - 3y) - 4(y² - 2) = y³ - 4y² - 3y + 8</p><p><strong>Step 6:</strong> Let g(y) = y³ - 4y² - 3y + 8 for y ≥ 2. Find g'(y) = 3y² - 8y - 3 = (3y + 1)(y - 3)</p><p><strong>Step 7:</strong> For y ≥ 2, the critical point is y = 3. Check: g(2) = 8 - 16 - 6 + 8 = -6 and g(3) = 27 - 36 - 9 + 8 = -10</p><p><strong>Step 8:</strong> Since g'(y) < 0 for 2 < y < 3 and g'(y) > 0 for y > 3, the minimum occurs at y = 3.</p><p>∴ Answer: B (Minimum value = -10)</p>
Correct Answer: B