<p>For \(x \ge 0\), the smallest value of the function \(f(x) = \dfrac{4x^2 + 8x + 13}{6(1+x)}\) is ______.</p>
Step-by-Step Solution
Key Concept: Rewrite the numerator in terms of the denominator's variable (1+x) to reveal a sum of terms whose minimum can be found using AM-GM inequality or calculus.
<p><strong>Step 1:</strong> Let u = 1 + x, so x = u - 1 and u ≥ 1 (since x ≥ 0).</p><p><strong>Step 2:</strong> Substitute into f(x):<br/>f(u) = [4(u-1)² + 8(u-1) + 13] / [6u]<br/>= [4(u² - 2u + 1) + 8u - 8 + 13] / [6u]<br/>= [4u² - 8u + 4 + 8u - 8 + 13] / [6u]<br/>= [4u² + 9] / [6u]<br/>= (4u/6) + (9/6u)<br/>= (2u/3) + (3/2u)</p><p><strong>Step 3:</strong> Apply AM-GM inequality:<br/>(2u/3) + (3/2u) ≥ 2√[(2u/3) · (3/2u)] = 2√1 = 2</p><p><strong>Step 4:</strong> Equality holds when (2u/3) = (3/2u), giving 4u² = 9, so u = 3/2 (since u > 0).<br/>This corresponds to x = u - 1 = 1/2 ≥ 0 ✓</p><p><strong>Verification:</strong> f(1/2) = [4(1/4) + 4 + 13] / [6(3/2)] = [1 + 4 + 13] / 9 = 18/9 = 2</p><p>∴ Answer: <strong>2</strong></p>
Correct Answer: 2