Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p>Determine the points of maxima and minima of the function \(f(x) = \frac{1}{8}\log x - bx + x^2,\, x > 0\) where \(b \geq 0\) is a constant.</p>
<p>If \(0 \leq b < 1\), \(f(x)\) will have no point of maxima or minima</p>
<p>If \(b > 1\), \(f(x)\) has a minimum at \(x = \frac{b - \sqrt{b^2-1}}{4}\) and maximum at \(x = \frac{b + \sqrt{b^2-1}}{4}\)</p>
<p>If \(b > 1\), \(f(x)\) has a maximum at \(x = \frac{b - \sqrt{b^2-1}}{4}\) and minimum at \(x = \frac{b + \sqrt{b^2-1}}{4}\)</p>
<p>If \(b = 1\), \(f(x)\) has a minimum at \(x = \frac{1}{4}\)</p>
Step-by-Step Solution
Key Concept: Find critical points by setting f'(x) = 0, then use the second derivative test to classify them as maxima or minima. The nature of critical points depends on the parameter b.
<p><strong>Step 1:</strong> Find the first derivative.</p><p>f'(x) = 1/(8x) - b + 2x</p><p><strong>Step 2:</strong> Set f'(x) = 0 to find critical points.</p><p>1/(8x) - b + 2x = 0</p><p>Multiply by 8x: 1 - 8bx + 16x² = 0</p><p>16x² - 8bx + 1 = 0</p><p><strong>Step 3:</strong> Apply the quadratic formula.</p><p>x = (8b ± √(64b² - 64))/32 = (b ± √(b² - 1))/4</p><p><strong>Step 4:</strong> Determine existence of real critical points.</p><p>Real critical points exist when b² - 1 ≥ 0, i.e., b ≥ 1</p><p>When 0 ≤ b < 1: No critical points (f is monotonically increasing)</p><p><strong>Step 5:</strong> Find the second derivative to classify critical points.</p><p>f''(x) = -1/(8x²) + 2</p><p><strong>Step 6:</strong> Evaluate f'' at critical points when b ≥ 1.</p><p>For x₁ = (b - √(b² - 1))/4: f''(x₁) > 0 → point of minima</p><p>For x₂ = (b + √(b² - 1))/4: f''(x₂) < 0 → point of maxima</p><p>∴ Answer: B</p>
Correct Answer: B