Applications of Derivatives
Local Maxima and Minima
Grade 12
Question:
<p>The total number of local maxima and local minima of the function \(f(x) = \begin{cases} (2x)^3, & 3 < x \leq 1 \\ 0, & 0 \leq x \\ x, & 3 < x \leq 2 \end{cases}\) is</p>
<p>(a) 0</p>
<p>(b) 1</p>
<p>(c) 2</p>
<p>(d) 3</p>
Step-by-Step Solution
Key Concept: A piecewise function's local extrema occur either at critical points within each piece (where f'(x) = 0) or at boundary points where the function's behavior changes. We must check continuity and differentiability at the junction point x = 3.
<p><strong>Step 1: Identify the structure</strong></p><p>We have a piecewise function with:</p><p>• First piece: f(x) = (2x)³ = 8x³ for x ≤ 3</p><p>• Second piece: Incomplete in the problem statement (starts with '3 < x' but the function definition is missing)</p><p><strong>Step 2: Analyze the first piece for critical points</strong></p><p>For f(x) = 8x³ where x ≤ 3:</p><p>f'(x) = 24x²</p><p>Setting f'(x) = 0: 24x² = 0 ⟹ x = 0</p><p>Since f''(x) = 48x, we have f''(0) = 0 (inconclusive test).</p><p>For x < 0: f'(x) > 0 (increasing); for x > 0: f'(x) > 0 (increasing)</p><p>So x = 0 is NOT a local extremum in the first piece (function is monotonically increasing).</p><p><strong>Step 3: Check boundary point x = 3</strong></p><p>f(3) = 8(3)³ = 216</p><p>Left derivative at x = 3: f'(3⁻) = 24(3)² = 216</p><p>Right derivative at x = 3: Cannot be determined (second piece undefined)</p><p><strong>Step 4: Conclusion</strong></p><p>Without the complete definition of the second piece of the function, we cannot determine whether x = 3 is a local extremum or identify any critical points in the second interval. The problem statement is incomplete.</p><p><strong>∴ Answer:</strong> Unknown</p>
Correct Answer: Unknown