Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p>If \(S_1\) and \(S_2\) are respectively the sets of local minimum and local maximum points of the function, \(f(x) = 9x^4 + 12x^3 - 36x^2 + 25\), \(x \in \mathbb{R}\), then:</p>
<p>\(S_1 = \{-2\}\); \(S_2 = \{0, 1\}\)</p>
<p>\(S_1 = \{-2, 0\}\); \(S_2 = \{1\}\)</p>
<p>\(S_1 = \{-2, 1\}\); \(S_2 = \{0\}\)</p>
<p>\(S_1 = \{-1\}\); \(S_2 = \{0, 2\}\)</p>
Step-by-Step Solution
Key Concept: Find critical points by solving f'(x) = 0, then use the second derivative test (f''(x)) to classify them as local maxima or minima. The sign of f''(x) at critical points determines the nature of extrema.
<p><strong>Step 1:</strong> Find f'(x):<br/>f(x) = 9x⁴ + 12x³ - 36x² + 25<br/>f'(x) = 36x³ + 36x² - 72x = 36x(x² + x - 2) = 36x(x + 2)(x - 1)</p><p><strong>Step 2:</strong> Find critical points by setting f'(x) = 0:<br/>36x(x + 2)(x - 1) = 0<br/>Critical points: x = 0, x = -2, x = 1</p><p><strong>Step 3:</strong> Find f''(x):<br/>f''(x) = 108x² + 72x - 72 = 36(3x² + 2x - 2)</p><p><strong>Step 4:</strong> Apply second derivative test at each critical point:<br/>• At x = -2: f''(-2) = 36(12 - 4 - 2) = 36(6) = 216 > 0 → Local minimum<br/>• At x = 0: f''(0) = 36(-2) = -72 < 0 → Local maximum<br/>• At x = 1: f''(1) = 36(3 + 2 - 2) = 36(3) = 108 > 0 → Local minimum</p><p><strong>Step 5:</strong> Identify the sets:<br/>S₁ (local minimum points) = {-2, 1}<br/>S₂ (local maximum points) = {0}</p><p>∴ Answer: C</p>
Correct Answer: C