Q.2 (B)(ii) Determine the minimum value of the function: f(x) = 2x³ - 21x² + 36x - 20.

Answer: Minimum value is -128, at x = 6. f'(x) = 6x² - 42x + 36 = 6(x² - 7x + 6) = 6(x-1)(x-6). For extreme values, f'(x) = 0 ⇒ x = 1 or x = 6. f''(x) = 12x - 42. At x = 1: f''(1) = -30 lt 0, so f has a maximum at x = 1. At x = 6: f''(6) = 72 - 42 = 30 gt 0, so f has a minimum at x = 6. Minimum value = f(6) = 2(216) - 21(36) + 36(6) - 20 = 432 - 756 + 216 - 20 = -128.

Chapter: Applications of Derivatives· Mathematics & Statistics (Commerce) Part 1

Determine the minimum value of the function : \(f(x) = 2x^3 - 21x^2 + 36x - 20\).

Q.2 (B)(ii) Determine the minimum value of the function : \(f(x) = 2x^3 - 21x^2 + 36x - 20\).

Doubtly Asked 29 Oct 2026 Board exam · March 2024 hsc 12th Save WhatsApp Report

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Answer: Minimum value is \(-128\), at \(x = 6\).

\(f'(x) = 6x^2 - 42x + 36 = 6(x^2 - 7x + 6) = 6(x-1)(x-6)\).

For extreme values, \(f'(x) = 0 \Rightarrow x = 1\) or \(x = 6\).

\(f''(x) = 12x - 42\).

At \(x = 1\): \(f''(1) = -30 \lt 0\), so \(f\) has a maximum at \(x = 1\).
At \(x = 6\): \(f''(6) = 72 - 42 = 30 \gt 0\), so \(f\) has a minimum at \(x = 6\).

Minimum value \(= f(6) = 2(216) - 21(36) + 36(6) - 20 = 432 - 756 + 216 - 20 = -128\).

Doubtly AI · 29 Oct 2026 AI-verified · Nov 2026 Report

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Determine the minimum value of the function : \(f(x) = 2x^3 - 21x^2 + 36x - 20\).

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