<p>Which of the following functions is increasing on \((-\infty, \infty)\)?<br>(1) \(f(x) = x^3 - 3x^2 + 3x + 3\)<br>(2) \(f(x) = 2x^3 - 3x^2 - 12x + 6\)<br>(3) \(f(x) = 3x^2 - 2x + 1\)</p>
Step-by-Step Solution
Key Concept: A function is increasing on (-∞, ∞) if and only if f'(x) ≥ 0 for all x, with equality at isolated points only. Check the discriminant of f'(x) to ensure it has no real roots or remains non-negative everywhere.
<p><strong>Step 1: Check function (1):</strong> f(x) = x³ - 3x² + 3x + 3</p><p>f'(x) = 3x² - 6x + 3 = 3(x² - 2x + 1) = 3(x - 1)²</p><p>Since f'(x) = 3(x - 1)² ≥ 0 for all x, and equals 0 only at x = 1 (isolated point), f is increasing on (-∞, ∞). ✓</p><p><strong>Step 2: Check function (2):</strong> f(x) = 2x³ - 3x² - 12x + 6</p><p>f'(x) = 6x² - 6x - 12 = 6(x² - x - 2) = 6(x - 2)(x + 1)</p><p>f'(x) < 0 for x ∈ (-1, 2), so f is NOT increasing on (-∞, ∞). ✗</p><p><strong>Step 3: Check function (3):</strong> f(x) = 3x² - 2x + 1</p><p>f'(x) = 6x - 2</p><p>f'(x) < 0 for x < 1/3, so f is NOT increasing on (-∞, ∞). ✗</p><p><strong>Conclusion:</strong> Only function (1) satisfies the condition because its derivative is a perfect square trinomial, guaranteeing non-negativity everywhere.</p><p>∴ Answer: A (function 1)</p>
Correct Answer: A