217.- Решите неравенство f'(x) < 0, если: a) f (x) = x 6x2 - 63x; 6) f (x) = 3x 5x2 + x³; B) f (x) = x²-8x; r) f (x) = 3x² - 9x - 3 3

Решение: a) f(x) = x³ - 6x² - 63x f'(x) = 3x² - 12x - 63 3x² - 12x - 63 < 0 x² - 4x - 21 < 0 D = (-4)² - 4 * 1 * (-21) = 16 + 84 = 100 x1 = (4 + √100) / 2 = (4 + 10) / 2 = 14 / 2 = 7 x2 = (4 - √100) / 2 = (4 - 10) / 2 = -6 / 2 = -3 -3 < x < 7 б) f(x) = 3x - 5x² + x³ f'(x) = 3 - 10x + 3x² 3x² - 10x + 3 < 0 D = (-10)² - 4 * 3 * 3 = 100 - 36 = 64 x1 = (10 + √64) / 6 = (10 + 8) / 6 = 18 / 6 = 3 x2 = (10 - √64) / 6 = (10 - 8) / 6 = 2 / 6 = 1/3 1/3 < x < 3 в) f(x) = 2/3 x³ - 8x f'(x) = 2x² - 8 2x² - 8 < 0 x² - 4 < 0 x² < 4 -2 < x < 2 г) f(x) = 3x² - 9x - (1/3)x³ f'(x) = 6x - 9 - x² -x² + 6x - 9 < 0 x² - 6x + 9 > 0 (x - 3)² > 0 x ≠ 3 Ответ: a) -3 < x < 7 б) 1/3 < x < 3 в) -2 < x < 2 г) x ≠ 3