213. Решите уравнение f'(x) = 0, если: a) f (x) = 2x²-x; 6) f (x) = x²+x²+12; 3 B) f (x) = x -1,5x²-4x; 3 r) f (x) = 2x 5x2.

Решим уравнение f'(x) = 0 для каждой функции: a) f(x) = 2x² - x f'(x) = 4x - 1 4x - 1 = 0 4x = 1 x = 1/4 б) f(x) = -2/3 x³ + x² + 12 f'(x) = -2x² + 2x -2x² + 2x = 0 2x(-x + 1) = 0 x = 0 или x = 1 в) f(x) = x³/3 - 1.5x² - 4x f'(x) = x² - 3x - 4 x² - 3x - 4 = 0 D = (-3)² - 4 * 1 * (-4) = 9 + 16 = 25 x1 = (3 + √25) / 2 = (3 + 5) / 2 = 8 / 2 = 4 x2 = (3 - √25) / 2 = (3 - 5) / 2 = -2 / 2 = -1 г) f(x) = 2x - 5x² f'(x) = 2 - 10x 2 - 10x = 0 10x = 2 x = 2/10 = 1/5 Ответ: a) x = 1/4 б) x = 0, x = 1 в) x = 4, x = -1 г) x = 1/5