Боковое ребро правильной четырехугольной призмы АВСDA1B1C1D1 равно 6√3 м, а сторона основания равна 6 м. Найдите угол между прямыми АВ1 и с подробным решением и рисунком,
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Пошаговое объяснение:
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ответ: а) промежутки возрастания: (-беск, -3) и (3, беск), промежуток убывания (-3, 3). Точка максимума: x = -3, Точка минимума: x = 3.
Пошаговое объяснение:Ищем первую производную: f'(x) = x^2 - 9 = (x-3)(x+3)
Нули производной: f'(x) = 0 = (x - 3)(x + 3); x = 3 или x = -3.
Проверим три промежутка: (-беск, -3), (-3, 3) и (3, беск)
1. (-беск, -3): f'(-5) = 16 - функция возрастает.
2. (-3, 3): f'(0) = -9, функция убывает.
3. (3, беск): f'(5) = 16 - функция возрастает.
Точка максимума: x = -3
Точка минимума: x = 3
Снизу график функции, подтверждающий строки сверху