+10
9 месяцев назад
Алгебра
Студенческий
Ответ:
f(X) = 1/3 X³ - X², Х€RR
f'(X) = X²- 2X, Х€R
0 = X² - 2X
X1 = 0; X2 = 2
(-∞; 0), (0; 2)
(0; 2), (2; +∞)
X1 = -1
X2 = 1
X3 = 1
X4 = 3
f'(-1) = (-1²) - 2(-1) = 1 + 3 = 4
f'(1) = 1² - 2*1 = 1 - 2 = -1
f'(3) = 3² - 2*3 = 9 - 6 = 3
f(X) = 1/3 * X³ - X², X=0
f(X) = 1/3 * X³ - X², X=22
f(0) = 1/3 * 0³ - 0² = 0
f(2) = 1/3 * 2³ - 2² = 1/3 * 8 - 4 = 8/3 - 4 = -4/3
Х max = 0
X min = -4/3
Объяснение:
Ответ:
f(X) = 1/3 X³ - X², Х€RR
f'(X) = X²- 2X, Х€R
0 = X² - 2X
X1 = 0; X2 = 2
(-∞; 0), (0; 2)
(0; 2), (2; +∞)
X1 = -1
X2 = 1
X3 = 1
X4 = 3
f'(-1) = (-1²) - 2(-1) = 1 + 3 = 4
f'(1) = 1² - 2*1 = 1 - 2 = -1
f'(1) = 1² - 2*1 = 1 - 2 = -1
f'(3) = 3² - 2*3 = 9 - 6 = 3
f(X) = 1/3 * X³ - X², X=0
f(X) = 1/3 * X³ - X², X=22
f(0) = 1/3 * 0³ - 0² = 0
f(2) = 1/3 * 2³ - 2² = 1/3 * 8 - 4 = 8/3 - 4 = -4/3
Х max = 0
X min = -4/3
Объяснение: