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Verified answer
Jawaby = 1/3 x³ - 3/2 x² - 4x + 2
y' = 0
x² - 3x - 4 = 0
(x - 4)(x +1)= 0
x= 4 atau x = - 1
x= 4 → y = 1/3 (4³) - 3/2 . 4² - 4(4) + 2
y = 64/3 - 24 - 16 + 2
y = - 16 ²/₃
x = -1 → y = 1/3(-1)³ - 3/2.(-1)² - 4(-1) + 2
y = -1/3 - 3/2 + 4 + 2
y = ²⁵/₄ = 4 ¹/₆
nilai maksimum y = 4 ¹/₆
Verified answer
F(x) = 1/3 x³ - 3/2 x² - 4x + 2Maksimum ---> f'(x) = 0
f'(x) = 0
x² - 3x - 4 = 0
(x - 4)(x + 1) = 0
x = 4 atau x = -1
f(4) = 1/3 .4³ - 3/2 .4² - 4.4 + 2
f(4) = 64/3 - 24 - 16 + 2
f(4) = 21 1/3 - 38
f(4) = -16 2/3
f(-1) = 1/3 . (-1)³ - 3/2 . (-1)² - 4(-1) + 2
f(-1) = -1/3 - 3/2 + 6
f(-1) = -11/6 + 6
f(-1) = -1 5/6 + 6
f(-1) = 4 1/6
NiLai maksimum = 4 1/6 = 25/6