nilai maksimum fungsi F(x) = 1/3 x pangkat 3 -3/2 xpangkat 2 - 4x + 2 adalah

jawab

y = 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

f(x) = 1/3 x³ - 3/2 x² - 4x + 2

Maksimum ---> 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