Nilai maksimum dari fungsi f(x) = 1/3x³ - 3/2x² + 2x + 9 pada interval 0 ≤ x ≤ 3
adalah ...
Nilai stasioner
f'(x) = 0
x² - 3x + 2 = 0
(x - 2)(x - 1) = 0
x₁ = 1 ; x₂ = 2
x = 0 ➙ f(0) = 0 - 0 + 0 + 9 = 9
x = 1 ➙ f(1) = 1(1)³ - 3(1)² + 2(1) + 9 = 59
3 2 6
x = 2 ➙ f(2) = 1(2)³ - 3(2)² + 2(2) + 9 = 9 2
3 2 3
x = 3 ➙ f(3) = 1(3)³ - 3(3)² + 2(3) + 9 = 10 1
3 2 2
Jadi, fungsi maksimum = 10 1
2