Question

Diketahui f(x) = ⅓x³ + x²-3x + 1. Titik-titik stasioner dari ƒ (x) adalah ​

Answer 1

Jawaban:

titik stasioner =(3, -7) dan (-1, 11/3)

Answer 2

aplikasi turunan

f(x) = 1/3 x³ + x² - 3x + 1

syarat stasioner → f'(x) = 0

f'(x) = 0

x² + 2x - 3 = 0

(x + 3)(x - 1) = 0

x = -3 atau x = 1

x = -3

f(-3) = 1/3 (-3)³ + (-3)² - 3(-3) + 1

f(-3) = -9 + 9 + 9 + 1

f(-3) = 10

x = 1

f(1) = 1/3 . 1³ + 1² - 3.1 + 1

f(1) = 1/3 + 1 - 3 + 1

f(1) = -2/3

titik stasioner :

(-3,10) dan (1,-2/3)