Titik stasioner 1/3x^2 – 2x^2+3x+5

Syarat titik stasioner f'(x) = 0
Langkah pertama mencari turunan f(x) = 1/3 x^2 – 2x^2 +3x + 5
f'(x) = 0
2/3x – 4x + 3 = 0
2/3x – 12/3 x + 3
-10/3 x + 3 = 0
-10/3 x = -3
x = -3 : -10/3
x = -3 x – 3/ 10
x = 9/10

Sub x = 9/10 ke f(x)
f (9/10) = 1/3 . (9/10)^2 – 2. (9/10)^2 + 3. 9/10 + 5
= (1/3. 81/100 ) – (2. 81/100) + 27/10 + 5
= 81/300 – 162/100 + 27/10 + 5
= 27/100 – 162/100 + 270/100 + 500/100
= 635/ 100 = 6 35/100 = 6 7/20
Titik stasioner ( 9/10, 6 7/20) atau (9/10, 127/ 20)