Penyelesaian:
f(x) = x³ – 3x² + 3x + 4
f'(x) = 3x² – 6x + 3
Kurva f(x) akan meraih stasioner jika f'(x) = 0.
f'(x) = 0
⇔ 3x² – 6x + 3 = 0
⇔ 3(x – 1)(x – 1) = 0
⇔ 3(x – 1)(x – 1) = 0
⇔ x = 1 atau x = 1
x = 1 ialah absis titik belok.
f(1) = 1³ – 3 × 1² + 3 × 1 + 4
= 1 – 3 + 3 + 4
= 5
Koordinat titik belok (1, 5).
Jadi, titik belok kurva f(x) = x³ – 3x² + 3x + 4 dicapai di titik (1, 5).