f'(2) + f"(2) + f"'(2) = 2

Pembahasan

f(x) = ½x³ – 3x² + 5x – 8

• f'(x) = (½x³ – 3x² + 5x – 8)'
  ⇔ f'(x) = 3·½x² – 2·3x + 5
  ⇔ f'(x) = (3/2)x² – 6x + 5
• f"(x) = [(3/2)x² – 6x + 5]'
  ⇔ f"(x) = 2·(3/2)x – 6
  ⇔ f"(x) = 3x – 6
• f"'(x) = (3x)'
  ⇔ f"'(x) = 3

Cara pertama: hitung langsung

f'(2) + f"(2) + f"'(2)
= (3/2)(2²) – 6(2) + 5 + 3(2) – 6 + 3
= (3/2)(4) – 12 + 5 + 6 – 6 + 3
= 6 – 12 + 5 + 3
= –6 + 8
= 2

Cara kedua: jumlahkan dulu turunan-turunannya menjadi fungsi lain

g(x) = f'(x) + f"(x) + f"'(x)
⇔ g(x) = (3/2)x² – 6x + 5 + 3x – 6 + 3
⇔ g(x) = (3/2)x² – 3x + 2

Substitusi x dengan 2.
g(2) = (3/2)(2²) – 3(2) + 2
⇔ g(2) = (3/2)(4) – 6 + 2
⇔ g(2) = 6 – 6 + 2
⇔ g(2) = 2
⇔ f'(2) + f"(2) + f"'(2) = 2

KESIMPULAN

∴ f'(2) + f"(2) + f"'(2) = 2