taylor series (deret taylor):
f(x) = f(a) + f'(a)•(x-a) + f''(a)/2! • (x-a)² + f"'(a)/3! • (x-a)³
(berhenti di 3 karena polinomnya berderajat 3).
f(x) = x³-18
f(2) = 2³-18 = -10
f'(x) = 3x²
f'(2) = 12
f"(x) = 6x
f"(2) = 12
f"'(x) = 6
f"'(2) = 6
berarti deret taylornya:
f(x) = f(2) + f'(2)•(x-2) + f"(2)/2! • (x-2)² + f"'(2)/3! • (x-2)³
f(x) = -10 + 12•(x-2) + 12/2 • (x-2)² + 6/6 • (x-2)³
f(x) = (x-2)³ + 6(x-2)² + 12(x-2) - 10
Jawab:
Penjelasan dengan langkah-langkah:
[tex]\displaystyle f(x) = x^3 - 18\\f(x) = x^3 - 3\cdot 2\cdot x^2 + 3\cdot 4\cdot x - 8 + 3\cdot 2\cdot x^2 - 3\cdot 4\cdot x + 8-18\\f(x) = (x-2)^3 + 6x^2 - 12x - 10 = (x-2)^3 + 3(x^2+x^2-4x) - 10\\f(x) = (x-2)^3 + 3(x^2-4x+4 - 4)+3x^2 - 10 \\f(x) = (x-2)^3 + 3(x-2)^2+3x^2 - 10 - 12\\f(x) = (x-2)^3 + 3(x-2)^2+3x^2-3\cdot 4\cdot x + 3\cdot 4\\ +3\cdot 4\cdot x - 3\cdot 4 - 22\\f(x) = (x-2)^3 + 6(x-2)^2 + 12x-34 \\[/tex]
[tex]f(x)= (x-2)^3 + 6(x-2)^2 + 12x-24 - 10 \\\\ \boxed{\boxed{f(x) = (x-2)^3 + 6(x-2)^2 + 12(x-2)-10}}\\\\[/tex]
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