The Taylor expansion of f for x→ 10 is f(x) = 10 + 3(x-10)² + 2(x-10)³ + 0((x a) ft (10) = -12 6) fr (10) = 12 c) f(0) = 12 f" (10)=2 e) f(0) = -12 how I solved f(x) = 6(x-10) + 6(x-10)² f"(x) = 6+12(2-10) f(2)=12 is it bor c ???

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