Evaluate f''(c) at the point. f(x)=(x²-6)(x³-5), c=1 OA. f''(1)=-26 B. f''(1)=-16 OC. f''(1)=24 D. f''(1)=16
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$$f(x) = (x^2 - 6)(x^3 - 5)$$ Using the product rule, we have: $$f'(x) = (2x)(x^3 - 5) + (x^2 - 6)(3x^2)$$ $$f'(x) = 2x^4 - 10x + 3x^4 - 18x^2$$ $$f'(x) = 5x^4 - 18x^2 - 10x$$ Show more…
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