For the function f(x)=x^3-6x^2-12x-3,x=2 is a | vedclass

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Question

(C) Given the function $f(x) = x^3 - 6x^2 - 12x - 3$. First,find the first derivative: $f'(x) = 3x^2 - 12x - 12$. Next,find the second derivative: $f'

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point of inflection

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(C) Given the function $f(x) = x^3 - 6x^2 - 12x - 3$. First,find the first derivative: $f'(x) = 3x^2 - 12x - 12$. Next,find the second derivative: $f''(x) = 6x - 12$. Evaluate the second derivative at $x = 2$: $f''(2) = 6(2) - 12 = 12 - 12 = 0$. Since $f'(2) = 3(2)^2 - 12(2) - 12 = 12 - 24 - 12 = -24 
eq 0$,$x=2$ is not a critical point. However,checking the options provided,the question implies evaluating the nature of the point where $f''(x)=0$. Since $f''(x)$ changes sign at $x=2$ (for $x  2$,$f''(x) > 0$),$x=2$ is a point of inflection.

For the function $f(x)=x^3-6x^2-12x-3$,$x=2$ is a

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