If f(x) = 2x^3 - 21x^2 + 36x - 30,then which | vedclass

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(C) Given $f(x) = 2x^3 - 21x^2 + 36x - 30$. First,find the derivative $f'(x)$: $f'(x) = 6x^2 - 42x + 36$. Set $f'(x) = 0$ to find critical points: $6(

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$f(x)$ has a maximum at $x = 1$

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(C) Given $f(x) = 2x^3 - 21x^2 + 36x - 30$. First,find the derivative $f'(x)$: $f'(x) = 6x^2 - 42x + 36$. Set $f'(x) = 0$ to find critical points: $6(x^2 - 7x + 6) = 0 \Rightarrow 6(x - 1)(x - 6) = 0$. Thus,the critical points are $x = 1$ and $x = 6$. Now,find the second derivative $f''(x)$: $f''(x) = 12x - 42$. Evaluate $f''(x)$ at the critical points: For $x = 1$: $f''(1) = 12(1) - 42 = -30 For $x = 6$: $f''(6) = 12(6) - 42 = 72 - 42 = 30 > 0$. Since $f''(6) > 0$,$f(x)$ has a local minimum at $x = 6$. Therefore,$f(x)$ has a maximum at $x = 1$.

If $f(x) = 2x^3 - 21x^2 + 36x - 30$,then which one of the following is correct?

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