The function f(x) = x^3 - 6x^2 + 9x + 2 has a | vedclass

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(A) Given function: $f(x) = x^3 - 6x^2 + 9x + 2$ First,find the derivative $f'(x)$: $f'(x) = 3x^2 - 12x + 9$ To find the critical points,set $f'(x) =

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(A) Given function: $f(x) = x^3 - 6x^2 + 9x + 2$ First,find the derivative $f'(x)$: $f'(x) = 3x^2 - 12x + 9$ To find the critical points,set $f'(x) = 0$: $3(x^2 - 4x + 3) = 0$ $3(x - 1)(x - 3) = 0$ So,the critical points are $x = 1$ and $x = 3$. Now,find the second derivative $f''(x)$: $f''(x) = 6x - 12$ Check the nature of the function at critical points using the second derivative test: At $x = 1$: $f''(1) = 6(1) - 12 = -6$. Since $f''(1) At $x = 3$: $f''(3) = 6(3) - 12 = 6$. Since $f''(3) > 0$,the function has a local minimum at $x = 3$. Therefore,the function has a maximum value when $x = 1$.

The function $f(x) = x^3 - 6x^2 + 9x + 2$ has a maximum value when $x$ is

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