The maximum and minimum values of {x^3} - 18{ | vedclass

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(C) Let $f(x) = {x^3} - 18{x^2} + 96x$. First,find the derivative: $f'(x) = 3{x^2} - 36x + 96$. Set $f'(x) = 0$ to find critical points: $3(x^2 - 12x

Vedclass · Accepted Answer

$160, 128$

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(C) Let $f(x) = {x^3} - 18{x^2} + 96x$. First,find the derivative: $f'(x) = 3{x^2} - 36x + 96$. Set $f'(x) = 0$ to find critical points: $3(x^2 - 12x + 32) = 0 \Rightarrow 3(x - 4)(x - 8) = 0$. Thus,the critical points are $x = 4$ and $x = 8$. Now,find the second derivative: $f''(x) = 6x - 36$. At $x = 4$,$f''(4) = 6(4) - 36 = -12 The maximum value is $f(4) = (4)^3 - 18(4)^2 + 96(4) = 64 - 288 + 384 = 160$. At $x = 8$,$f''(8) = 6(8) - 36 = 12 > 0$,so $f(x)$ has a local minimum at $x = 8$. The minimum value is $f(8) = (8)^3 - 18(8)^2 + 96(8) = 512 - 1152 + 768 = 128$. Therefore,the maximum and minimum values are $160$ and $128$ respectively.

The maximum and minimum values of ${x^3} - 18{x^2} + 96x$ in the interval $(0, 9)$ are

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