What is the maximum value of f(x) = x^3 - 18x | vedclass

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(C) Given function is $f(x) = x^3 - 18x^2 + 96x$. Find the first derivative: $f'(x) = 3x^2 - 36x + 96$. Set $f'(x) = 0$ to find critical points: $3(x^

Vedclass · Accepted Answer

$160$

Vedclass · Answer

(C) Given function is $f(x) = x^3 - 18x^2 + 96x$. Find the first derivative: $f'(x) = 3x^2 - 36x + 96$. Set $f'(x) = 0$ to find critical points: $3(x^2 - 12x + 32) = 0$. Factor the quadratic: $3(x - 4)(x - 8) = 0$. Thus,the critical points are $x = 4$ and $x = 8$. Find the second derivative: $f''(x) = 6x - 36$. Check for local maxima at $x = 4$: $f''(4) = 6(4) - 36 = 24 - 36 = -12 Since $f''(4) Calculate the maximum value: $f(4) = (4)^3 - 18(4)^2 + 96(4) = 64 - 18(16) + 384 = 64 - 288 + 384 = 160$. Therefore,the maximum value in the interval $(0, 9)$ is $160$.

What is the maximum value of $f(x) = x^3 - 18x^2 + 96x$ in the interval $(0, 9)$?

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