The absolute minimum value of the function f( | vedclass

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(A) To find the absolute minimum value of the function $f(x) = x^3 - 18x^2 + 96x$ on the interval $[0, 9]$, we first find the derivative $f'(x)$.$f'(x

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$0$

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(A) To find the absolute minimum value of the function $f(x) = x^3 - 18x^2 + 96x$ on the interval $[0, 9]$, we first find the derivative $f'(x)$.$f'(x) = 3x^2 - 36x + 96$.Setting $f'(x) = 0$ to find critical points:$3(x^2 - 12x + 32) = 0$$3(x - 4)(x - 8) = 0$So, the critical points are $x = 4$ and $x = 8$.Now, we evaluate the function $f(x)$ at the critical points and the endpoints of the interval $[0, 9]$:$f(0) = 0^3 - 18(0)^2 + 96(0) = 0$$f(4) = 4^3 - 18(4)^2 + 96(4) = 64 - 288 + 384 = 160$$f(8) = 8^3 - 18(8)^2 + 96(8) = 512 - 1152 + 768 = 128$$f(9) = 9^3 - 18(9)^2 + 96(9) = 729 - 1458 + 864 = 135$Comparing these values $(0, 160, 128, 135)$, the absolute minimum value is $0$.

The absolute minimum value of the function $f(x) = x^3 - 18x^2 + 96x$ for $x \in [0, 9]$ is:

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