The minimum value of the function f(x) = 2 x^ | vedclass

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(A) First,determine the set $A$ by solving the inequality $x^2 + 20 \le 9 x$. $x^2 - 9 x + 20 \le 0$ $(x - 4)(x - 5) \le 0$ Thus,$A = [4, 5]$. Next,an

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$-16$

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(A) First,determine the set $A$ by solving the inequality $x^2 + 20 \le 9 x$. $x^2 - 9 x + 20 \le 0$ $(x - 4)(x - 5) \le 0$ Thus,$A = [4, 5]$. Next,analyze the function $f(x) = 2 x^3 - 15 x^2 + 36 x - 48$. Find the derivative: $f'(x) = 6 x^2 - 30 x + 36 = 6(x^2 - 5 x + 6) = 6(x - 2)(x - 3)$. For $x \in [4, 5]$,both $(x - 2)$ and $(x - 3)$ are positive,so $f'(x) > 0$. Since $f'(x) > 0$ on the interval $[4, 5]$,the function $f(x)$ is strictly increasing on $A = [4, 5]$. The minimum value occurs at the left endpoint $x = 4$. $f(4) = 2(4)^3 - 15(4)^2 + 36(4) - 48 = 2(64) - 15(16) + 144 - 48 = 128 - 240 + 144 - 48 = -16$.

The minimum value of the function $f(x) = 2 x^3 - 15 x^2 + 36 x - 48$ on the set $A = \{x \mid x^2 + 20 \le 9 x\}$ is

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