The maximum value of the function f(x) = 3x^3 | vedclass

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(C) First,we determine the interval $S$ by solving the inequality $x^2 + 30 \le 11x$.$x^2 - 11x + 30 \le 0$$(x - 5)(x - 6) \le 0$This implies $x \in [

Vedclass · Accepted Answer

$122$

Vedclass · Answer

(C) First,we determine the interval $S$ by solving the inequality $x^2 + 30 \le 11x$.$x^2 - 11x + 30 \le 0$$(x - 5)(x - 6) \le 0$This implies $x \in [5, 6]$.Now,we find the critical points of $f(x) = 3x^3 - 18x^2 + 27x - 40$.$f'(x) = 9x^2 - 36x + 27 = 9(x^2 - 4x + 3) = 9(x - 1)(x - 3)$.The critical points are $x = 1$ and $x = 3$. Neither of these points lies in the interval $[5, 6]$.Since $f'(x) > 0$ for all $x > 3$,the function $f(x)$ is strictly increasing on the interval $[5, 6]$.Therefore,the maximum value occurs at the right endpoint $x = 6$.$f(6) = 3(6)^3 - 18(6)^2 + 27(6) - 40$$f(6) = 3(216) - 18(36) + 162 - 40$$f(6) = 648 - 648 + 162 - 40 = 122$.

The maximum value of the function $f(x) = 3x^3 - 18x^2 + 27x - 40$ on the set $S = \{x \in R : x^2 + 30 \le 11x\}$ is

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