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

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(A) First,we determine the set $S$ by solving the inequality $x^2 + 30 \leq 11x$. $x^2 - 11x + 30 \leq 0$ $(x - 5)(x - 6) \leq 0$ Thus,$x \in [5, 6]$.

Vedclass · Accepted Answer

$122$

Vedclass · Answer

(A) First,we determine the set $S$ by solving the inequality $x^2 + 30 \leq 11x$. $x^2 - 11x + 30 \leq 0$ $(x - 5)(x - 6) \leq 0$ Thus,$x \in [5, 6]$. Now,consider the function $f(x) = 3x^3 - 18x^2 + 27x - 40$. Find the derivative $f'(x)$: $f'(x) = 9x^2 - 36x + 27 = 9(x^2 - 4x + 3) = 9(x - 1)(x - 3)$. For $x \in [5, 6]$,both $(x - 1)$ and $(x - 3)$ are positive,so $f'(x) > 0$. Since $f'(x) > 0$ for all $x \in [5, 6]$,the function $f(x)$ is strictly increasing on the interval $[5, 6]$. 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 $S = \{x \in R : x^2 + 30 \leq 11x\}$ is

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