The maximum value of the function f(x)=2x^3-1 | vedclass

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(D) Given the function $f(x)=2x^3-15x^2+36x-48$.First,we determine the set $A$ by solving the inequality $x^2-9x+20 \leq 0$.Factoring the quadratic,we

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$7$

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(D) Given the function $f(x)=2x^3-15x^2+36x-48$.First,we determine the set $A$ by solving the inequality $x^2-9x+20 \leq 0$.Factoring the quadratic,we get $(x-4)(x-5) \leq 0$,which implies $x \in [4, 5]$.Next,we find the critical points of $f(x)$ by setting $f'(x)=0$:$f'(x)=6x^2-30x+36=6(x^2-5x+6)=6(x-2)(x-3)=0$.The critical points are $x=2$ and $x=3$.Since both $x=2$ and $x=3$ are outside the interval $[4, 5]$,the function $f(x)$ is monotonic on the interval $[4, 5]$.We evaluate $f(x)$ at the endpoints of the interval $A=[4, 5]$:For $x=4$: $f(4)=2(4)^3-15(4)^2+36(4)-48 = 128-240+144-48 = -16$.For $x=5$: $f(5)=2(5)^3-15(5)^2+36(5)-48 = 250-375+180-48 = 7$.Comparing the values,the maximum value of $f(x)$ on the set $A$ is $7$.

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

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