If A={x : 9x geq x^2+20} and f: A rightarrow | vedclass

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(B) Given $A=\{x : 9x \geq x^2+20\}$.Solving the inequality $x^2-9x+20 \leq 0$,we get $(x-4)(x-5) \leq 0$,which implies $x \in [4, 5]$.Thus,$A=[4, 5]$

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$7$

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(B) Given $A=\{x : 9x \geq x^2+20\}$.Solving the inequality $x^2-9x+20 \leq 0$,we get $(x-4)(x-5) \leq 0$,which implies $x \in [4, 5]$.Thus,$A=[4, 5]$.Given $f(x)=2x^3-15x^2+36x-48$.Finding the derivative,$f'(x)=6x^2-30x+36=6(x^2-5x+6)=6(x-3)(x-2)$.For critical points,set $f'(x)=0$,which gives $x=2$ and $x=3$.Since both critical points $x=2$ and $x=3$ lie outside the interval $A=[4, 5]$,the function $f(x)$ is monotonic in this interval.Checking the sign of $f'(x)$ for $x \in [4, 5]$: $f'(4)=6(4-3)(4-2)=12 > 0$.Since $f'(x) > 0$ for all $x \in [4, 5]$,$f(x)$ is an increasing function on $[4, 5]$.Therefore,the maximum value occurs at the right endpoint $x=5$.$f(5)=2(5)^3-15(5)^2+36(5)-48 = 2(125)-15(25)+180-48 = 250-375+180-48 = 7$.

If $A=\{x : 9x \geq x^2+20\}$ and $f: A \rightarrow R$ is defined by $f(x)=2x^3-15x^2+36x-48$,then the maximum value of $f(x)$ is

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