The maximum value of the function $f(x)=3 x^3-18 x^2+27 x-40$ on the set $\mathrm{S}=\left\{x \in \mathbb{R} / x^2+30 \leq 11 x\right\}$ is

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{S}=\left\{x \in \mathbb{R} / x^2+30 \leq 11 xight\} \ & x^2+30 \leq 11 x \ & x^2-11 x+30 \leq 0 \ & (x-5)(x-6) \leq 0 \ & x \in[5,6] \end{aligned}$ Now, $\mathrm{f}(x)=3 x^3-18 x^2+27 x-40$ $\begin{aligned} \mathrm{f}^{\prime}(x) & =9 x^2-36 x+27 \ \mathrm{f}^{\prime}(x) & =9\left(x^2-4 x+3ight) \ & =9\left[\left(x^2-4 x+4ight)-1ight] \ & =9(x-2)^2-9 \end{aligned}$ $	herefore \quad \mathrm{f}^{\prime}(x)>0 \forall x \in[5,6]$ $	herefore \mathrm{f}(x)$ is strictly increasing in the interval $[5,6]$ $	herefore \quad$ Maximum value of $\mathrm{f}(x)$ when $x \in[5,6]$ is $f(6)=122$

The maximum value of the function $f(x)=3 x^3-18 x^2+27 x-40$ on the set $\mathrm{S}=\left\{x \in \mathbb{R}…

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