The minimum value of the function $\mathrm{f}(x)=2 x^3-15 x^2+36 x-48 \quad$ on the set $\mathrm{A}=\left\{x \mid x^2+20 \leqslant 9 x\right\}$ is

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Aicharya · Accepted Answer

$\begin{aligned} \mathrm{A} & =\left\{x \mid x^2+20 \leq 9 xight\} \ & =\left\{x \mid x^2-9 x+20 \leq 0ight\} \ & =\{x \mid(x-4)(x-5) \leq 0\} \ \mathrm{A} & =\{4,5\} \ \mathrm{f}(x) & =2 x^3-15 x^2+36 x-48 \ 	herefore \quad \mathrm{f}^{\prime}(x) & =6 x^2-30 x+36 \ & =6\left(x^2-5 x+6ight) \ & =6(x-2)(x-3) \lt 0 \forall x \in(4,5) \end{aligned}$ $	herefore \quad \mathrm{f}(x)$ is strictly increasing in the interval $(4,5)$. $	herefore \quad$ Minimum value of $\mathrm{f}(x)$ when $x \in(4,5)$ is $f(4)=-16$

The minimum value of the function $\mathrm{f}(x)=2 x^3-15 x^2+36 x-48 \quad$ on the set $\mathrm{A}=\left\{x…

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