Let the function $f(x)=\frac{x}{3}+\frac{3}{x}+3, x
eq 0$ be strictly increasing in $\left(-\infty, \alpha_1 ight) \mathrm{U}\left(\alpha_2, \infty ight)$ and strictly decreasing in $\left(\alpha_3, \alpha_4 ight) \mathrm{U}\left(\alpha_4, \alpha_5 ight)$. Then $\sum_{\mathrm{i}=1}^5…

Question

Let the function $f(x)=\frac{x}{3}+\frac{3}{x}+3, x 
eq 0$ be strictly increasing in $\left(-\infty, \alpha_1ight) \mathrm{U}\left(\alpha_2, \inftyight)$ and strictly decreasing in $\left(\alpha_3, \alpha_4ight) \mathrm{U}\left(\alpha_4, \alpha_5ight)$. Then $\sum_{\mathrm{i}=1}^5 \alpha_{\mathrm{i}}^2$ is equal to :-

Aicharya · Accepted Answer

$\begin{aligned} & f(x)=\frac{x}{3}+\frac{3}{x}+3, x 
eq 0 \ & f^{\prime}(x)=\frac{1}{3}-\frac{3}{x^2}=0 \quad \Rightarrow x= \pm 3 \ & f^{\prime}(x)=\frac{x^2-3}{3 x^2}\end{aligned}$ $\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{x}) \gt 0 \forall(-\infty,-3) \cup(3, \infty) ightarrow 	ext { increasing } \ & \mathrm{f}^{\prime}(\mathrm{x}) \lt 0 \forall(-3,0) \cup(0,3) ightarrow 	ext { decreasing }\end{aligned}$ $\sum_{i=1}^5 \alpha_i^2=(-3)^2+(3)^2+(-3)^2+(0)^2+(3)^2$ $=36$

Let the function $f(x)=\frac{x}{3}+\frac{3}{x}+3, x \neq 0$ be strictly increasing in $\left(-\infty,…

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