Let the function $ f(x) = \frac{x}{3} + \frac{3}{x} + 3, x \neq 0 $ be strictly increasing in $(-\infty, \alpha_1) \cup (\alpha_2, \infty)$ and strictly decreasing in $(\alpha_3, \alpha_4) \cup (\alpha_4, \alpha_5)$. Then $ \sum\limits_{i=1}^{5} \alpha_i^2 $ is equal to
Solution
<p>$$\begin{aligned}
\mathrm{f}(\mathrm{x}) & =\frac{\mathrm{x}}{3}+\frac{3}{\mathrm{x}}+3, \mathrm{x} \neq 0 \\
\mathrm{f}^{\prime}(\mathrm{x}) & =\frac{1}{3}-\frac{3}{\mathrm{x}^2}=0 \quad \Rightarrow \mathrm{x}= \pm 3 \\
\mathrm{f}^{\prime}(\mathrm{x}) & =\frac{\mathrm{x}^2-3}{3 \mathrm{x}^2} \\
\mathrm{f}^{\prime}(\mathrm{x}) & >0 \forall(-\infty,-3) \cup(3, \infty) \rightarrow \text { increasing } \\
\mathrm{f}^{\prime}(\mathrm{x}) & <0 \forall(-3,0) \cup(0,3) \rightarrow \text { decreasing } \\
\sum_{\mathrm{i}=1}^5 \alpha_{\mathrm{i}}^2 & =(-3)^2+(3)^2+(-3)^2+(0)^2+(3)^2 \\
& =36
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Application of Derivatives · Topic: Tangents and Normals
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