Let the function f(x)=x/3+3/x+3, x ≠ 0 be strictly increasing in (-∞, α_1) ∪(α_2, ∞) and strictly decreasing in (α_3, α_4) ∪(α_4, α_5). Then Σ_i=1^5…

Let the function $\displaystyle f(x)=\frac{x}{3}+\frac{3}{x}+3, x \neq 0$ be strictly increasing in $\displaystyle \left(-\infty, \alpha_1\right) \cup\left(\alpha_2, \infty\right)$ and strictly decreasing in $\displaystyle \left(\alpha_3, \alpha_4\right) \cup\left(\alpha_4, \alpha_5\right)$. Then $\displaystyle \sum_{i=1}^5 \alpha_i^2$ is equal to