1. Given function

We have $f(x)=\frac{x}{3}+\frac{3}{x}+3, \qquad x\ne 0.$

To determine where $f(x)$ is increasing/decreasing, we compute the derivative.

2. Differentiate

$f'(x)=\frac{d}{dx}\left(\frac{x}{3}+\frac{3}{x}+3\right)=\frac{1}{3}-\frac{3}{x^2}.$

Simplify:

$f'(x)=\frac{x^2-9}{3x^2}.$

3. Critical points and sign of derivative

Since $x\ne 0$, the denominator $3x^2>0$ for all $x\ne 0$. Hence the sign of $f'(x)$ depends only on $x^2-9=(x-3)(x+3).$

So:

• $f'(x)>0$ when $x^2>9 \iff |x|>3$, i.e. on $(-\infty,-3)\cup(3,\infty)$.
• $f'(x)<0$ when $x^2<9$, i.e. on $(-3,0)\cup(0,3)$.

Thus:

• strictly increasing on $(-\infty,-3)\cup(3,\infty)$,
• strictly decreasing on $(-3,0)\cup(0,3)$.

4. Match with the intervals in the question

Given:

• increasing in $(-\infty,\alpha_1)\cup(\alpha_2,\infty)$,
• decreasing in $(\alpha_3,\alpha_4)\cup(\alpha_4,\alpha_5)$.

Comparing, $\alpha_1=-3, \quad \alpha_2=3,$ and for decreasing intervals, $\alpha_3=-3, \quad \alpha_4=0, \quad \alpha_5=3.$

5. Compute $\sum_{i=1}^5 \alpha_i^2$