Question

Let the function $f(x) = \frac{x}{3} + \frac{3}{x} + 3, x \ne 0$ 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 $\sum_{i=1}^5 \alpha_i^2$ is equal to :-

Answer 1

36

Answer 2

The derivative of $f(x)$ is $f'(x) = \frac{1}{3} - \frac{3}{x^2}$. Setting $f'(x) = 0$ gives $x^2 = 9$, so $x = \pm 3$. The function is undefined at $x=0$. Analyzing the sign of $f'(x)$ shows that $f(x)$ is strictly increasing on $(-\infty, -3) \cup (3, \infty)$ and strictly decreasing on $(-3, 0) \cup (0, 3)$. Comparing with the given intervals, we have $\alpha_1 = -3$, $\alpha_2 = 3$, $\alpha_3 = -3$, $\alpha_4 = 0$, and $\alpha_5 = 3$. Therefore, $\sum_{i=1}^5 \alpha_i^2 = (-3)^2 + (3)^2 + (-3)^2 + (0)^2 + (3)^2 = 9 + 9 + 9 + 0 + 9 = 36$.