Let the function f(x) = frac{x}{3} + frac{3}{ | vedclass

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(D) Given $f(x) = \frac{x}{3} + \frac{3}{x} + 3$,$x 
eq 0$. To find the intervals of increase and decrease,we calculate the derivative: $f'(x) = \fra

Vedclass · Accepted Answer

$36$

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(D) Given $f(x) = \frac{x}{3} + \frac{3}{x} + 3$,$x 
eq 0$. To find the intervals of increase and decrease,we calculate the derivative: $f'(x) = \frac{1}{3} - \frac{3}{x^2} = \frac{x^2 - 9}{3x^2} = \frac{(x-3)(x+3)}{3x^2}$. Setting $f'(x) = 0$,we get $x = 3$ and $x = -3$. The critical points are $x = -3, 0, 3$. Testing the intervals: For $x \in (-\infty, -3)$,$f'(x) > 0$,so $f(x)$ is strictly increasing. For $x \in (-3, 0)$,$f'(x) For $x \in (0, 3)$,$f'(x) For $x \in (3, \infty)$,$f'(x) > 0$,so $f(x)$ is strictly increasing. Comparing with the given intervals: Increasing in $(-\infty, -3) \cup (3, \infty) \Rightarrow \alpha_1 = -3, \alpha_2 = 3$. Decreasing in $(-3, 0) \cup (0, 3) \Rightarrow \alpha_3 = -3, \alpha_4 = 0, \alpha_5 = 3$. Thus,$\sum_{i=1}^5 \alpha_i^2 = (-3)^2 + (3)^2 + (-3)^2 + (0)^2 + (3)^2 = 9 + 9 + 9 + 0 + 9 = 36$.

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 $\sum_{i=1}^5 \alpha_i^2$ is equal to :-

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