Question

Let $ f $ be defined as $ R \setminus \{0\} \to R $ such that $ f(x) = \frac{x}{3} + \frac{3}{x} + 3 $ If $ f(x) $ is 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:

• A. 28
• B. 36
• C. 48
• D. 40

Hint:

To determine the behavior of a function (increasing or decreasing), find the critical points using the first derivative and analyze the sign of the derivative on each interval.

Answer 1

The Correct Option is B

The function \( f(x) = \frac{x}{3} + \frac{3}{x} + 3 \) is observed to be strictly increasing on \( (-\infty, \alpha_1) \cup (\alpha_2, \infty) \) and strictly decreasing on \( (\alpha_3, \alpha_4) \cup (\alpha_4, \alpha_5) \).

Step 1: Compute the derivative of \( f(x) \)
To identify intervals of increase and decrease, the derivative is calculated as:\[f'(x) = \frac{d}{dx} \left( \frac{x}{3} + \frac{3}{x} + 3 \right)\]\[f'(x) = \frac{1}{3} - \frac{3}{x^2}\]
Step 2: Determine critical points
Critical points are found by setting the derivative to zero:\[f'(x) = 0 \quad \Rightarrow \quad \frac{1}{3} - \frac{3}{x^2} = 0\]\[\Rightarrow \quad \frac{3}{x^2} = \frac{1}{3}\]\[\Rightarrow \quad x^2 = 9 \quad \Rightarrow \quad x = 3 \text{ or } x = -3\]The critical points are therefore \( x = 3 \) and \( x = -3 \).
Step 3: Analyze function behavior over intervals
For \( x<-3 \), \( f'(x)>0 \), indicating strictly increasing behavior.
For \( -3<x<3 \), \( f'(x)<0 \), indicating strictly decreasing behavior.
For \( x>3 \), \( f'(x)>0 \), indicating strictly increasing behavior.
Consequently, \( \alpha_1 = -3 \) and \( \alpha_2 = 3 \).
Step 4: Calculate the sum of squares of critical points
The objective is to compute \( \sum_{i=1}^5 (\alpha_i)^2 \).
The identified critical points are \( \alpha_1 = -3 \) and \( \alpha_2 = 3 \). The values \( \alpha_3, \alpha_4, \alpha_5 \) are associated with intervals of decrease, which are determined by the critical points.Thus, the sum is calculated as:\[(-3)^2 + (3)^2 = 9 + 9 = 36\]