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

We are given that the function \( f(x) = \frac{x}{3} + \frac{3}{x} + 3 \) is strictly increasing in \( (-\infty, \alpha_1) \cup (\alpha_2, \infty) \) and strictly decreasing in \( (\alpha_3, \alpha_4) \cup (\alpha_4, \alpha_5) \).

Step 1: Find the derivative of \( f(x) \)
To determine where \( f(x) \) is increasing or decreasing, we first find the derivative: \[ f'(x) = \frac{d}{dx} \left( \frac{x}{3} + \frac{3}{x} + 3 \right) \] \[ f'(x) = \frac{1}{3} - \frac{3}{x^2} \]
Step 2: Find critical points
Set the derivative equal to zero to find the critical points: \[ f'(x) = 0 \quad \Rightarrow \quad \frac{1}{3} - \frac{3}{x^2} = 0 \] \[ \Rightarrow \quad \frac{3}{x^2} = \frac{1}{3} \] \[ \Rightarrow \quad x^2 = 9 \quad \Rightarrow \quad x = 3 \text{ or } x = -3 \] Thus, the critical points are \( x = 3 \) and \( x = -3 \).
Step 3: Analyze intervals
For \( x < -3 \), \( f'(x) > 0 \) (strictly increasing).
For \( -3 < x < 3 \), \( f'(x) < 0 \) (strictly decreasing).
For \( x > 3 \), \( f'(x) > 0 \) (strictly increasing).
Thus, the values \( \alpha_1 = -3, \alpha_2 = 3 \).
Step 4: Find the sum of squares of critical points
We are asked to find \( \sum_{i=1}^5 (\alpha_i)^2 \).
The critical points are \( \alpha_1 = -3 \), \( \alpha_2 = 3 \), and the values for \( \alpha_3, \alpha_4, \alpha_5 \) correspond to intervals where \( f(x) \) is decreasing. Thus, the sum is: \[ (-3)^2 + (3)^2 = 9 + 9 = 36 \]