Question

The interval, in which the function \( f(x) = \frac{3}{x} + \frac{x}{3} \) is strictly decreasing, is :

• A. \( (-\infty, -3) \cup (3, \infty) \)
• B. \( (-3, 3) \)
• C. \( (-3, 0) \cup (0, 3) \)
• D. \( \mathbb{R} - \{0\} \)

Hint:

Always look at the denominator of the original function. If it contains \( x \), any answer that is a single continuous interval crossing 0 (like Option 2) is almost certainly wrong.
Also, note that for \( x > 0 \), this is \( \text{AM} \ge \text{GM} \) territory. The minimum value occurs at \( x=3 \). For \( 0 < x < 3 \), the function must be decreasing towards that minimum.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Monotonicity of a function refers to whether it is increasing or decreasing over a specific domain. A differentiable function \( f(x) \) is strictly decreasing on an interval if its first derivative \( f'(x) \) is strictly negative (\( < 0 \)) for every \( x \) in that interval.
An important consideration for this specific function, \( f(x) = \frac{3}{x} + \frac{x}{3} \), is its domain. Because of the \( \frac{3}{x} \) term, the function is undefined at \( x = 0 \). Therefore, any interval describing the behavior of the function must explicitly exclude 0. This is a common point where errors occur in multiple-choice questions.

Step 2: Key Formula or Approach:
1. Define the domain: \( x \in \mathbb{R}, x \neq 0 \).
2. Find the first derivative \( f'(x) \) using the power rule.
3. Set up the inequality \( f'(x) < 0 \).
4. Solve the inequality for \( x \), keeping in mind that when taking square roots or multiplying by variables, the sign of the variable matters. However, since we deal with \( x^2 \) in the denominator, it will always be positive in the domain.

Step 3: Detailed Explanation:
Step 3.1: Differentiating the function.
We can write the function as \( f(x) = 3x^{-1} + \frac{1}{3}x \).
Using \( \frac{d}{dx}x^n = nx^{n-1} \):
\[ f'(x) = 3(-1)x^{-2} + \frac{1}{3}(1) \]
\[ f'(x) = -\frac{3}{x^2} + \frac{1}{3} \]
Step 3.2: Solving the inequality for decreasing behavior.
We need \( f'(x) < 0 \):
\[ -\frac{3}{x^2} + \frac{1}{3} < 0 \]
Add \( \frac{3}{x^2} \) to both sides:
\[ \frac{1}{3} < \frac{3}{x^2} \]
Since \( x^2 > 0 \) for all \( x \) in the domain, we can safely multiply both sides by \( 3x^2 \) without changing the inequality sign:
\[ x^2 < 9 \]
Step 3.3: Interpreting the result.
The inequality \( x^2 < 9 \) is satisfied when the magnitude of \( x \) is less than 3:
\[ |x| < 3 \implies -3 < x < 3 \]
Step 3.4: Applying domain constraints.
Recall from Step 1 that \( x \neq 0 \). Therefore, we must remove 0 from the interval \( (-3, 3) \).
This results in two sub-intervals: \( (-3, 0) \) and \( (0, 3) \).
Written in union notation, this is \( (-3, 0) \cup (0, 3) \).
Checking the options:
Option (1) is where the function is increasing.
Option (2) incorrectly includes 0.
Option (3) is the correct representation.

Step 4: Final Answer:
The function's derivative is negative when \( x^2 < 9 \), excluding zero. Thus, the interval of strict decrease is \( (-3, 0) \cup (0, 3) \). This corresponds to Option (3).