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: Note the domain.
The function $f(x) = \dfrac{3}{x} + \dfrac{x}{3}$ has $x$ in a denominator, so $x = 0$ is not allowed. Keep that in mind.

Step 2: Differentiate.
\[ f'(x) = -\frac{3}{x^2} + \frac{1}{3} \]

Step 3: Set up the decreasing condition.
Strictly decreasing means $f'(x) < 0$.
\[ \frac{1}{3} - \frac{3}{x^2} < 0 \]
Bring terms together over $3x^2$, which is positive, so the sign stays.
\[ \frac{x^2 - 9}{3x^2} < 0 \implies x^2 - 9 < 0 \]

Step 4: Solve and remove zero.
$x^2 < 9$ means $-3 < x < 3$. But $x = 0$ is barred, so we cut it out.
\[ \boxed{(-3, 0)\cup(0, 3)} \]
That is option 3.