Question

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

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

Answer 1

The Correct Option is B

Start by computing the derivative of \( f(x) \):

\[ f(x) = \frac{3}{x} + \frac{x}{3}. \]

The derivative is:

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

Simplify the derivative:

\[ f'(x) = \frac{-9 + x^2}{3x^2}. \]

Set \( f'(x) = 0 \) to find critical points:

\[ \frac{-9 + x^2}{3x^2} = 0 \implies -9 + x^2 = 0 \implies x^2 = 9 \implies x = \pm 3. \]

Now, analyze the sign of \( f'(x) \) in the intervals \((-\infty, -3)\), \((-3, 0)\), \((0, 3)\), and \((3, \infty)\):

• For \( x \in (-\infty, -3) \), \( x^2 > 9 \) and \( -9 + x^2 > 0 \). Thus, \( f'(x) > 0 \) (increasing).
• For \( x \in (-3, 0) \), \( x^2 < 9 \) and \( -9 + x^2 < 0 \). Thus, \( f'(x) < 0 \) (decreasing).
• For \( x \in (0, 3) \), \( x^2 < 9 \) and \( -9 + x^2 < 0 \). Thus, \( f'(x) < 0 \) (decreasing).
• For \( x \in (3, \infty) \), \( x^2 > 9 \) and \( -9 + x^2 > 0 \). Thus, \( f'(x) > 0 \) (increasing).

From this analysis, \( f(x) \) is strictly decreasing in the intervals \((-3, 0) \cup (0, 3)\).