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

Answer: (B)

$(-3, 0) \cup (0, 3)$

Answer 2

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 $(-3, 0) \cup (0, 3)$.