Question

Determine the interval(s) in which \[ f(x)=\frac{x}{3}+\frac{3}{x}, \quad x\neq 0 \] is increasing.

Hint:

When solving inequalities with fractions, always keep track of the domain (\( x \neq 0 \) here). Squaring or multiplying by denominators is only safe if you know the denominator's sign.

Answer 1

Concept:
• A function \( f(x) \) is increasing in an interval where its first derivative is greater than or equal to zero (\( f'(x) \geq 0 \)).

Step 1: Find the first derivative \( f'(x) \)
The function is \( f(x) = \frac{1}{3}x + 3x^{-1} \).
Differentiating with respect to \( x \):
\[ f'(x) = \frac{1}{3} - 3x^{-2} \]
\[ f'(x) = \frac{1}{3} - \frac{3}{x^2} \]

Step 2: Set the derivative for increasing condition
For \( f(x) \) to be increasing, \( f'(x) \geq 0 \):
\[ \frac{1}{3} - \frac{3}{x^2} \geq 0 \]
\[ \frac{x^2 - 9}{3x^2} \geq 0 \]

Step 3: Solve the inequality
Since \( 3x^2 > 0 \) for all \( x \neq 0 \), the sign depends only on the numerator:
\[ x^2 - 9 \geq 0 \implies (x - 3)(x + 3) \geq 0 \]
Using the wavy curve method, the solution is:
\[ x \in (-\infty, -3] \cup [3, \infty) \]