The function $f(x)=\frac{x}{3}+\frac{3}{x}$ is increasing in the interval:

$(-∞,-3) ∪ (3,∞)$

The correct answer is Option (1) → $(-∞,-3) ∪ (3,∞)$ **

$f(x)=\frac{x}{3}+\frac{3}{x}$

$f'(x)=\frac{1}{3}-\frac{3}{x^2}$

For increasing function: $f'(x)>0$

$\frac{1}{3}-\frac{3}{x^2}>0$

$\frac{1}{3}>\frac{3}{x^2}$

$x^2>9$

$x<-3$ or $x>3$

The function is increasing on $(-\infty,-3)\cup(3,\infty)$.