The function f(x)=displaystylex/3+3/x is increasing in the interval:

The correct answer is Option (1) → \((-\infty,-3) \cup (3,\infty)\)

The function is defined for \(x \ne 0\). It is increasing where \(f'(x) \gt 0\):

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

    \( \displaystyle = \frac{x^2 - 9}{3x^2} \)

Since \(3x^2 \gt 0\), the sign of \(f'(x)\) is the sign of \(x^2 - 9\):

    \(x^2 - 9 \gt 0\)
\(|x| \gt 3\)

So f is increasing on \((-\infty,-3) \cup (3,\infty)\), and decreasing on \((-3,0) \cup (0,3)\).