Question:

Let $\tan x = t$ where $t \in (0, \infty)$. Then $f(x) = 3t + \frac{1}{t}$. For $y = 3t + \frac{1}{t}$, we have $\frac{dy}{dt} = 3 - \frac{1}{t^2}$. Setting $\frac{dy}{dt} = 0$ gives $t^2 = \frac{1}{3}$, so $t = \frac{1}{\sqrt{3}}$ (taking positive value). Since $y \in [0, \frac{1}{\sqrt{3}}]$ gives a minimum, we verify $y \geq 0$ for all positive $t$. Thus, the minimum value of the function is 0.