What is the local minimum value of $f(x) = x + \frac{9}{x}$ for $x > 0$?

Explanation

1. 01

   $f'(x) = 1 - \frac{9}{x^2} = 0$ at $x = 3$, and $f''(x) = \frac{18}{x^3} > 0$ there: a local minimum.

2. 02

   The value is $f(3) = 3 + 3 = 6$.