What is the local maximum value of $f(x) = x + \frac{1}{x}$?

Explanation

1. 01

   $f'(x) = 1 - \frac{1}{x^{2}} = 0$ at $x = \pm 1$, and $f''(x) = \frac{2}{x^{3}}$.

2. 02

   $f''(-1) = -2 < 0$: a local maximum, with value $f(-1) = -2$. (At $x = 1$, the local minimum value is 2.)