Finding and classifying critical points using the Second Derivative Test

First compute $f'(x) = 1 - \dfrac{1}{x^2}$. Setting $f'(x) = 0$ gives $x^2 = 1$, so $x = 1$ or $x = -1$. Note that $x = 0$ is excluded from the domain of $f$, so it is not a critical number even though $f'$ is undefined there.

Next compute $f''(x) = \dfrac{2}{x^3}$. At $x = 1$, $f''(1) = 2 > 0$, so by the Second Derivative Test $f$ has a local minimum at $x = 1$, with value $f(1) = 2$. At $x = -1$, $f''(-1) = -2 < 0$, so $f$ has a local maximum at $x = -1$, with value $f(-1) = -2$.

Since $f''(x) = \dfrac{2}{x^3}$, the sign of $f''$ matches the sign of $x^3$, which matches the sign of $x$ itself (since cubing preserves sign).

For $x > 0$, $f''(x) > 0$, so $f$ is concave up on $(0, \infty)$. For $x < 0$, $f''(x) < 0$, so $f$ is concave down on $(-\infty, 0)$. Thus the graph is concave down exactly on $(-\infty, 0)$.