## Calculus: Early Transcendentals (2nd Edition)

$f$ is decreasing on $(0, 1)$ and increasing on $(1,∞)$.
$f'(x) = 2x − \frac{2}{x} = \frac{2(x^2−1)}{x}$ which is $0$ for $x =\pm1$. Note that the domain of $f$ is $(0,∞)$ and that $f'(1/2) = −3 < 0$ and $f'(2) = 3 > 0$, so $f$ is decreasing on $(0, 1)$ and increasing on $(1,∞)$.