Answer
$f$ is increasing on $(1,\infty)$,
$f$ is decreasing on $(-\infty,1)$.
Work Step by Step
$f'(x) = 2(x−1)$, which is zero exactly when $x = 1$. On $(−∞, 1)$ we note that $f' < 0$, so that $f$ is decreasing on this interval. On $(1,∞)$, we note that $f' > 0$, so $f$ is increasing on this interval.