## Calculus: Early Transcendentals (2nd Edition)

$f$ is increasing on $(0,1) \cup (2,\infty)$, $f$ is decreasing on $(-\infty,0) \cup (1,2)$.
$f'(x) = 4x^3 − 12x^2 + 8x = 4x(x^2 − 3x + 2) = 4x(x − 2)(x − 1)$, which is $0$ when $x$ is $0$, $1$, or $2$. On $(−∞, 0)$ , $f' < 0$ so $f$ is decreasing. On $(0, 1)$, $f' > 0$ so $f$ is increasing. On $(1, 2)$, $f' < 0$ so $f$ is decreasing, and on $(2,∞)$, $f' > 0$ so $f$ is increasing.