Calculus: Early Transcendentals (2nd Edition)

$f$ is increasing on $(1, 4)$ and is decreasing on $(−∞, 1)$ and on $(4,∞)$.
$f'(x) = −60x^4+300x^3−240x^2$ $= −60x^2(x^2−5x+4)$ = $−60x^2(x−4)(x−1)$. This is $0$ for $x = 0$, $x = 1$, and $x = 4$. Note that $f'(−1) = −600 < 0$, $f'(1/2) ≈ −26.25 < 0$, $f'(2) = 480 > 0$, and $f'(5) = −6000 < 0$. Thus $f$ is increasing on $(1, 4)$ and is decreasing on $(−∞, 1)$ and on $(4,∞)$.