Answer
$f$ is concave up on $(−∞, 0)$, concave down on $(0, 2)$, and concave up on $(2,∞)$. There are inflection points at $x = 0$ and $x = 2$.
Work Step by Step
$f'(x) = 20x^3 −60x^2$, and $f''(x) = 60x^2 −120x = 60x(x−2)$.
This is $0$ for $x = 0$ and for $x = 2$. Note that $f(−1) > 0$, $f(1) < 0$, and $f(3) > 0$. So $f$ is concave up on $(−∞, 0)$, concave down on $(0, 2)$, and concave up on $(2,∞)$. There are inflection points at $x = 0$ and $x = 2$.