Find inflection points and describe concavity

First compute the derivatives: $f'(x) = 4x^3 - 12x^2$ and $f''(x) = 12x^2 - 24x = 12x(x-2)$. Setting $f''(x) = 0$ gives $x = 0$ and $x = 2$ as candidate inflection points, since $f''$ is a polynomial and defined everywhere.

Testing signs of $f''(x) = 12x(x-2)$ on the three intervals: for $x < 0$, both factors are negative so $f''(x) > 0$ (concave up); for $0 < x < 2$, $x>0$ and $x-2<0$ so $f''(x) < 0$ (concave down); for $x > 2$, both factors positive so $f''(x) > 0$ (concave up).

Since the sign of $f''$ changes at both $x=0$ (from + to -) and $x=2$ (from - to +), both are genuine inflection points. Evaluating $f$ there: $f(0) = 0$ and $f(2) = 16 - 32 = -16$, giving inflection points $(0,0)$ and $(2,-16)$.