Concavity and inflection points of a quartic

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

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

Since the concavity changes at both $x=0$ and $x=2$, both are inflection points. Thus $p$ is concave up on $(-\infty, 0)$ and $(2, \infty)$, and concave down on $(0, 2)$, with inflection points at $x = 0$ and $x = 2$.

The two inflection points occur at x equals zero and x equals two. On the interval between them, from part a, the second derivative is negative, meaning the curve is concave down. This matches the sign analysis of p double prime on the interval zero to two.