Answer
$g$ is concave down on $(−∞, 0)$ and on $(2, 4)$, and $g$ is concave up on $(0, 2)$ and on $(4,∞)$. There are inflection points at $t = 0, 2$, and $4$.
Work Step by Step
$g'(t) = 15t^4 − 120t^3 + 240t^2$, and $g''(t) = 60t^3 − 360t^2 + 480t = 60t(t − 2)(t − 4)$.
Note that $g$ is $0$ for $t = 0, 2$, and $4$. Note also that $g' < 0$ on $(−∞, 0)$ and on $(2, 4)$, so $g$ is concave down on those intervals, while $g' > 0$ on $(0, 2)$ and on $(4,∞)$, so $g$ is concave up there. There are inflection points at $t = 0, 2$, and $4$.
