Answer
$h$ is concave down on $(−π,−3π/4)$, and on $(−π/4, π/4)$, and on $(3π/4, π)$, and $h$ is concave up on $(−3π/4,−π/4)$ and on $(π/4, 3π/4)$. There are inflection points at $t = −3π/4$, $t = −π/4$, $t = π/4$, and $t = 3π/4$.
Work Step by Step
$h'(t) = −2 sin 2t$ and $h''(t) = −4 cos 2t$, which on the stated domain is $0$ when $2t = −3π/2,−π/2, π/2$, and $3π/2$, which means for $t = −3π/4,−π/4, π/4$, and $3π/4$. $h' < 0$ on $(−π,−3π/4)$, and on $(−π/4, π/4)$, and on $(3π/4, π)$, so $h$ is concave down those intervals, while $h' > 0$ on $(−3π/4,−π/4)$ and on $(π/4, 3π/4)$, so $h$ is concave up on those intervals. There are inflection points at $t = −3π/4$, $t = −π/4$, $t = π/4$, and $t = 3π/4$.
