For $v > 1$, $g'(v) < 0$, so $g(v)$ is decreasing on $(1, 2]$. Hence, maximum at $v \to 1^+$? But let’s check value at endpoint. - AMAZONAWS

📅 March 6, 2026 👤 scraface

Mar 06, 2026

$For $v > 1$, $g'(v) < 0$, so $g(v)$ is decreasing on $(1, 2]$. Hence, maximum at $v \to 1^+$? But let’s check value at endpoint.$