But $ d(t) = t^3 $ has $ d'(t) = 3t^2 \geq 0 $, zero only at $ t=0 $, but second derivative $ 6t $, so $ d''(0)=0 $ — not a strict minimum. - AMAZONAWS

📅 March 11, 2026 👤 scraface

Mar 11, 2026

$But $ d(t) = t^3 $ has $ d'(t) = 3t^2 \geq 0 $, zero only at $ t=0 $, but second derivative $ 6t $, so $ d''(0)=0 $ — not a strict minimum.$