As $t \to 0^+$, $h(t) \to \infty$, and $h'(t) = \frac2t^2 - (1 + 2t)(2t)t^4 = \frac2t - 2(1 + 2t)t^3 = \frac2t - 2 - 4tt^3 = \frac-2t - 2t^3 < 0$, so decreasing. - AMAZONAWS

📅 March 6, 2026 👤 scraface

Mar 06, 2026

$As $t \to 0^+$, $h(t) \to \infty$, and $h'(t) = \frac{2t^2 - (1 + 2t)(2t)}{t^4} = \frac{2t - 2(1 + 2t)}{t^3} = \frac{2t - 2 - 4t}{t^3} = \frac{-2t - 2}{t^3} < 0$, so decreasing.$

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