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Question Number 915 by 112358 last updated on 24/Apr/15

Show that ∀t≥0 , x≤1 where x=(e^(−t) /2)(t^2 +2t+2) , t∈R.

$${Show}\:{that}\:\forall{t}\geqslant\mathrm{0}\:,\:{x}\leqslant\mathrm{1}\:{where} \\ $$$${x}=\frac{{e}^{−{t}} }{\mathrm{2}}\left({t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{2}\right)\:\:,\:{t}\in\mathbb{R}.\: \\ $$

Commented by 123456 last updated on 24/Apr/15

t≥0⇔−t≤0⇔e^(−t) ≤e^0 =1 x(t)=((e^(−t) (t^2 +2t+2))/2)≤((t^2 +2t+2)/2)

$${t}\geqslant\mathrm{0}\Leftrightarrow−{t}\leqslant\mathrm{0}\Leftrightarrow{e}^{−{t}} \leqslant{e}^{\mathrm{0}} =\mathrm{1} \\ $$$${x}\left({t}\right)=\frac{{e}^{−{t}} \left({t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{2}\right)}{\mathrm{2}}\leqslant\frac{{t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{2}}{\mathrm{2}} \\ $$

Commented by 123456 last updated on 24/Apr/15

x=(((t^2 +2t+2)e^(−t) )/2) (dx/dt)=−((t^2 e^(−t) )/2) t≥0⇔e^(−t) ≤1 (dx/dt)=−((t^2 e^(−t) )/2)≤−(t^2 /2)≤0

$${x}=\frac{\left({t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{2}\right){e}^{−{t}} }{\mathrm{2}} \\ $$$$\frac{{dx}}{{dt}}=−\frac{{t}^{\mathrm{2}} {e}^{−{t}} }{\mathrm{2}} \\ $$$${t}\geqslant\mathrm{0}\Leftrightarrow{e}^{−{t}} \leqslant\mathrm{1} \\ $$$$\frac{{dx}}{{dt}}=−\frac{{t}^{\mathrm{2}} {e}^{−{t}} }{\mathrm{2}}\leqslant−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\leqslant\mathrm{0} \\ $$

Answered by 123456 last updated on 24/Apr/15

as shown in the coments t≥0,(dx/dt)≤0 so x(t) is decreasing into t∈[0,+∞) so for all t≥0 x(t)≤x(0)=(((0^2 +2×0+2)e^(−0) )/2)=1 x(t)≤1 as desired ■

$$\mathrm{as}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{the}\:\mathrm{coments} \\ $$$${t}\geqslant\mathrm{0},\frac{{dx}}{{dt}}\leqslant\mathrm{0} \\ $$$$\mathrm{so}\:{x}\left({t}\right)\:\mathrm{is}\:\mathrm{decreasing}\:\mathrm{into}\:{t}\in\left[\mathrm{0},+\infty\right) \\ $$$$\mathrm{so}\:\mathrm{for}\:\mathrm{all}\:{t}\geqslant\mathrm{0} \\ $$$${x}\left({t}\right)\leqslant{x}\left(\mathrm{0}\right)=\frac{\left(\mathrm{0}^{\mathrm{2}} +\mathrm{2}×\mathrm{0}+\mathrm{2}\right){e}^{−\mathrm{0}} }{\mathrm{2}}=\mathrm{1} \\ $$$${x}\left({t}\right)\leqslant\mathrm{1} \\ $$$$\mathrm{as}\:\mathrm{desired}\:\blacksquare \\ $$

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