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Question Number 208431 by alcohol last updated on 15/Jun/24

$$h_a\left(x\right)=e^{-x}+{ax}^2$$$$showthat{h}_{a}admitsaminimumin\mathbb{R}$$

Answered by mathzup last updated on 15/Jun/24

$$cas1a>0$$$${lim}_{x\to -\mathrm{\infty }}h\left(x\right)=+\mathrm{\infty }$$$${lim}_{x\to +\mathrm{\infty }}h\left(x\right)=+\mathrm{\infty }$$$$h^{{}^{\prime }}\left(x\right)=2ax-e^{-x}$$$${lim}_{x\to -\mathrm{\infty }}{h}^{{}^{\prime }}\left(x\right)=-\mathrm{\infty }$$$${lim}_{x\to +\mathrm{\infty }}{h}^{{}^{\prime }}\left(x\right)=+\mathrm{\infty }$$$$h^{\left(2\right)}\left(x\right)=2a+e^{-x}>0\Rightarrow h^{{}^{\prime }}eststrictementcroissante$$$$x-\mathrm{\infty }0\alpha +\mathrm{\infty }$$$$h^{{}^{\prime }}\left(x\right)-\mathrm{\infty }\to -1\to 0$$$$+\mathrm{\infty }$$$$\Rightarrow \mathrm{\exists }\alpha /h^{{}^{\prime }}\left(\alpha \right)=0$$$$x-\mathrm{\infty }\alpha +\mathrm{\infty }$$$$h^{{}^{\prime }}-0+$$$$hdecrh\left(\alpha \right)croi$$$$donchpossedeunminimunquiest$$$$h\left(\alpha \right)=e^{-\alpha }+a{\alpha }^2$$$$h\left(0\right)=-1andh\left(1\right)=e^{-1}+a>0\Rightarrow$$$$\alpha \in ]0,1[$$$$restaetudierlescasa<0eta=0$$

Answered by Frix last updated on 16/Jun/24

$${\mathrm{It}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{true}\mathrm{for}-\frac{\mathrm{e}}{2}⩽a⩽0$$$$a<-\frac{\mathrm{e}}{2}\Rightarrow 1\mathrm{local}\min \mathrm{plus}1\mathrm{local}\max$$$$a=-\frac{\mathrm{e}}{2}\Rightarrow 1\mathrm{saddle}\mathrm{point}\left[{}_{\mathrm{horizontal}\mathrm{tangent}}^{=1\mathrm{point}\mathrm{with}\mathrm{a}}\right]$$$$-\frac{\mathrm{e}}{2}<a⩽0\Rightarrow \mathrm{no}\min /\max \mathrm{at}\mathrm{all}$$$$a>0\Rightarrow 1\mathrm{absolute}\min$$

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