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h-a-x-e-x-ax-2-show-that-h-a-admits-a-minimum-in-R-

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Question Number 208431 by alcohol last updated on 15/Jun/24
h_a (x) = e^(−x) + ax^2 show that h_a admits a minimum in R
ha(x)=e−x+ax2showthathaadmitsaminimuminR
Answered by mathzup last updated on 15/Jun/24
cas 1 a>0 lim_(x→−∞) h(x)=+∞ lim_(x→+∞) h(x)=+∞ h^′ (x)=2ax−e^(−x) lim_(x→−∞) h^′ (x)=−∞ lim_(x→+∞) h^′ (x)=+∞ h^((2)) (x)=2a+e^(−x) >0 ⇒h^′ est strictement croissante x −∞ 0 α +∞ h^′ (x) −∞ → −1 → 0 +∞ ⇒∃ α / h^′ (α)=0 x −∞ α +∞ h^′ − 0 + h decr h(α) croi donc h possede un minimun qui est h(α)=e^(−α) +aα^2 h(0)=−1 and h(1)=e^(−1) +a>0 ⇒ α ∈]0,1[ rest a etudier les cas a<0 et a=0
cas1a>0limx→−∞h(x)=+∞limx→+∞h(x)=+∞h′(x)=2ax−e−xlimx→−∞h′(x)=−∞limx→+∞h′(x)=+∞h(2)(x)=2a+e−x>0⇒h′eststrictementcroissantex−∞0α+∞h′(x)−∞→−1→0+∞⇒∃α/h′(α)=0x−∞α+∞h′−0+hdecrh(α)croidonchpossedeunminimunquiesth(α)=e−α+aα2h(0)=−1andh(1)=e−1+a>0⇒α∈]0,1[restaetudierlescasa<0eta=0
Answered by Frix last updated on 16/Jun/24
It′s not true for −(e/2)≤a≤0 a<−(e/2) ⇒ 1 local min plus 1 local max a=−(e/2) ⇒ 1 saddle point [ _(horizontal tangent)^(=1 point with a) ] −(e/2)<a≤0 ⇒ no min/max at all a>0 ⇒ 1 absolute min
It′snottruefor−e2⩽a⩽0a<−e2⇒1localminplus1localmaxa=−e2⇒1saddlepoint[horizontaltangent=1pointwitha]−e2<a⩽0⇒nomin/maxatalla>0⇒1absolutemin

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