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Question Number 100388 by Ar Brandon last updated on 26/Jun/20

       Given f:[0,2]→R , f(x) is twice derivable and   f(0)=f(1)=f(2)=0  i-Show that there exist c_1 , c_2 , such that f′(c_1 )=0   and f′(c_2 )=0  ii-Show that there exist c_3  such that f′′(c_3 )=0

$$\:\:\:\:\:\:\:\mathcal{G}\mathrm{iven}\:\mathrm{f}:\left[\mathrm{0},\mathrm{2}\right]\rightarrow\mathbb{R}\:,\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{twice}\:\mathrm{derivable}\:\mathrm{and}\: \\ $$$$\mathrm{f}\left(\mathrm{0}\right)=\mathrm{f}\left(\mathrm{1}\right)=\mathrm{f}\left(\mathrm{2}\right)=\mathrm{0} \\ $$$${i}-\mathcal{S}\mathrm{how}\:\mathrm{that}\:\mathrm{there}\:\mathrm{exist}\:\mathrm{c}_{\mathrm{1}} ,\:\mathrm{c}_{\mathrm{2}} ,\:\mathrm{such}\:\mathrm{that}\:\mathrm{f}'\left(\mathrm{c}_{\mathrm{1}} \right)=\mathrm{0}\: \\ $$$$\mathrm{and}\:\mathrm{f}'\left(\mathrm{c}_{\mathrm{2}} \right)=\mathrm{0} \\ $$$${ii}-\mathcal{S}\mathrm{how}\:\mathrm{that}\:\mathrm{there}\:\mathrm{exist}\:\mathrm{c}_{\mathrm{3}} \:\mathrm{such}\:\mathrm{that}\:\mathrm{f}''\left(\mathrm{c}_{\mathrm{3}} \right)=\mathrm{0} \\ $$

Answered by maths mind last updated on 26/Jun/20

f(0)=f(1)⇒∃c∈]0,1[,  f′(c_1 )=0  f(1)=f(2)⇒∃c_2 ∈]1,2[ f′(c_2 )=0  let f′(x) over [c_1 ,c_2 ]   f′(c_1 )=f′(c_2 )⇒∃c_3 ∈[c_1 ,c_2 ]such f′′(c_3 )=0  i used if f continus differentiabl over [a,b] such  f(a)=f(b)⇒∃c∈[a,b] such f′(c)=0.Roll theorem

$$\left.{f}\left(\mathrm{0}\right)={f}\left(\mathrm{1}\right)\Rightarrow\exists{c}\in\right]\mathrm{0},\mathrm{1}\left[,\:\:{f}'\left({c}_{\mathrm{1}} \right)=\mathrm{0}\right. \\ $$$$\left.{f}\left(\mathrm{1}\right)={f}\left(\mathrm{2}\right)\Rightarrow\exists{c}_{\mathrm{2}} \in\right]\mathrm{1},\mathrm{2}\left[\:{f}'\left({c}_{\mathrm{2}} \right)=\mathrm{0}\right. \\ $$$${let}\:{f}'\left({x}\right)\:{over}\:\left[{c}_{\mathrm{1}} ,{c}_{\mathrm{2}} \right]\: \\ $$$${f}'\left({c}_{\mathrm{1}} \right)={f}'\left({c}_{\mathrm{2}} \right)\Rightarrow\exists{c}_{\mathrm{3}} \in\left[{c}_{\mathrm{1}} ,{c}_{\mathrm{2}} \right]{such}\:{f}''\left({c}_{\mathrm{3}} \right)=\mathrm{0} \\ $$$${i}\:{used}\:{if}\:{f}\:{continus}\:{differentiabl}\:{over}\:\left[{a},{b}\right]\:{such} \\ $$$${f}\left({a}\right)={f}\left({b}\right)\Rightarrow\exists{c}\in\left[{a},{b}\right]\:{such}\:{f}'\left({c}\right)=\mathrm{0}.{Roll}\:{theorem} \\ $$$$ \\ $$$$ \\ $$

Commented by DGmichael last updated on 26/Jun/20

��very good.

Commented by Ar Brandon last updated on 26/Jun/20

Thanks ��

Commented by Ar Brandon last updated on 26/Jun/20

Ouaye DG, dès que ce monsieur se connecte oulala !�� On dirait une machine qui venait d'être activée.����

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