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](https://www.tinkutara.com/question/Q100388.png)
$$\:\:\:\:\:\:\:\mathcal{G}\mathrm{iven}\:\mathrm{f}:\left[\mathrm{0},\mathrm{2}\right]\rightarrow\mathbb{R}\:,\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{twice}\:\mathrm{derivable}\:\mathrm{and}\: \\ $$$$\mathrm{f}\left(\mathrm{0}\right)=\mathrm{f}\left(\mathrm{1}\right)=\mathrm{f}\left(\mathrm{2}\right)=\mathrm{0} \\ $$$${i}-\mathcal{S}\mathrm{how}\:\mathrm{that}\:\mathrm{there}\:\mathrm{exist}\:\mathrm{c}_{\mathrm{1}} ,\:\mathrm{c}_{\mathrm{2}} ,\:\mathrm{such}\:\mathrm{that}\:\mathrm{f}'\left(\mathrm{c}_{\mathrm{1}} \right)=\mathrm{0}\: \\ $$$$\mathrm{and}\:\mathrm{f}'\left(\mathrm{c}_{\mathrm{2}} \right)=\mathrm{0} \\ $$$${ii}-\mathcal{S}\mathrm{how}\:\mathrm{that}\:\mathrm{there}\:\mathrm{exist}\:\mathrm{c}_{\mathrm{3}} \:\mathrm{such}\:\mathrm{that}\:\mathrm{f}''\left(\mathrm{c}_{\mathrm{3}} \right)=\mathrm{0} \\ $$
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](https://www.tinkutara.com/question/Q100398.png)
$$\left.{f}\left(\mathrm{0}\right)={f}\left(\mathrm{1}\right)\Rightarrow\exists{c}\in\right]\mathrm{0},\mathrm{1}\left[,\:\:{f}'\left({c}_{\mathrm{1}} \right)=\mathrm{0}\right. \\ $$$$\left.{f}\left(\mathrm{1}\right)={f}\left(\mathrm{2}\right)\Rightarrow\exists{c}_{\mathrm{2}} \in\right]\mathrm{1},\mathrm{2}\left[\:{f}'\left({c}_{\mathrm{2}} \right)=\mathrm{0}\right. \\ $$$${let}\:{f}'\left({x}\right)\:{over}\:\left[{c}_{\mathrm{1}} ,{c}_{\mathrm{2}} \right]\: \\ $$$${f}'\left({c}_{\mathrm{1}} \right)={f}'\left({c}_{\mathrm{2}} \right)\Rightarrow\exists{c}_{\mathrm{3}} \in\left[{c}_{\mathrm{1}} ,{c}_{\mathrm{2}} \right]{such}\:{f}''\left({c}_{\mathrm{3}} \right)=\mathrm{0} \\ $$$${i}\:{used}\:{if}\:{f}\:{continus}\:{differentiabl}\:{over}\:\left[{a},{b}\right]\:{such} \\ $$$${f}\left({a}\right)={f}\left({b}\right)\Rightarrow\exists{c}\in\left[{a},{b}\right]\:{such}\:{f}'\left({c}\right)=\mathrm{0}.{Roll}\:{theorem} \\ $$$$ \\ $$$$ \\ $$
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.����