Question Number 199694 by Mingma last updated on 07/Nov/23
Answered by witcher3 last updated on 07/Nov/23
![g(x)=f(((x+b)/2))−((f(x)+f(b))/2) g(a)=0,g(b)=0,rohll theorem ⇒∃c∈]a,b[∣g′(c)=0; g′(x)=(1/2)f′(((x+b)/2))−(1/2)f′(x) g′(c)=0⇔f′(((c+b)/2))=f′(c) applie Rohll to x→f′(x) over [c,((c+b)/2)] f′′(c)=f′′(((c+b)/2))⇒∃d∈]c,((c+d)/2)[⊂]a,b[ f′′(d)=0](https://www.tinkutara.com/question/Q199696.png)
$$\mathrm{g}\left(\mathrm{x}\right)=\mathrm{f}\left(\frac{\mathrm{x}+\mathrm{b}}{\mathrm{2}}\right)−\frac{\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{b}\right)}{\mathrm{2}} \\ $$$$\mathrm{g}\left(\mathrm{a}\right)=\mathrm{0},\mathrm{g}\left(\mathrm{b}\right)=\mathrm{0},\mathrm{rohll}\:\mathrm{theorem} \\ $$$$\left.\Rightarrow\exists\mathrm{c}\in\right]\mathrm{a},\mathrm{b}\left[\mid\mathrm{g}'\left(\mathrm{c}\right)=\mathrm{0};\right. \\ $$$$\mathrm{g}'\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{f}'\left(\frac{\mathrm{x}+\mathrm{b}}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{f}'\left(\mathrm{x}\right) \\ $$$$\mathrm{g}'\left(\mathrm{c}\right)=\mathrm{0}\Leftrightarrow\mathrm{f}'\left(\frac{\mathrm{c}+\mathrm{b}}{\mathrm{2}}\right)=\mathrm{f}'\left(\mathrm{c}\right)\:\mathrm{applie}\:\mathrm{Rohll}\:\mathrm{to} \\ $$$$\left.\mathrm{x}\rightarrow\mathrm{f}'\left(\mathrm{x}\right)\:\mathrm{over}\:\left[\mathrm{c},\frac{\mathrm{c}+\mathrm{b}}{\mathrm{2}}\right]\:\mathrm{f}''\left(\mathrm{c}\right)=\mathrm{f}''\left(\frac{\mathrm{c}+\mathrm{b}}{\mathrm{2}}\right)\Rightarrow\exists\mathrm{d}\in\right]\mathrm{c},\frac{\mathrm{c}+\mathrm{d}}{\mathrm{2}}\left[\subset\right]\mathrm{a},\mathrm{b}\left[\right. \\ $$$$\mathrm{f}''\left(\mathrm{d}\right)=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by MM42 last updated on 07/Nov/23
$${very}\:{nice}\:\:\:\underline{\underbrace{\lesseqgtr}} \\ $$
Commented by Mingma last updated on 08/Nov/23
Perfect ��
Commented by witcher3 last updated on 08/Nov/23
$$\mathrm{Thank}\:\mathrm{You} \\ $$