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Question Number 199694 by Mingma last updated on 07/Nov/23

Answered by witcher3 last updated on 07/Nov/23

$$\mathrm{g}\left(\mathrm{x}\right)=\mathrm{f}\left(\frac{\mathrm{x}+\mathrm{b}}{2}\right)-\frac{\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{b}\right)}{2}$$$$\mathrm{g}\left(\mathrm{a}\right)=0,\mathrm{g}\left(\mathrm{b}\right)=0,\mathrm{rohll}\mathrm{theorem}$$$$\Rightarrow \mathrm{\exists }\mathrm{c}\in ]\mathrm{a},\mathrm{b}[\mid {\mathrm{g}}^{\prime }\left(\mathrm{c}\right)=0;$$$${\mathrm{g}}^{\prime }\left(\mathrm{x}\right)=\frac{1}{2}{\mathrm{f}}^{\prime }\left(\frac{\mathrm{x}+\mathrm{b}}{2}\right)-\frac{1}{2}{\mathrm{f}}^{\prime }\left(\mathrm{x}\right)$$$${\mathrm{g}}^{\prime }\left(\mathrm{c}\right)=0\iff {\mathrm{f}}^{\prime }\left(\frac{\mathrm{c}+\mathrm{b}}{2}\right)={\mathrm{f}}^{\prime }\left(\mathrm{c}\right)\mathrm{applie}\mathrm{Rohll}\mathrm{to}$$$$\mathrm{x}\to {\mathrm{f}}^{\prime }\left(\mathrm{x}\right)\mathrm{over}[\mathrm{c},\frac{\mathrm{c}+\mathrm{b}}{2}]{\mathrm{f}}^{″}\left(\mathrm{c}\right)={\mathrm{f}}^{″}\left(\frac{\mathrm{c}+\mathrm{b}}{2}\right)\Rightarrow \mathrm{\exists }\mathrm{d}\in ]\mathrm{c},\frac{\mathrm{c}+\mathrm{d}}{2}[\subset ]\mathrm{a},\mathrm{b}[$$$${\mathrm{f}}^{″}\left(\mathrm{d}\right)=0$$$$$$$$$$$$$$$$$$$$$$$$$$

Commented by MM42 last updated on 07/Nov/23

$$verynice\underset{―}{\underbrace{⋚}}$$

Commented by Mingma last updated on 08/Nov/23

Perfect ��

Commented by witcher3 last updated on 08/Nov/23

$$\mathrm{Thank}\mathrm{You}$$

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