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Category: Algebra

Question-199731

Question Number 199731 by Rupesh123 last updated on 08/Nov/23 Answered by des_ last updated on 08/Nov/23 $$\left({x}\:+\:\frac{\mathrm{1}}{{x}}\:\right)^{\mathrm{3}} =\:\mathrm{3}\sqrt{\mathrm{3}}\:\Rightarrow \\ $$$$\Rightarrow\:{x}^{\mathrm{3}} \:+\:\mathrm{3}\left({x}\:+\:\frac{\mathrm{1}}{{x}}\:\right)\:+\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:=\:\mathrm{3}\sqrt{\mathrm{3}}\:\:\Rightarrow \\ $$$$\Rightarrow\:{x}^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{3}}…

Question-199694

Question Number 199694 by Mingma last updated on 07/Nov/23 Answered by witcher3 last updated on 07/Nov/23 $$\mathrm{g}\left(\mathrm{x}\right)=\mathrm{f}\left(\frac{\mathrm{x}+\mathrm{b}}{\mathrm{2}}\right)−\frac{\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{b}\right)}{\mathrm{2}} \\ $$$$\mathrm{g}\left(\mathrm{a}\right)=\mathrm{0},\mathrm{g}\left(\mathrm{b}\right)=\mathrm{0},\mathrm{rohll}\:\mathrm{theorem} \\ $$$$\left.\Rightarrow\exists\mathrm{c}\in\right]\mathrm{a},\mathrm{b}\left[\mid\mathrm{g}'\left(\mathrm{c}\right)=\mathrm{0};\right. \\ $$$$\mathrm{g}'\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{f}'\left(\frac{\mathrm{x}+\mathrm{b}}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{f}'\left(\mathrm{x}\right) \\ $$$$\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}…

1-3-lt-2x-1-lt-7-find-x-x-Z-2-4-x-2-lt-5-find-x-x-Z-

Question Number 199608 by mathlove last updated on 06/Nov/23 $$\left.\mathrm{1}\right)\:\:\:\mathrm{3}<\mid\mathrm{2}{x}−\mathrm{1}\mid<\mathrm{7}\:\:{find}\:\Sigma{x}\:\:\:\:;{x}\in{Z} \\ $$$$\left.\mathrm{2}\right)\:\:\:\mathrm{4}\leqslant\mid{x}−\mathrm{2}\mid<\mathrm{5}\:\:{find}\:\Sigma{x}\:\:\:\:;{x}\in{Z} \\ $$$$ \\ $$ Answered by mr W last updated on 06/Nov/23 $$\left.\mathrm{1}\right)…

Question-199620

Question Number 199620 by Mingma last updated on 06/Nov/23 Answered by AST last updated on 06/Nov/23 $${f}\left({x}\right)=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{cx};{f}\left(\mathrm{4}\right)=\mathrm{10}\Rightarrow{c}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{f}\left({x}\right)=\frac{{x}\left({x}+\mathrm{1}\right)}{\mathrm{2}} \\ $$ Commented by Mingma last updated…