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Category: Relation and Functions

The-range-of-f-x-x-2-x-1-x-2-x-1-where-denotes-fractional-function-is-

Question Number 15093 by Tinkutara last updated on 07/Jun/17 $$\mathrm{The}\:\mathrm{range}\:\mathrm{of}\:{f}\left({x}\right)\:=\:\frac{\left\{{x}\right\}^{\mathrm{2}} \:−\:\left\{{x}\right\}\:+\:\mathrm{1}}{\left\{{x}\right\}^{\mathrm{2}} \:+\:\left\{{x}\right\}\:+\:\mathrm{1}}; \\ $$$$\left(\mathrm{where}\:\left\{\centerdot\right\}\:\mathrm{denotes}\:\mathrm{fractional}\:\mathrm{function}\right) \\ $$$$\mathrm{is}? \\ $$ Answered by mrW1 last updated on 07/Jun/17…

The-range-of-f-x-10-x-10-4-10-x-10-2-is-

Question Number 15086 by Tinkutara last updated on 07/Jun/17 $$\mathrm{The}\:\mathrm{range}\:\mathrm{of}\:{f}\left({x}\right)\:=\:\sqrt{\frac{\mathrm{10}^{{x}} \:−\:\mathrm{10}^{\mathrm{4}} }{\mathrm{10}^{{x}} \:+\:\mathrm{10}^{\mathrm{2}} }}\:\mathrm{is}? \\ $$ Answered by ajfour last updated on 07/Jun/17 $$\:{f}\left({x}\right)=\sqrt{\mathrm{1}−\left(\frac{\mathrm{10}^{\mathrm{4}} +\mathrm{10}^{\mathrm{2}}…

The-domain-of-f-x-x-2-x-where-denotes-fractional-part-of-x-is-

Question Number 15082 by Tinkutara last updated on 07/Jun/17 $$\mathrm{The}\:\mathrm{domain}\:\mathrm{of}\:{f}\left({x}\right)\:=\:\sqrt{{x}\:−\:\mathrm{2}\left\{{x}\right\}}.\:\left(\mathrm{where}\right. \\ $$$$\left.\left\{\centerdot\right\}\:\mathrm{denotes}\:\mathrm{fractional}\:\mathrm{part}\:\mathrm{of}\:{x}\right)\:\mathrm{is}? \\ $$ Answered by mrW1 last updated on 07/Jun/17 $$\mathrm{x}=\mathrm{n}+\mathrm{f} \\ $$$$\left\{\mathrm{x}\right\}=\mathrm{f} \\…

The-domain-of-f-x-1-x-2-x-where-denotes-fractional-part-of-x-is-

Question Number 15084 by Tinkutara last updated on 07/Jun/17 $$\mathrm{The}\:\mathrm{domain}\:\mathrm{of}\:{f}\left({x}\right)\:=\:\frac{\mathrm{1}}{\:\sqrt{−{x}^{\mathrm{2}} \:+\:\left\{{x}\right\}}}; \\ $$$$\left(\mathrm{where}\:\left\{\centerdot\right\}\:\mathrm{denotes}\:\mathrm{fractional}\:\mathrm{part}\:\mathrm{of}\:{x}\right) \\ $$$$\mathrm{is}? \\ $$ Answered by mrW1 last updated on 07/Jun/17 $$\mathrm{x}=\mathrm{n}+\mathrm{f}…

f-x-y-x-x-2y-1-condition-on-x-and-y-to-have-f-symetric-2-find-f-x-f-y-2-f-x-y-2-f-y-x-3-find-2-f-2-x-and-2-f-2-y-

Question Number 146085 by mathmax by abdo last updated on 10/Jul/21 $$\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{x}−\sqrt{\mathrm{x}+\mathrm{2y}} \\ $$$$\left.\mathrm{1}\right)\mathrm{condition}\:\mathrm{on}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y}\:\mathrm{to}\:\mathrm{have}\:\mathrm{f}\:\mathrm{symetric} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{find}\:\frac{\partial\mathrm{f}}{\partial\mathrm{x}}\:,\frac{\partial\mathrm{f}}{\partial\mathrm{y}}\:,\frac{\partial^{\mathrm{2}} \mathrm{f}}{\partial\mathrm{x}\partial\mathrm{y}}\:,\frac{\partial^{\mathrm{2}} \mathrm{f}}{\partial\mathrm{y}\partial\mathrm{x}} \\ $$$$\left.\mathrm{3}\right)\:\mathrm{find}\:\frac{\partial^{\mathrm{2}} \mathrm{f}}{\partial^{\mathrm{2}} \mathrm{x}}\:\mathrm{and}\:\frac{\partial^{\mathrm{2}} \mathrm{f}}{\partial^{\mathrm{2}} \mathrm{y}} \\…

F-x-x-n-e-in-1-roots-of-F-x-2-factorize-F-x-inside-C-x-

Question Number 146083 by mathmax by abdo last updated on 10/Jul/21 $$\mathrm{F}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{n}} \:−\mathrm{e}^{\mathrm{in}\alpha} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{roots}\:\mathrm{of}\:\mathrm{F}\left(\mathrm{x}\right)? \\ $$$$\left.\mathrm{2}\right)\:\mathrm{factorize}\:\mathrm{F}\left(\mathrm{x}\right)\:\mathrm{inside}\:\mathrm{C}\left[\mathrm{x}\right] \\ $$ Answered by Olaf_Thorendsen last updated on…

p-x-x-2-x-1-n-x-2-x-1-n-1-roots-of-p-x-2-factorize-p-x-inside-C-x-

Question Number 146082 by mathmax by abdo last updated on 10/Jul/21 $$\mathrm{p}\left(\mathrm{x}\right)=\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)^{\mathrm{n}} −\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)^{\mathrm{n}} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{roots}\:\mathrm{of}\:\mathrm{p}\left(\mathrm{x}\right)? \\ $$$$\left.\mathrm{2}\right)\:\mathrm{factorize}\:\mathrm{p}\left(\mathrm{x}\right)\:\mathrm{inside}\:\mathrm{C}\left[\mathrm{x}\right] \\ $$ Terms of Service Privacy…