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Category: Relation and Functions

let-U-n-1-2-2-2-3-2-n-2-1-4-2-4-3-4-n-4-1-find-lim-n-U-n-2-calculate-n-1-U-n-

Question Number 58354 by maxmathsup by imad last updated on 21/Apr/19 $${let}\:{U}_{{n}} =\frac{\mathrm{1}^{\mathrm{2}} \:+\mathrm{2}^{\mathrm{2}} \:+\mathrm{3}^{\mathrm{2}} \:+….+{n}^{\mathrm{2}} }{\mathrm{1}^{\mathrm{4}} \:+\mathrm{2}^{\mathrm{4}} \:+\mathrm{3}^{\mathrm{4}} \:+….+{n}^{\mathrm{4}} } \\ $$$$\left.\mathrm{1}\right){find}\:{lim}_{{n}\rightarrow+\infty} {U}_{{n}} \\…

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Question Number 123706 by I want to learn more last updated on 27/Nov/20 $$\mathrm{If}\:\:\:\:\mathrm{h}\left(\mathrm{x}\right)\:\:=\:\:\mathrm{f}^{\mathrm{2}} \left(\mathrm{x}\right)\:\:−\:\:\mathrm{g}^{\mathrm{2}} \left(\mathrm{x}\right),\:\:\:\:\:\:\:\:\mathrm{f}\:'\left(\mathrm{x}\right)\:\:\:=\:\:\:−\:\mathrm{g}\left(\mathrm{x}\right)\:\:\:\:\mathrm{and}\:\:\:\:\:\mathrm{g}\:'\left(\mathrm{x}\right)\:\:=\:\:\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\:\:\:\mathrm{then}\:\:\:\:\mathrm{h}\:'\left(\mathrm{x}\right)\:\:=\:\:\:??? \\ $$$$ \\ $$$$\left(\mathrm{a}\right)\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\left(\mathrm{b}\right)\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{c}\right)\:\:\:\:\:−\:\:\mathrm{4}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{g}\left(\mathrm{x}\right)\:\:\:\:\:\:\:\:\:\:\left(\mathrm{d}\right)\:\:\:\:\:\left[−\mathrm{g}\left(\mathrm{x}\right)\right]^{\mathrm{2}} \:\:−\:\:\left[\mathrm{f}\left(\mathrm{x}\right)\right]^{\mathrm{2}} \\ $$…

calculste-lim-x-0-arctan-x-2-x-1-arctan-x-1-x-3-

Question Number 123675 by mathmax by abdo last updated on 27/Nov/20 $$\mathrm{calculste}\:\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\:\frac{\mathrm{arctan}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)−\mathrm{arctan}\left(\mathrm{x}+\mathrm{1}\right)}{\mathrm{x}^{\mathrm{3}} } \\ $$ Answered by benjo_mathlover last updated on 27/Nov/20 $$\underset{{x}\rightarrow\mathrm{0}}…

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Question Number 189053 by mathlove last updated on 11/Mar/23 $${find}\:{f}\left({x}\right) \\ $$$$\mathrm{1}:{f}\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)={x}+\mathrm{3};\:{x}\neq\mathrm{1} \\ $$$$\mathrm{2}:{f}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)={x}^{\mathrm{2}} +\mathrm{2}{x}\:;{x}\neq\mathrm{1} \\ $$$$\mathrm{3}:{f}\left({x}+\mathrm{1}\right)+{f}\left({x}−{y}\right)=\mathrm{2}{f}\left({x}\right){cosy}\:\forall{x},{y} \\ $$$${f}\left(\mathrm{0}\right)={f}\left(\frac{\pi}{\mathrm{2}}\right)=\mathrm{1} \\ $$ Answered by cortano12 last…