Question Number 207390 by alcohol last updated on 13/May/24 Commented by alcohol last updated on 13/May/24 $${please}\:{i}\:{really}\:{need}\:{help}\:{here} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 207314 by universe last updated on 11/May/24 $$\:\mathrm{let}\:\mathrm{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{be}\:\mathrm{a}\:\mathrm{continuous}\:\mathrm{function}\:\mathrm{then} \\ $$$$\:\mathrm{show}\:\mathrm{that} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{if}\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{f}\left(\mathrm{x}^{\mathrm{2}} \right)\:\forall\:\mathrm{x}\:\in\mathbb{R}\:\mathrm{then}\:\mathrm{f}\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{constant} \\ $$$$\:\:\mathrm{function} \\ $$$$\:\left(\mathrm{2}\right)\:\mathrm{if}\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{f}\left(\mathrm{2x}+\mathrm{1}\right)\:\forall\mathrm{x}\in\mathbb{R}\:\mathrm{then}\:\mathrm{f}\:\:\mathrm{is}\:\mathrm{a}\: \\ $$$$\:\:\mathrm{constant}\:\mathrm{function}\: \\ $$ Answered by…
Question Number 207085 by necx122 last updated on 06/May/24 $${Let}\:{f}\:{be}\:{a}\:{function}\:{with}\:{the}\:{following} \\ $$$${properties}:\:\left({i}\right)\:{f}\left(\mathrm{1}\right)\:=\mathrm{1}\:\left({ii}\right)\:{f}\left(\mathrm{2}{n}\right)={n}.{f}\left({n}\right)\:{for} \\ $$$${any}\:{positive}\:{integer}\:{n}.\:{Find}\:{the}\:{value} \\ $$$${of}\:{f}\left(\mathrm{2}^{\mathrm{10}} \right) \\ $$$$\left.{a}\left.\right)\left.\mathrm{1}\left.\:{b}\right)\:\mathrm{2}^{\mathrm{10}\:} {c}\right)\:\mathrm{2}^{\mathrm{35}} \:{d}\right)\:\mathrm{2}^{\mathrm{45}} \\ $$ Answered by…
Question Number 206993 by necx122 last updated on 03/May/24 $${If}\:{the}\:{nth}\:{term}\:{of}\:{a}\:{sequence}\:{is} \\ $$$${given}\:{by}\:\frac{{n}^{\mathrm{2}} −\mathrm{2}{n}}{\mathrm{4}}\:,{what}\:{is}\:{the}\:{sum}\:{of}\:{n} \\ $$$${terms}\:{of}\:{the}\:{sequence}? \\ $$ Answered by Rasheed.Sindhi last updated on 03/May/24 $$\underset{{k}=\mathrm{1}}…
Question Number 206442 by universe last updated on 14/Apr/24 Answered by Berbere last updated on 14/Apr/24 $${let}\:{f}\left({x}\right)={e}^{{g}\left({x}\right)} ;\:{particular}\:{Solution}\:{just}\:{to}\:{simplifie} \\ $$$${the}\:{problems};{f}\left(\mathrm{0}\right)=\mathrm{1}={e}^{{g}\left(\mathrm{0}\right)} \Rightarrow{g}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\Rightarrow{f}'\left({x}\right)={g}'{e}^{{g}\left({x}\right)} ;{f}''\left({x}\right)=\left({g}'^{\mathrm{2}} +{g}''\right){e}^{{g}\left({x}\right)}…
Question Number 206151 by cortano21 last updated on 08/Apr/24 Answered by dimentri last updated on 08/Apr/24 $$\:\:\mathrm{x}=\mathrm{2}\Rightarrow\mathrm{f}\left(\mathrm{3}\right)=\mathrm{4}\: \\ $$$$\:\:\mathrm{x}=\mathrm{1}\Rightarrow\mathrm{f}\left(\mathrm{3}\right)=\:\mathrm{g}^{−\mathrm{1}} \left(\mathrm{1}\right) \\ $$$$\:\:\mathrm{g}\left(\mathrm{4}\right)=\:\mathrm{g}\left(\mathrm{f}\left(\mathrm{3}\right)\right)\:=\:\mathrm{g}\left(\mathrm{g}^{−\mathrm{1}} \left(\mathrm{1}\right)\right)\:=\:\mathrm{1} \\ $$…
Question Number 206136 by cortano21 last updated on 07/Apr/24 Answered by A5T last updated on 07/Apr/24 $${f}\left({f}\left({f}\left({f}\left({x}\right)\right)\right)\right)=\left[{f}\left({x}\right)\right]^{\mathrm{2}} −\mathrm{3}\left[{f}\left({x}\right)\right]+\mathrm{4}={f}\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}\right) \\ $$$$\left[{f}\left(\mathrm{2}\right)\right]^{\mathrm{2}} −\mathrm{3}\left[{f}\left(\mathrm{2}\right)\right]+\mathrm{4}={f}\left(\mathrm{2}\right)\Rightarrow\left[{f}\left(\mathrm{2}\right)\right]^{\mathrm{2}} −\mathrm{4}{f}\left(\mathrm{2}\right)+\mathrm{4}=\mathrm{0} \\ $$$$\left[{f}\left(\mathrm{2}\right)−\mathrm{2}\right]^{\mathrm{2}}…
Question Number 205919 by Red1ight last updated on 03/Apr/24 $$\mathrm{write}\:\mathrm{the}\:\mathrm{following}\:\mathrm{recursive}\:\mathrm{function}\:\mathrm{in}\:\mathrm{explicit}\:\mathrm{form} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$${f}\left({n}+\mathrm{1}\right)=\left({n}+\mathrm{1}\right){f}\left({n}\right)+{n}! \\ $$ Answered by Tinku Tara last updated on 03/Apr/24 $${f}\left({n}\right)=\left({n}\right){f}\left({n}−\mathrm{1}\right)+\left({n}−\mathrm{1}\right)!…
Question Number 205262 by mathzup last updated on 13/Mar/24 $${nature}\:{of}\:{the}\:{serie}\:\sum_{{n}\geqslant\mathrm{1}} \:\frac{{ln}\left({n}\right)}{{n}} \\ $$ Answered by Berbere last updated on 13/Mar/24 $$\forall{n}\geqslant\mathrm{2}\:{ln}\left({n}\right)\geqslant{ln}\left(\mathrm{2}\right)>\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{ln}\left({n}\right)}{{n}}\geqslant\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}}\rightarrow+\infty\:{serie}\:{dv}…
Question Number 204478 by a.lgnaoui last updated on 18/Feb/24 $$\mathrm{soit}\:\boldsymbol{\mathrm{f}}:\:\:\mathbb{R}^{\mathrm{3}} \rightarrow\mathbb{R}^{\mathrm{3}} \:\:\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}},\boldsymbol{\mathrm{z}}\right)=\left(\mathrm{x}+\mathrm{y},\mathrm{2x}−\mathrm{y},\mathrm{x}+\mathrm{z}\right) \\ $$$$\bullet\mathrm{1}\:\:\mathrm{Ecrire}\:\mathrm{la}\:\mathrm{matrice}\:\mathrm{M}\:\mathrm{de}\:\mathrm{cette}\:\mathrm{application} \\ $$$$\:\:\:\mathrm{dans}\:\mathrm{la}\:\mathrm{base}\:\mathrm{canonique}\:{B}\:\mathrm{de}\:\:\mathbb{R}^{\mathrm{3}} \: \\ $$$$\bullet\mathrm{2}\:\:\mathrm{Calculer}\:\boldsymbol{\mathrm{f}}\left(\mathrm{1},\mathrm{2},\mathrm{3}\right)\mathrm{de}\:\mathrm{2}\:\mathrm{manieres}\:\mathrm{differentes} \\ $$$$\:−\mathrm{en}\:\mathrm{utilisant}\:\mathrm{la}\:\mathrm{definition}\:\mathrm{de}\:\mathrm{f} \\ $$$$−\mathrm{en}\:\mathrm{utilisant}\:\mathrm{la}\:\mathrm{matrice}\:{M}\: \\ $$$$\bullet\mathrm{3}\:\:\mathrm{determiner}\:\mathrm{bsse}\:\mathrm{de}\:\mathrm{Ker}\left(\:\boldsymbol{\mathrm{f}}\right)\:\mathrm{et}\:\mathrm{de}\:{I}\mathrm{m}\left(\boldsymbol{\mathrm{f}}\right)…