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) \\ $$$$\bullet\mathrm{1}\:\:\mathrm{Ecrire}\:\mathrm{la}\:\mathrm{matrice}\:\mathrm{M}\:\mathrm{de}\:\mathrm{cette}\:\mathrm{application} \\ $$$$\:\:\:\mathrm{dans}\:\mathrm{la}\:\mathrm{base}\:\mathrm{canonique}\:{B}\:\mathrm{de}\:\:\mathbb{R}^{\mathrm{3}} \: \\ $$$$\bullet\mathrm{2}\:\:\mathrm{Calculer}\:\boldsymbol{\mathrm{f}}\left(\mathrm{1},\mathrm{2},\mathrm{3}\right)\mathrm{de}\:\mathrm{2}\:\mathrm{manieres}\:\mathrm{differentes} \\ $$$$\:−\mathrm{en}\:\mathrm{utilisant}\:\mathrm{la}\:\mathrm{definition}\:\mathrm{de}\:\mathrm{f} \\ $$$$−\mathrm{en}\:\mathrm{utilisant}\:\mathrm{la}\:\mathrm{matrice}\:{M}\: \\ $$$$\bullet\mathrm{3}\:\:\mathrm{determiner}\:\mathrm{bsse}\:\mathrm{de}\:\mathrm{Ker}\left(\:\boldsymbol{\mathrm{f}}\right)\:\mathrm{et}\:\mathrm{de}\:{I}\mathrm{m}\left(\boldsymbol{\mathrm{f}}\right)…
Question Number 204426 by universe last updated on 17/Feb/24 $$\:\:\mathrm{let}\:\mathrm{a}\:,\:\mathrm{b}\:>\mathrm{0}\:\:\mathrm{find}\:\mathrm{all}\:\mathrm{differentiable}\:\mathrm{function} \\ $$$$\:\:\:\mathrm{f}:\left(\mathrm{0},\infty\right)\rightarrow\left(\mathrm{0},\infty\right)\:\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\:\:\:\:\mathrm{f}'\left(\frac{{a}}{\mathrm{x}}\right)\:\:=\:\:\frac{\mathrm{bx}}{\mathrm{f}\left(\mathrm{x}\right)}\:\:\:\:,\:\:\:\forall\:\mathrm{x}>\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 203905 by sniper237 last updated on 01/Feb/24 $${let}\:{ABC}\:{a}\:{given}\:{triangle}.\:{Can}\:{we}\:{find} \\ $$$${three}\:{positions}\:{I},{J},{K}\:{on}\:{the}\:{side}\:{AB},{AC},{BC} \\ $$$${Such}\:{that}\:{IJK}\:{is}\:{equilateral}? \\ $$ Commented by mr W last updated on 01/Feb/24 $${yes},\:{we}\:{can}\:{find}\:{infinite}\:{many}\:{such}…