Question Number 93414 by abdomathmax last updated on 13/May/20 $${let}\:{p}\left({x}\right)=\frac{{x}^{{n}} \left(\mathrm{4}−\mathrm{2}{x}\right)^{{n}} }{{n}!} \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:\:{p}^{\left({k}\right)} \left(\mathrm{0}\right)={p}^{\left({k}\right)} \left(\mathrm{2}\right)=\mathrm{0}\:{for}\:{all}\:{k}\in\left[\mathrm{1},{n}−\mathrm{1}\right] \\ $$$$\left.\mathrm{2}\right)\:\:{prove}\:{that}\:\:\forall{m}\in{N}\:\:\:\:{p}^{\left({m}\right)} \left(\mathrm{0}\right)\:{and}\:{p}^{\left({m}\right)} \left(\mathrm{2}\right)\:{are}\:{integrs} \\ $$ Terms of Service…
Question Number 27847 by NECx last updated on 15/Jan/18 $${Find}\:{the}\:{range}\:{of}\:{f}\left({x}\right)=\frac{{x}+\mathrm{5}}{{x}^{\mathrm{2}} −\mathrm{4}} \\ $$ Commented by NECx last updated on 16/Jan/18 $${please}\:{help} \\ $$ Answered by…
Question Number 93331 by john santu last updated on 12/May/20 $$\mathrm{If}\:\mathrm{f}\left(\mathrm{x}+\mathrm{y}\right)\:=\:\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{y}\right)\:\mathrm{and} \\ $$$$\mathrm{f}\left(\mathrm{1}\right)=\mathrm{5}\:\mathrm{then}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{of}\:\mathrm{f}\left(\mathrm{2020}\right)\:?\: \\ $$ Commented by mr W last updated on 12/May/20…
Question Number 27790 by abdo imad last updated on 14/Jan/18 $${let}\:{give}\:\:{f}\left({x}\right)=\:{x}^{{n}−\mathrm{1}} \:{ln}\left(\mathrm{1}+{x}\right)\:\:{with}\:{n}\:{fromN}^{\ast} \:\:\:{find}\:{f}^{\left({n}\right)} \left({x}\right) \\ $$ Commented by abdo imad last updated on 15/Jan/18 $$\:{by}\:{leibnitz}\:{formule}\:{f}^{\left({n}\right)}…
Question Number 158862 by mathlove last updated on 09/Nov/21 Answered by gsk2684 last updated on 09/Nov/21 $${f}\left({x}\right)+\mathrm{2}{f}\left(\frac{\mathrm{1}}{{x}}\right)={x}…\left(\mathrm{1}\right) \\ $$$${replace}\:{x}\:{by}\:\frac{\mathrm{1}}{{x}} \\ $$$${f}\left(\frac{\mathrm{1}}{{x}}\right)+\mathrm{2}{f}\left({x}\right)=\frac{\mathrm{1}}{{x}}\Rightarrow\frac{{x}−{f}\left({x}\right)}{\mathrm{2}}+\mathrm{2}{f}\left({x}\right)=\frac{\mathrm{1}}{{x}} \\ $$$$\frac{{x}−{f}\left({x}\right)+\mathrm{4}{f}\left({x}\right)}{\mathrm{2}}=\frac{\mathrm{1}}{{x}}\Rightarrow{x}+\mathrm{3}{f}\left({x}\right)=\frac{\mathrm{2}}{{x}} \\ $$$$\mathrm{3}{f}\left({x}\right)=\frac{\mathrm{2}}{{x}}−{x}\Rightarrow{f}\left({x}\right)=\frac{\mathrm{2}−{x}^{\mathrm{2}}…
Question Number 27785 by abdo imad last updated on 14/Jan/18 $${find}\:{nature}\:{of}\:{the}\:{serie}\:\sum_{{n}=\mathrm{1}} ^{\propto} \:\:\xi\left({n}\right)\:{x}^{{n}} \\ $$$${with}\:\xi\left({x}\right)=\:\sum_{{n}=\mathrm{1}} ^{\propto} \:\:\:\frac{\mathrm{1}}{{n}^{{x}} }\:\:\:\:\:{and}\:{x}>\mathrm{1}\:. \\ $$ Commented by abdo imad last…
Question Number 27784 by abdo imad last updated on 14/Jan/18 $${let}\:{give}\:\:{f}\left({x}\right)\:=\:{e}^{−\frac{\mathrm{1}}{{x}}} \:\:\:{with}\:{f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{is}\:{f}\:{derivable}\:{in}\:{point}\:\mathrm{0}? \\ $$$$\left.\mathrm{2}\right){prove}\:{that}\:\:{f}^{\left({n}\right)} =\:{F}_{{n}} \left({x}\right)\:{e}^{−\frac{\mathrm{1}}{{x}}} \:\:{with}\:{F}_{{n}\:} {is}\:{rational}\:{function} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\:\:{f}^{\left(\mathrm{6}\right)} \:\left({x}\right)\:{and}\:\:{f}^{\left(\mathrm{9}\right)} \left({x}\right)\:\:. \\…
Question Number 158805 by tounghoungko last updated on 09/Nov/21 $$\left(\mathrm{1}\right){F}\left({x}\right)=\:{x}^{\mathrm{3}} \:\left[\:{x}\:\right]\:\Rightarrow\begin{cases}{{F}\:'\left(\mathrm{0}\right)=?}\\{{F}\:'\left(\mathrm{1}\right)=?}\end{cases} \\ $$$$\:\left(\mathrm{2}\right)\:{F}\left({x}\right)=\:\left[\:{x}\:\right]−\mid{x}\mid\:\Rightarrow{F}\:'\left(−\frac{\mathrm{5}}{\mathrm{2}}\right)=? \\ $$$$\:{where}\:\left[\:\right]\::\:{floor}\:{function} \\ $$$$\:\mid\:\mid\:{absolute}\:{function}\: \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 27618 by NECx last updated on 11/Jan/18 $${Find}\:{the}\:{range}\:{of}\:{y}={x}\left({x}^{\mathrm{6}} −\mathrm{1}\right).{For} \\ $$$${which}\:{y}=\mathrm{0} \\ $$ Answered by mrW2 last updated on 13/Jan/18 $${y}={x}\left({x}^{\mathrm{6}} −\mathrm{1}\right)=\mathrm{0} \\…
Question Number 158668 by cortano last updated on 07/Nov/21 $$\:{f}\left({f}\left({x}\right)\right)=\:\left(\mathrm{9}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{2}\right){f}\left({x}\right) \\ $$$$\:{f}\left({x}\right)=? \\ $$ Answered by ajfour last updated on 07/Nov/21 $${Assuming}\:{f}\left({x}\right)\:{is}\:{a}\:{polynome}.. \\ $$$${n}^{\mathrm{2}}…