Question Number 203291 by zahaku last updated on 14/Jan/24 $${f}\left({x}\right)=\left\{\mathrm{1}+\frac{\sqrt{{x}^{\mathrm{2}} }}{{x}}\:\:\:{if}\:{x}#\mathrm{0}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:{if}\:\:{x}=\mathrm{0} \\ $$$${study}\:{the}\:{continuty}\:{of}\:{f}\:{in}\:\mathrm{0} \\ $$ Answered by esmaeil last updated on 15/Jan/24 $${f}\left({x}\right)=\begin{cases}{\mathrm{1}+\frac{\mid{x}\mid}{{x}}\:\:\:\:{if}\:\:{x}\neq\mathrm{0}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{if}\:\:{x}=\mathrm{0}}\end{cases}\:\:\:…
Question Number 202865 by Mathspace last updated on 04/Jan/24 $${find}\:{the}\:{seauence}\:{u}_{{n}} {wich}\:{verify} \\ $$$${u}_{\mathrm{0}} =\mathrm{1}\:{and}\:{u}_{{n}} +{u}_{{n}+\mathrm{1}} =\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}} \\ $$$${for}\:{n}\geqslant\mathrm{1} \\ $$ Commented by mr W…
Question Number 202795 by mathlove last updated on 03/Jan/24 $${if}\:{f}\left(−\mathrm{1}\right)={f}\left(\mathrm{0}\right)={f}\left(\mathrm{2}\right)=\mathrm{0}\:{and}\:{f}\left(\mathrm{1}\right)=\mathrm{6} \\ $$$${then}\:{find}\:{f}\left({x}\right)=? \\ $$ Answered by AST last updated on 03/Jan/24 $${If}\:{f}\left({x}\right)\:{is}\:{a}\:{polynomial}:\:{x}\left({x}+\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{4}\right) \\ $$ Answered…
Question Number 202536 by cortano12 last updated on 29/Dec/23 $$\:\:\: \\ $$ Answered by taguim001 last updated on 29/Dec/23 $$ \\ $$$${f}\left({xy}\right){f}\left({x}\right)−{f}\left({xy}\right){f}\left({y}\right)={xf}\left({x}\right){f}\left({y}\right)−{yf}\left({x}\right){f}\left({y}\right) \\ $$$$\left\{_{{f}\left({xy}\right){f}\left({y}\right)={yf}\left({x}\right){f}\left({y}\right)} ^{{f}\left({xy}\right){f}\left({x}\right)={xf}\left({x}\right){f}\left({y}\right)}…
Question Number 201848 by cortano12 last updated on 14/Dec/23 Answered by dimentri last updated on 14/Dec/23 $$\:\:\mathrm{1820}\:=\:\mathrm{2}^{\mathrm{2}} ×\mathrm{5}×\mathrm{7}×\mathrm{13} \\ $$$$\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{positive}\:\mathrm{factors} \\ $$$$\:\mathrm{from}\:\mathrm{1820}\:=\:\underset{\mathrm{i}=\mathrm{0}} {\overset{\mathrm{5}} {\sum}}\begin{pmatrix}{\mathrm{5}}\\{\mathrm{i}}\end{pmatrix}\:=\:\mathrm{2}^{\mathrm{5}} =\:\mathrm{32}…
Question Number 201681 by jabarsing last updated on 10/Dec/23 $${f}\left({x}+\mathrm{1}\right)−{f}\left({x}\right)=\mathrm{3}{f}\left({x}\right)×{f}\left({x}+\mathrm{1}\right) \\ $$$${D}_{{f}} ={N} \\ $$$$\mathrm{2023}×{f}\left(\mathrm{1402}\right)=\mathrm{1} \\ $$$${have}\:{equation}\:{f}\left({x}\right)=\mathrm{1}\:{solution}? \\ $$ Answered by Frix last updated on…
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Question Number 200722 by cortano12 last updated on 22/Nov/23 $$\:\:\mathrm{Find}\:\mathrm{all}\:\mathrm{polynomials}\:\mathrm{P}\left(\mathrm{x}\right)\:\mathrm{with} \\ $$$$\:\mathrm{real}\:\mathrm{coefficients}\:\mathrm{such}\:\mathrm{that}\:\mathrm{for} \\ $$$$\:\mathrm{all}\:\mathrm{nonzero}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{x},\: \\ $$$$\:\:\:\:\:\mathrm{P}\left(\mathrm{x}\right)+\mathrm{P}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)=\frac{\mathrm{P}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right)+\mathrm{P}\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)}{\mathrm{2}}\:\:\: \\ $$ Commented by Frix last updated on 22/Nov/23…
Question Number 199781 by cortano12 last updated on 09/Nov/23 $$\:\:\: \\ $$ Answered by qaz last updated on 09/Nov/23 $${f}\left({x}\right)=\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{2}{x}\right)}{{x}\left(\mathrm{1}−{x}\right)}−{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right) \\ $$$$=\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{2}{x}\right)}{{x}\left(\mathrm{1}−{x}\right)}−\left(\frac{\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{1}−{x}}\right)}{\frac{\mathrm{1}}{\mathrm{1}−{x}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)}−{f}\left(\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{x}}}\right)\right) \\ $$$$=\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{2}{x}\right)}{{x}\left(\mathrm{1}−{x}\right)}−\frac{\mathrm{2}\left(\mathrm{1}+{x}\right)\left(\mathrm{1}−{x}\right)}{{x}}+{f}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right) \\…
Question Number 199167 by tri26112004 last updated on 28/Oct/23 $${Give}\:{a}\:{function}\: \\ $$$${f}:\:{R}\rightarrow\left(\mathrm{0};+\infty\right)\:{continous}\:{on}\:{R}\:{and}\:{such}\:{that} \\ $$$${f}\left({x}+{y}\right)\:=\:{f}\left({x}\right).{f}\left({y}\right) \\ $$$${a}.\:{Prove}\:{f}\left(\mathrm{0}\right)\:=\:\mathrm{1} \\ $$$${b}.\:{Let}\:{h}\left({x}\right)\:=\:{ln}\left[{f}\left({x}\right)\right].\:{Prove}\:{that}: \\ $$$$\:{h}\left({x}+{y}\right)\:=\:{h}\left({x}\right)\:+\:{h}\left({y}\right) \\ $$$${c}.\:{Find}\:{all}\:{the}\:{function}\:{f}\:{such}\:{that}\:{problem}\:{request} \\ $$$$\:\:\: \\…