Question Number 116376 by bemath last updated on 03/Oct/20 $$\mathrm{Let}\:\mathrm{f}\:\mathrm{be}\:\mathrm{a}\:\mathrm{function}\:\mathrm{defined}\:\mathrm{on}\:\mathrm{non}\:\mathrm{zero}\:\: \\ $$$$\mathrm{real}\:\mathrm{numbers}\:\mathrm{such}\:\mathrm{that}\:\frac{\mathrm{27}\:\mathrm{f}\left(−\mathrm{x}\right)}{\mathrm{x}}\:−\mathrm{x}^{\mathrm{2}} \:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\:=−\mathrm{2x}^{\mathrm{2}} \\ $$$$\mathrm{for}\:\forall\mathrm{x}\:\neq\:\mathrm{0}\:.\:\mathrm{Find}\:\rightarrow\begin{cases}{\mathrm{f}\left(\mathrm{x}\right)}\\{\mathrm{f}\left(\mathrm{3}\right)}\end{cases}\:?\: \\ $$ Answered by bobhans last updated on 03/Oct/20 $$\mathrm{Letting}\:\mathrm{x}\:=\:−\mathrm{y},\:\mathrm{we}\:\mathrm{get}\:…
Question Number 116365 by soumyasaha last updated on 03/Oct/20 Answered by Olaf last updated on 03/Oct/20 $${f}\left({x}\right)\:=\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{3}} {\sum}}\frac{{f}^{\left({k}\right)} \left({a}\right)}{{k}!}\left({x}−{a}\right)^{{k}} +{R}_{\mathrm{3}} \left({x}\right) \\ $$$${f}^{\left(\mathrm{0}\right)} \left({a}\right)\:=\:{f}\left(\frac{\pi}{\mathrm{2}}\right)\:=\:\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}\right)\:=\:\mathrm{0}…
Question Number 115997 by bemath last updated on 30/Sep/20 Answered by bobhans last updated on 30/Sep/20 $$\left(\mathrm{1}\right){f}\left(\frac{\mathrm{1}}{{x}}\right)+{f}\left(\mathrm{1}−{x}\right)\:=\:{x}\rightarrow{f}\left(\mathrm{1}−{x}\right)={x}−{f}\left(\frac{\mathrm{1}}{{x}}\right) \\ $$$${replace}\:\mathrm{1}−{x}\:{by}\:{x} \\ $$$$\left(\mathrm{2}\right)\:{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)\:+\:{f}\left({x}\right)\:=\:\mathrm{1}−{x} \\ $$$${replace}\:\frac{\mathrm{1}}{{x}}\:{by}\:\mathrm{1}−{x}\: \\ $$$$\left(\mathrm{3}\right)\:{f}\left(\mathrm{1}−{x}\right)+{f}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{x}}…
Question Number 115960 by Fikret last updated on 29/Sep/20 $${f}^{\mathrm{2}} \left({x}\right)={f}\left(\mathrm{2}{x}\right)+\mathrm{2}{f}\left({x}\right)−\mathrm{2}\: \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{3}\:\:\:\Rightarrow\:{f}\left(\mathrm{6}\right)=? \\ $$ Commented by prakash jain last updated on 29/Sep/20 $$\mathrm{power}\:\mathrm{converted}\:\mathrm{into}\:\mathrm{linear} \\…
Question Number 50426 by Abdo msup. last updated on 16/Dec/18 $${study}\:{the}\:{convergence}\:{of}\: \\ $$$${U}_{{n}} =\left(\frac{\mathrm{2}}{\pi}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({sinx}\right)^{\frac{\mathrm{1}}{{n}}} \right)^{{n}} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on…
Question Number 50411 by Abdo msup. last updated on 16/Dec/18 $${find}\:{all}\:{function}\:{f}\:\:{continues}\:{from}\:{R}\:{to}\:{R}\:/ \\ $$$$\forall\left({x},{h}\right)\in{R}^{\mathrm{2}} \:\:\:{f}\left({x}\right).{f}\left({y}\right)=\int_{{x}−{y}} ^{{x}+{y}} \:{f}\left({t}\right){dt}\:. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 50409 by Abdo msup. last updated on 16/Dec/18 $${find}\:\left[\sum_{{k}=\mathrm{1}} ^{\mathrm{10}^{\mathrm{4}} } \:\:\frac{\mathrm{1}}{\:\sqrt{{k}}}\right] \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 50408 by Abdo msup. last updated on 16/Dec/18 $${find}\:{the}\:{value}\:{of}\:{lim}_{{n}\rightarrow+\infty} \sum_{{i}=\mathrm{1}} ^{{n}} \:\sum_{{j}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{i}+{j}} }{{i}+{j}} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 50405 by Abdo msup. last updated on 16/Dec/18 $${let}\:\:{V}_{{n}} =\:\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{3}}\:+…+\frac{\mathrm{1}}{\mathrm{4}{n}−\mathrm{1}} \\ $$$${determine}\:{lim}_{{n}\rightarrow+\infty} \:{V}_{{n}} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 17/Dec/18 $${V}_{{n}}…
Question Number 50403 by Abdo msup. last updated on 16/Dec/18 $${study}\:{and}\:{give}\:{the}\:{graph}\:{for}\: \\ $$$${f}\left({x}\right)={x}^{\mathrm{2}} \:{e}^{\left(−\mathrm{2}{x}\:+\mathrm{1}\right){ln}\left({x}\right)} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com