Question Number 79644 by loveineq. last updated on 26/Jan/20 $${f}\left({x}\right)\:=\:\frac{\mathrm{4}\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{3}\right)}{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{5}} \\ $$$$\mathrm{prove}\:\mathrm{that}\:{f}\left({x}\right)\:\geqslant\:\mathrm{2}\:. \\ $$ Commented by john santu last updated on 27/Jan/20 $$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)<\mathrm{2}…
Question Number 145164 by mim24 last updated on 02/Jul/21 Answered by mathmax by abdo last updated on 02/Jul/21 $$\mathrm{this}\:\mathrm{question}\:\mathrm{is}\:\mathrm{solved}\:\mathrm{see}\:\mathrm{the}\:\mathrm{platform} \\ $$ Commented by mim24 last…
Question Number 145163 by mim24 last updated on 02/Jul/21 Answered by mathmax by abdo last updated on 02/Jul/21 $$\mathrm{y}=\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+…+\infty}}}\:\Rightarrow\mathrm{y}=\sqrt{\mathrm{x}+\mathrm{y}}\:\Rightarrow\mathrm{y}^{\mathrm{2}} \:=\mathrm{x}+\mathrm{y}\:\Rightarrow \\ $$$$\mathrm{y}^{\mathrm{2}} −\mathrm{y}−\mathrm{x}=\mathrm{0}\:\rightarrow\Delta=\mathrm{1}+\mathrm{4x}\:\Rightarrow\mathrm{y}=\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4x}}}{\mathrm{2}}\:\mathrm{or}\:\mathrm{y}=\frac{\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{4x}}}{\mathrm{2}} \\ $$$$\mathrm{but}\:\mathrm{y}\geqslant\mathrm{0}\:\Rightarrow\mathrm{y}=\frac{\mathrm{1}+\sqrt{\mathrm{4x}+\mathrm{1}}}{\mathrm{2}}\:\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{4}}{\mathrm{2}\sqrt{\mathrm{4x}+\mathrm{1}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{4x}+\mathrm{1}}}\:\Rightarrow…
Question Number 79588 by loveineq. last updated on 26/Jan/20 $${f}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}{x}}+\mathrm{2}{x}}+\frac{{x}}{\mathrm{4}} \\ $$$$\mathrm{prove}\:\mathrm{that}\:{f}\left({x}\right)\:\geqslant\:\frac{\mathrm{3}}{\mathrm{8}}\:. \\ $$ Commented by MJS last updated on 26/Jan/20 $$\mathrm{2}\sqrt{\mathrm{2}{x}}\:\mathrm{or}\:\mathrm{2}\sqrt{\mathrm{2}}{x}\:? \\ $$ Commented…
Question Number 13955 by ajfour last updated on 25/May/17 $${y}\left({x}\right)=\begin{cases}{\mathrm{4}+\mathrm{6}{x}−\mathrm{3}{x}^{\mathrm{2}} \:\:\:\:\:\:\:;\:\:{x}\:<\:\mathrm{2}}\\{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{14}{x}+\mathrm{20}\:\:;\:\:{x}\:\geqslant\:\mathrm{2}}\end{cases} \\ $$$${Is}\:{the}\:{function}\:{y}\left({x}\right)\: \\ $$$${differentiable}\:{with}\:{respect}\:{to}\:{x} \\ $$$${at}\:{x}=\mathrm{2}\:? \\ $$ Commented by prakash jain last…
Question Number 145021 by mim24 last updated on 01/Jul/21 Answered by phally last updated on 01/Jul/21 $$\:\mathrm{Derivative}\:\mathrm{formula} \\ $$$$\:\left(\mathrm{Cos}^{−\mathrm{1}} \left(\mathrm{u}\right)\right)^{'} =\frac{−\mathrm{u}'}{\:\sqrt{\mathrm{1}−\mathrm{u}^{\mathrm{2}} }} \\ $$ Answered…
Question Number 79480 by TawaTawa last updated on 25/Jan/20 Commented by mathmax by abdo last updated on 25/Jan/20 $${let}\:{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{ln}\left({tanx}\right)}{{cos}\left(\mathrm{2}{x}\right)}{dx}\:\:{changement}\:{tanx}\:={t}\:{give} \\ $$$${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left({t}\right)}{\frac{\mathrm{1}−{t}^{\mathrm{2}}…
Question Number 144995 by liberty last updated on 01/Jul/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{distance}\: \\ $$$$\mathrm{between}\:\mathrm{two}\:\mathrm{points}\:\mathrm{on}\:\mathrm{the}\: \\ $$$$\:\mathrm{curve}\:\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{a}^{\mathrm{4}} }\:+\:\frac{\mathrm{y}^{\mathrm{4}} }{\mathrm{b}^{\mathrm{4}} }\:=\:\mathrm{1}\:. \\ $$ Answered by mr W last…
Question Number 13871 by Nayon last updated on 24/May/17 $${why}\:{the}\:{function},{sin}\left({x}\right)\:{is}\:{a}\:{power} \\ $$$${series}?? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 79398 by ubaydulla last updated on 24/Jan/20 $$\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$ Commented by john santu last updated on 24/Jan/20 $$\int\:\frac{{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx}\:=\:\int\:\mathrm{1}\:{dx}−\int\frac{\mathrm{2}{dx}}{{x}^{\mathrm{2}}…