Question Number 213291 by MathematicalUser2357 last updated on 02/Nov/24 $${Find}\:{domain}\:{of}\:{y}_{\mathrm{213291}} : \\ $$$${y}_{\mathrm{213291}} =\frac{\mathrm{3}+{e}^{\frac{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}{{x}−\mathrm{6}}} }{\mathrm{log}_{\frac{\mathrm{3}}{\mathrm{4}}} \sqrt{{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}} \\ $$ Answered by MrGaster last updated…
Question Number 212518 by efronzo1 last updated on 16/Oct/24 $$\:\:\:\cancel{\underbrace{\downharpoonleft}\underline{}\:} \\ $$ Answered by golsendro last updated on 16/Oct/24 $$\:\:\:\cancel{\underline{\underbrace{\pm}} }=\:\left(\mathrm{x}−\mathrm{a}\right)^{\mathrm{2}} +\mathrm{a}\: \\ $$$$\:\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)=\:\sqrt{\mathrm{x}−\mathrm{a}}\:+\mathrm{a}\:…
Question Number 210200 by Erico last updated on 02/Aug/24 $$\mathrm{Prove}\:\mathrm{that} \\ $$$$\forall\mathrm{n}\geqslant\mathrm{4}\:\:\:\:\:\:\:\mathrm{n}!\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}}}\right)\:\leqslant\:\underset{\:\mathrm{0}} {\int}^{\:\mathrm{n}} \mathrm{t}^{\mathrm{n}} \mathrm{e}^{−\mathrm{t}} \mathrm{dt}\:\leqslant\:\frac{\mathrm{n}!}{\mathrm{2}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 209707 by alcohol last updated on 19/Jul/24 $${a},{b}\:\in\mathbb{C}\::\:{a}\overset{−} {{b}}\:+\:{b}\:=\:\mathrm{0}\:{f}\::\:{z}'\:=\:{a}\overset{−} {{z}}\:+\:{b}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{such}\:{that}\:{f}\left({M}\right)\:=\:{M}' \\ $$$$\mathrm{1}.\:{let}\:{z}_{{A}} \:=\:{z}\:{and}\:{z}_{{A}'} \:=\:{z}'\:{and}\:{f}\left({A}\right)\:=\:{A} \\ $$$${show}\:{that}\:\mathrm{2}{Re}\left(\overset{−} {{b}z}\right)\:=\:{b}\overset{−} {{b}} \\ $$$$\left({A}\:{is}\:{the}\:{set}\:{of}\:{invariant}\:{points}\:{and}\right. \\…
Question Number 208533 by alcohol last updated on 18/Jun/24 $${z}'\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({z}+\frac{\mathrm{1}}{{z}}\right) \\ $$$${z}\:{and}\:{z}'\:{are}\:{complex}\:{numbers} \\ $$$${show}\:{that}\:{z}\:=\:\mathrm{2}{e}^{{i}\theta} \\ $$$${show}\:{that}\:{M}'\:{describes}\:{a}\:{conic}\:{section} \\ $$ Answered by Berbere last updated on 18/Jun/24…
Question Number 208431 by alcohol last updated on 15/Jun/24 $${h}_{{a}} \left({x}\right)\:=\:{e}^{−{x}} \:+\:{ax}^{\mathrm{2}} \\ $$$${show}\:{that}\:{h}_{{a}} \:{admits}\:{a}\:{minimum}\:{in}\:\mathbb{R} \\ $$ Answered by mathzup last updated on 15/Jun/24 $${cas}\:\mathrm{1}\:\:\:\:\:\:{a}>\mathrm{0}…
Question Number 208418 by alcohol last updated on 16/Jun/24 $$\left.{u}_{{n}+\mathrm{1}} \:=\:{u}_{{n}} −{u}_{{n}} ^{\mathrm{3}} \:;\:{u}_{\mathrm{0}} \:\in\:\right]\mathrm{0},\:\mathrm{1}\left[\right. \\ $$$$\left..\:{show}\:{that}\:{u}_{{n}} \:\in\:\right]\mathrm{0},\:\mathrm{1}\left[\right. \\ $$$$.\:{show}\:{that}\:{u}_{{n}} \:{converges}\:{to}\:\mathrm{0} \\ $$$${v}_{{n}} \:=\:\frac{\mathrm{1}}{{u}_{{n}+\mathrm{1}} ^{\mathrm{2}}…
Question Number 208103 by depressiveshrek last updated on 05/Jun/24 $$\mathrm{Find}\:\mathrm{inf}\left\{\frac{{m}}{{n}}\:\mid\:{m},\:{n}\:\in\:\mathbb{N},\:{m}<{n}−\mathrm{2}\right\} \\ $$ Answered by A5T last updated on 05/Jun/24 $${Since}\:{m},{n}\in\mathbb{N},\:\frac{{m}}{{n}}>\mathrm{0},{so}\:\mathrm{0}\:{is}\:{a}\:{lower}\:{bound}. \\ $$$$\frac{{m}}{{n}}\geqslant\frac{\mathrm{1}}{{n}}\:{and}\:\frac{\mathrm{1}}{{n}}\:{is}\:{in}\:{the}\:{set}. \\ $$$${For}\:{any}\:\epsilon>\mathrm{0},\:{one}\:{can}\:{always}\:{find}\:{n}\:{such}\:{that} \\…
Question Number 207390 by alcohol last updated on 13/May/24 Commented by alcohol last updated on 13/May/24 $${please}\:{i}\:{really}\:{need}\:{help}\:{here} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 207314 by universe last updated on 11/May/24 $$\:\mathrm{let}\:\mathrm{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{be}\:\mathrm{a}\:\mathrm{continuous}\:\mathrm{function}\:\mathrm{then} \\ $$$$\:\mathrm{show}\:\mathrm{that} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{if}\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{f}\left(\mathrm{x}^{\mathrm{2}} \right)\:\forall\:\mathrm{x}\:\in\mathbb{R}\:\mathrm{then}\:\mathrm{f}\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{constant} \\ $$$$\:\:\mathrm{function} \\ $$$$\:\left(\mathrm{2}\right)\:\mathrm{if}\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{f}\left(\mathrm{2x}+\mathrm{1}\right)\:\forall\mathrm{x}\in\mathbb{R}\:\mathrm{then}\:\mathrm{f}\:\:\mathrm{is}\:\mathrm{a}\: \\ $$$$\:\:\mathrm{constant}\:\mathrm{function}\: \\ $$ Answered by…