Question Number 208646 by hardmath last updated on 20/Jun/24 $$\mathrm{y}\:=\:\mid\mathrm{x}\:−\:\mathrm{2}\mid\:+\:\mid\mathrm{x}\:+\:\mathrm{4}\mid \\ $$$$\mathrm{Find}:\:\:\:\mathrm{min}\left(\mathrm{y}\right)\:\:\:\mathrm{and}\:\:\:\mathrm{max}\left(\mathrm{y}\right) \\ $$ Answered by A5T last updated on 20/Jun/24 $${y}\:{has}\:{no}\:{global}\:{maximum},{y}\rightarrow+\infty\:{as}\:{x}\rightarrow\underset{−} {+}\infty \\ $$$$\mid{x}−\mathrm{2}\mid+\mid−\mathrm{4}−{x}\mid\geqslant\mid{x}−\mathrm{2}−\mathrm{4}−{x}\mid=\mathrm{6}\Rightarrow{min}\left({y}\right)=\mathrm{6}…
Question Number 208647 by hardmath last updated on 20/Jun/24 $$\left(\mathrm{tan}^{\mathrm{2}} \boldsymbol{\mathrm{x}}\:−\:\mathrm{3}\right)\:\centerdot\:\mathrm{sin}\boldsymbol{\mathrm{x}}\:=\:\mathrm{0} \\ $$$$\mathrm{Find}:\:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Answered by Frix last updated on 20/Jun/24 $${f}\left({x}\right)×{g}\left({x}\right)=\mathrm{0}\:\Leftrightarrow\:{f}\left({x}\right)=\mathrm{0}\vee{g}\left({x}\right)=\mathrm{0} \\ $$$$\mathrm{sin}\:{x}\:=\mathrm{0}\:\Rightarrow\:{x}={n}\pi…
Question Number 208623 by MWSuSon last updated on 19/Jun/24 $$\mathrm{solve}\:\mathrm{for}\:\mathrm{x},\:\mathrm{3}^{\mathrm{x}} −\mathrm{2}^{\mathrm{x}} =\mathrm{65} \\ $$ Answered by Berbere last updated on 19/Jun/24 $${if}\:{x}<\mathrm{0} \\ $$$$\mathrm{3}^{{x}} <\mathrm{1\&2}^{{x}}…
Question Number 208632 by necx122 last updated on 19/Jun/24 $$\int{e}^{−{x}^{\mathrm{2}} } {dx} \\ $$$${could}\:{this}\:{be}\:{integrated}\:{by}\:{part}?\:{What} \\ $$$${approach}\:{would}\:{most}\:{likely}\:{be}\:{suitable} \\ $$$${for}\:{this}\:{integral}? \\ $$$$ \\ $$ Commented by Tinku…
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Question Number 208624 by vipin last updated on 19/Jun/24 Answered by Berbere last updated on 19/Jun/24 $$\frac{\mathrm{1}}{{cosec}^{−} \left(−\sqrt{\mathrm{2}}\right)}=\mathrm{sin}^{−\mathrm{1}} \left(−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)=−\frac{\pi}{\mathrm{4}} \\ $$$${f}\left({x}\right).\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{1}+{x}}−\sqrt{\mathrm{1}−{x}}}{\:\sqrt{\mathrm{1}+{x}}+\sqrt{\mathrm{1}−{x}}}\right)=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}}{\mathrm{1}+\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}}\right)…{E} \\ $$$${g}\left({y}\right)=\mathrm{tan}^{−\mathrm{1}}…
Question Number 208569 by pticantor last updated on 18/Jun/24 $$\boldsymbol{{help}}\:\boldsymbol{{me}}\:\boldsymbol{{to}}\:\boldsymbol{{solve}}\:\boldsymbol{{this}}\:\boldsymbol{{please}} \\ $$$$\:\:\boldsymbol{{y}}''−\sqrt{\mathrm{1}+\boldsymbol{{y}}'^{\mathrm{2}} }=\boldsymbol{{x}}^{\mathrm{2}} \\ $$$$\boldsymbol{{solve}}\:\boldsymbol{{this}}\:\boldsymbol{{differential}}\:\boldsymbol{{equation}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$ Terms of…
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Question Number 208567 by Tawa11 last updated on 18/Jun/24 Answered by kgmxdd last updated on 18/Jun/24 Commented by Tawa11 last updated on 18/Jun/24 $$\mathrm{Any}\:\mathrm{workings}\:\mathrm{sir}? \\…