Question Number 78046 by jagoll last updated on 13/Jan/20
$${minimum}\:{of}\: \\ $$$${function}\:{y}\:=\:\sqrt{{x}^{\mathrm{2}} +{e}^{\mathrm{2}{x}} }\:\:{is} \\ $$
Answered by john santu last updated on 13/Jan/20
$${y}'\:=\:\frac{\mathrm{2}{x}+\mathrm{2}{e}^{\mathrm{2}{x}} }{\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{e}^{\mathrm{2}{x}} }}\:=\:\mathrm{0} \\ $$$${e}^{\mathrm{2}{x}} \:=\:−\:{x}\:,\:{x}<\mathrm{0}\: \\ $$$$\mathrm{1}\:=\:−{xe}^{−\mathrm{2}{x}} \:\Rightarrow\:{use}\:{Lambert}\:{W}\:\:{function} \\ $$$$ \\ $$
Commented by john santu last updated on 13/Jan/20
$$\mathrm{2}\:=−\mathrm{2}{xe}^{−\mathrm{2}{x}} \:\Rightarrow{W}\left(\mathrm{2}\right)=\:{W}\left(−\mathrm{2}{xe}^{−\mathrm{2}{x}} \right) \\ $$$${W}\left(\mathrm{2}\right)=\:−\mathrm{2}{x}\:\Rightarrow\:{x}=−\frac{{W}\left(\mathrm{2}\right)}{\mathrm{2}} \\ $$